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KidMath Introduction to Geometry
KidMath Sept. 04
Prof. T Parker
KidMath — Introduction to Geometry
Geometry is a game of logic played with shapes. The shapes lie in a plane. They are
constructed from the following basic pieces.
• Point — specifies a location (has no thickness)
• Line — extends infinitely in both directions (also has no thickness)
• Ray — "half-line". Part of a line; has an endpoint and extends infinitely in one direction.
• Line segment — the part of a line between two points (called the endpoints).
Fact 1 : Two points determine a line
Given two different points, there is 1 and only 1 line containing both of them.
Notation: We name points by capital letters. Two points A and B determine
←→
• a line AB A B
−−→ −
−→ B A
• rays AB and BA A B
A
• a segment AB B
Length: We measure length in meters, centimeters, millimeters, and kilometers.
1 millimeter 1 mm
1 centimeter 1 cm=10 mm
1 meter 1 m = 100 cm
1 kilometer 1 km = 1000 m.
Because these units are related by factors of 10, all conversions can be made by just shifting the
decimal point!
Exercise 1 Complete the following expressions.
18 cm mm 3.5 m = cm
2.4 km = m 85 mm = cm
860 cm = m 63.2 m = mm
268 mm = cm
Angles: Two rays with the same endpoint separate the plane into two regions
We can distinguish these regions by drawing small arcs, An angle is two such rays with such an
arc. We can name them by naming the arc, or by naming three points in order.
P
x
Q R
These are pictures of ∠x and ∠P QR.
Angle Terms:
a right angle is half a straight angle (90◦ ).
a straight angle (180◦ )
Small squares mark right angles.
x y x
y
∠x and ∠y are complementary
∠x and ∠y are supplementary
(they add to 90◦ )
(they add to 180◦ )
Opposite sides of crossed lines, such as ∠x and
∠z, are called vertical angles. Notice that
y z
∠x = 180◦ − ∠y
x ∠z = 180◦ − ∠y.
so ∠x = ∠z (both equal to same thing). Thus
Fact 2 : Vertical angles are equal
Angles in a Triangle
Every triangle has 3 vertices and 3 interior angles. It is an amazing fact that if you know the
measurement of two of those angles, you can figure out what the third is.
Question: How many lines can be determined by two different points?
Answer: 1 | 677.169 | 1 |
The triangles, in blue, are only of the 7:11:13 near-pi near-right variety, though scaling between them is as necessary . All share the same line, (which I'll call the NW ordinate), and with the exception of those triangles used to position the center of the surviving stone circle, all rest their hypotenuse, (or long side), upon it . Of the four principal triangles, one, (the inner of the lower), starts and ends on the '11' circle, (numbers are ratios, not measures) . This was likely the most holy of the triangles, as it crosses the '7' circle at its diameter, (on the line I'll call the NE ordinate), then corners again on the '11' before returning and ending on same ; 6 of the ten stones shown touch this triangle . The outer lower triangle begins on or near the '13' circle, corners and returns from numerically unsolved places ; 4 of the ten stones touch it . The inner upper triangle begins on the stone circle itself, where this crosses the common centerline - at or very close to the radius of a '10' circle, not shown and returns from a second grounding on the stone circle line . 3 stones touch this triangle . The outer upper triangle begins on the eleven circle and returns from a 9 circle, (also not shown) . 3 stones touch this .
The Stone circle itself is centered at or near where a triangle with its short side on the common centerline meets a circle of scale '1', (all the small circles except for the blue ones are '1's), which is centered where the NE ordinate meets the '7' circle . Along this triangle's long side, that spot is apparently chosen which is perpendicular to the midpoint of the inner upper triangle's long side on the common line .
The pale turquoise traces represent the area of light, (my supposing), ordinarily the crescent moon, but on the rarest of occasions a crescent sun . Ratios between the string of turquoise circles have yet to be resolved, but they seem to recall later Celtic design work . The monthly cycle diameter for this area is anchored by a pair of stones . The northern of these is hard to explain in this context other than that it's there for the purpose . However it is well placed within the lozenge of panel 1, ('Lochbuie') .
That's about it, obviously a picture is worth a thousand words, though I wasn't counting . A good tool for viewing the diagrams is the Opera web browser, ( which lets the user drag a trace-copy of an image across itself with the mouse - facilitating a search for new lines . There are often more
Lastly, if the circle did represent shadow in this tradition, it implies that most of the people would want to be outside of it most of the time ... a reversal of emphasis
Question: What is the likely significance of the triangle that starts and ends on the '11' circle? Answer: It was likely the most holy of the triangles
Question: What is the purpose of the northern stone that anchors the monthly cycle diameter? Answer: It's there for the purpose, but its exact significance is not explained in this context | 677.169 | 1 |
In our first panoramas, we find the radius. You'll be able to see yarn stretching from the center of our classroom to the front of the room. The radius is a line segment extending from the center of a circle out to the edge of the circle. To navigate through the picture, just click and hold down on the mouse, then move it from left to right or right to left!
If these have trouble loading on your computer, you can also access our panoramas by clicking here. You can load any of the panoramas by simply clicking on them.
Our next panoramas show us the diameter. The diameter of a circle is a line segment that stretches from one side of the circle to the other. The diameter has to pass through the middle point of the circle. In the image, you'll be able to see yarn stretching across our image to represent our diameter.
The next group of panoramas show the circumference of our circle. The circumference is the distance around a circle. Look for a black piece of yarn stretching all the way around our classroom to mark our circumference.
Our fourth group of panoramas represent the many angles within a circle. We often hear the term of someone doing a 360 in basketball, skateboarding, or snowboarding. That refers to 360 degrees, or 360° which represents the distance all the way around a circle. That means that half way around a circle would be 180°. Start with 0° in the front of the room. Work your way around clockwise until you end up back at 360°. You'll be able to see the many angles along the way! If you can't see an angle well, you can zoom in by hitting the plus button on the left side of the panorama or by using the scroll bar on your mouse.
For our final panorama, we decided just to have a little bit of fun. Students were able to pose however they wanted as long as they were still along the circumference.
*What else do you think we could use a panorama for to help us learn?*
I hope everyone has enjoyed their summer! I know mine has certainly been busy getting ready for the new school year. This year will definitely be a little bit different though. I'm no longer teaching fourth grade and have moved up to sixth grade! I also have a new teaching partner, Ms. Girard. This will be her first year at our school and I'm very excited to work with her!
In changing grade levels, that also means changing rooms. Today was the first opportunity I had to get into my new classroom. One of my favorite parts of the room is a giant blank wall that I have. My old classroom barely had any blank wall space! I decided that instead of using cloth for a green screen next year, we'll have our own green screen wall! Below you can find some before and after pictures as I introduce you to "The Green Monster."
Question: What is the circumference of a circle? Answer: The circumference is the distance around a circle.
Question: What is the author's new classroom wall used for? Answer: The author's new classroom wall is used as a green screen for various learning activities. | 677.169 | 1 |
Definitions
Century Dictionary and Cyclopedia
n. In geometry, a solid of thirty-two faces formed by cutting down the corners of the icosahedron parallel to the faces of the coaxial regular dodecahedron until the new faces just touch at the angles, thus leaving 20 triangular and 12 pentagonal faces. It is one of the thirteen Archimedean solids.
Wiktionary
Examples
"These are simply extraordinary: hand-drawn images of immensely complex geometrical bodies, among them the 72-sided hebdomicontadissaedron and the very first representation in history of icosidodecahedron, that is, a three-dimensional solid with 20 triangular faces and 12 pentagonal faces."
"Then there are the 13 Archimedean solids (PolyhedronData [ "Archimedean"]; the cuboctahedron, icosidodecahedron, truncated cube, etc.), constructed by requiring the same configuration at each vertex, but allowing more than one kind of regular face."
Question: How many faces does an icosahedron have? Answer: 20 (triangular faces) | 677.169 | 1 |
This is a great problem for students because it shows the way visual thinking and sheer logic (the core of geometric reasoning) can support each other. There is a nice "high" involved in tracking down the crucial points that, no matter which color they are, yield the needed results.
The way this problem is received will depend enormously on the nature of your students. Those who love arguing their parents into tight corners may be delighted with this exploration. There's a sort of perversity about the logical prowling that can be very satisfying, but finding the layout of points on which to operate can be very difficult, so use your judgment about whether to provide the worksheets.
If you're doing this with the class as a whole, you may prefer to use the worksheets to make transparencies. Putting dots on the transparency using red and blue markers will speed things along. For Tasks (a) and (b), the points are called lattice points, if we assume that they have integer coordinates. (But note that coordinates are not needed.) The discussion about density of points is worth having, and is interesting.
An easy extension of the problem, or at least a variant using the same drawings, is to assume that some particular point has coordinates (0, 0), then to specify the distance to a neighboring point and figure out the coordinates of all the remaining points in the drawing. For instance, in the top drawing on Worksheet 2, let A be at (0, 0) and B at (2, 0). Then F is at (1, sq rt 3), and so on. An interesting pattern develops.
I suggest giving students time to cudgel their brains over each task. Some students will not want to let go of it.
Question: In the top drawing on Worksheet 2, what are the coordinates of point A? Answer: (0, 0) | 677.169 | 1 |
Problem 117.
Area of Triangles, Incenter, Excircles.
Level: High School, SAT Prep, College
In the figure below, given a triangle
ABC, construct the incenter I and the excircles with
excenters P and Q. Let be D and E the tangent
points of triangle ABC with its excircles. IE and BC meet at H,
and ID and AB meet at G. If S1, S2, S3,
and S4 are the areas of the triangles AIG, BEH, BDG, and
CHI
respectively, prove that S1 + S2 = S3
+ S4.
Question: What is the name of the point where the incenter of triangle ABC is located? Answer: I | 677.169 | 1 |
Page:The Elements of Euclid for the Use of Schools and Colleges - 1872.djvu/245
9. A solid angle is that which is made by more than two plane angles, which are not in the same plane, meeting at one point.
10. Equal and similar solid figures are such as are contained by similar planes equal in number and magni- tude. [See the Notes.]
11. Similar solid figures are such as have all their solid angles equal, each to each, and are contained by the same number of similar planes.
12. A pyramid is a solid figure contained by planes which are constructed between one plane and one point above it at which they meet.
13. A prism is a solid figure contained by plane figures, of which two that are opposite are equal, similar, and par- allel to one another ; and the others are parallelograms.
14. A sphere is a solid figure described by the revolu- tion of a semicircle about its diameter, which remains fixed.
15. The axis of a sphere is the fixed straight line about which the semicircle revolves.
16. The centre of a sphere is the same with that of the semicircle.
17. The diameter of a sphere is any straight line which passes through the centre, and is terminated both ways by the superficies of the sphere.
18. A cone is a solid figure described by the revolution of a right-angled triangle about one of the sides containing the right angle, which side remains fixed.
If the fixed side be equal to the other side containing the right angle, the cone is called a right-angled cone ; if it be less than the other side, an obtuse-angled cone ; and if greater, an acute-angled cone.
19. The axis of a cone is the fixed straight line about which the triangle revolves.
Question: What is the axis of a cone? Answer: The axis of a cone is the fixed straight line about which the triangle revolves.
Question: How are equal and similar solid figures defined? Answer: Equal and similar solid figures are such as are contained by similar planes equal in number and magnitude. | 677.169 | 1 |
Geometry If two six-sided dice are rolled, the probability that they both show the same number can be expressed as a b where a and b are coprime positive integers. What is the value of a+b ?
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Thursday, April 11, 2013 at 6:56Thursday, April 11, 2013 at 6:54pm
Geometry ABC is a triangle with circumcenter O, obtuse angle BAC and AB less than AC. M and N are the midpoints of BC and AO respectively. Let D be the intersection of MN with AC. If 2AD=(AB+AC), what is the measure of angle BAC ?
Thursday, April 11, 2013 at 11:27am
Geometry 2x^2-9x-5=0 Is it possible to combine 2x squared with 9x? Not sure how to tackle solving this one.
Wednesday, April 10, 2013 at 9:26pm
geometry Which description does NOT guarantee that a trapezoid is isosceles? A. congruent bases B. congruent legs C. both pairs of base angles congruent D. congruent diagonals
Tuesday, April 9, 2013 at 9:31Monday, April 8, 2013 at 10:29pm
Analytic Geometry/Calculus We didn't go over the perpendicular form in class, only the parallel. Find the general form of the equation of the plane passing through the point and perpendicular to the specified vector or line.
Monday, April 8, 2013 at 7:40pm
geometry On my last question, just posted, I asked what percent of the area of the target was shaded. The answer options were 5%, 10%, 20% or 80%?
Monday, April 8, 2013 at 6:45pm
geometry There is a square with a star inside. The square has 25cm to show the length of one side. On the right, outside the square, it says 125cm squared, with a line pointing the center of the star in the square. The question is; The target of a shooting game features a shaded star ...
Monday, April 8, 2013 at 6:42pm
Question: What is the measure of angle BAC in triangle ABC, given that 2AD = AB + AC?
Answer: 120 degrees (since AD is the median to the hypotenuse in a right-angled triangle, and 2AD = AB + AC implies AD is half the perimeter, making triangle ADC a 30-60-90 triangle) | 677.169 | 1 |
Wednesday, March 13, 2013 at 11:01pm
geometry Alexa has 200 square inches of wrapping paper left. Which is the side length of a cube she could not cover with the paper?
Wednesday, March 13, 2013 at 10:01pm
geometry What is 2(1x1/2)+2(1x1/4)+2(1/2x1/4)
Wednesday, March 13, 2013 at 8:40pm
Geometry If EF=2x-19, FG=3x-15, and EG=26, find the values of x, EF, And FG
Wednesday, March 13, 2013 at 6:46pm
geometry help!!Wednesday, March 13, 2013 at 4:36pm
geometry If the measure of ÐDAB = 50°, and ÐDAC = 20°, what is ÐCAB?
Wednesday, March 13, 2013 at 3:37pm
geometry prove that rectangles do not exist in hyperbolic geometry
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Math/Geometry Two circles of radius r1 and r2 are extenally tangential to each other and are also externally tangential to a staight line l. Another circle of some unknown radius is externally tangential to both the circles and to the straight line l. Find the adius of that circle.
Wednesday, March 13, 2013 at 6:12am
geometry! Γ is a circle with center O. A and B are points on Γ such that the sector AOB has a perimeter of 40. Amongst all circular sectors with a perimeter of 40, what is the central measure of ∠AOB (in radians) of the sector with the largest area?
Wednesday, March 13, 2013 at 1:18am
Geometry 7 points are placed in a regular hexagon with side length 20. Let m denote the distance between the two closest points. What is the maximum possible value of m?
Tuesday, March 12, 2013 at 10:15pm
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Tuesday, March 12, 2013 at 10:15pm
Geometry Determine the largest positive integer N, such that given any N-gon (not necessarily convex), there exists a line (infinitely extended in both directions) that contains exactly 1 edge of the N-gon. The figure in blue is an example of a 20-gon that doesn't satisfy the ...
Tuesday, March 12, 2013 at 10:05pm
Question: What is the sum of the given expressions? Answer: 2(1x1/2)+2(1x1/4)+2(1/2x1/4) = 2(1/2) + 2(1/4) + 2(1/8) = 1 + 1/2 + 1/4 = 2.5
Question: What is the maximum possible value of m in the given hexagon? Answer: The maximum possible value of m is the side length of the hexagon, which is 20.
Question: What is the central angle of the circular sector with the largest area and a perimeter of 40? Answer: The central angle of the sector with the largest area is π radians, as this corresponds to a full circle, which has the largest area for a given perimeter.
Question: What is the maximum number of houses in odd-town? Answer: The maximum number of houses in odd-town is 999, as there are 900 possible 3-digit odd numbers with distinct digits (99 odd numbers for each of the first two digits and 9 for the last digit). | 677.169 | 1 |
coplanar mean all the points lie on a plane or a flat 2-dimensional surface the points can be spread out but have to lie in the plane collinear means are the point lie in one straight lineif points are all collinear then they're also coplanar
Simple.Coplaner: Like the teacher said, Co = together, planer = plane. They are together in the same plane, whatever figure in one planeCollinear: Co = together, linear, sounds like line, right? Three points on the same line. That's collinear.Hope it helped.Have funCheers.
Question: Are coplanar points spread out in three dimensions? Answer: No, coplanar points are spread out in a two-dimensional plane. | 677.169 | 1 |
As for the question, Statement A alone could be true of a trapezoid, and Statement B could just as well be a rectangle, so you need both statements ("Both statements 1 and 2 together are sufficient to answer the question but neither statement is sufficient alone"), which means the answer is C.
I hope that helps, feel free to let me know if you have any other questions! _________________
Think of it this way: Stmnt 1. Even if all 4 sides are equal, it still may not be a square. It may be a rhombus (angles are not 90 degrees). Clearly not sufficient. Stmnt 2. 3 angles 90 degrees means the fourth angle has to be 90 degrees too (since the sum of all interior angles of a quadrilateral is 360 degrees). But if a quadrilateral has all four angles 90, does it mean it is a square? No. It could be a rectangle. Only opposite sides are equal in length in a rectangle. Not sufficient.
Using both statements: The quadrilateral must have all four angles 90 degrees. Since 3 sides are equal, the fourth one must be equal to these 3 since the pair of opposite sides must be equal if all four angles are 90 degrees. Hence, the quadrilateral must be a square.
As for: Hi, I need some clarity on quadrilaterals concepts/rules.
I would suggest you to check out the Veritas Geometry book. It covers lines, angles, triangles, quadrilaterals, circles, 3 D figures, co-ordinate ... pretty much everything you might see on GMAT.
_________________
Question: Which shape could Statement A represent if it's not a square? Answer: A rhombus. | 677.169 | 1 |
'Euclid? No, but Carol, Yes' printed from
In the preface to A New Geometry for Schools, parts i and ii (1961), Durrell stated that his aim was:
..."to provide a treatment which lends itself both to class teaching and to individual use by the pupil."
Then, Durell's modus operandi was to "develop each group of geometrical facts by the following successive stages:"
examples for oral discussion
an exercise of (graded) numerical examples
formal proofs of the corresponding theorems
an exercise of readers.
It was the author's conviction that discussion not only facilitated "the learning of formal proofs of theorems" but was "the best method of strengthening the power of the pupil to tackle riders by showing... the types of construction most often required."
The material presented in the book had been faithful to the recommendations of the Second Report on the Teaching of Geometry, which in its day (c. 1955) was considered a milestone of sound educational practice.
Dipping into one "group of geometrical facts" - loci, can be found exercise 46 (Oral)
What loci are described by the following:
4. The top of your head if you slide downstairs on a tea tray?
10. The centre of a marble which rolls about inside a spherical bowl?
15. A variable point P that is due north of a fixed point A?
24. A, B are fixed points; P is a variable point on a given circle, centre A. ABPQ is a parallelogram. Find the locus of Q.
38. A circular cone rolls on a plane. What is the locus of the centre of the base of the cone?
There were plenty of questions to choose from before consideration was given to:
Theorem 29 (1): "Any point on the perpendicular bisector of the line joining two given points is equidistant from the given points."
Theorem 29 (2): "A point which was equidistant from two given points lies on the perpendicular bisector of the straight line joining the given points."
Then, more opportunity for oral discussion followed which in turn laid the foundations for:
Theorem 30: "The perpendicular bisectors of the three sides of a triangle, are concurrent." (i.e. The Circumcentre theorem), and
Theorem 31: "The altitudes of a triangle are concurrent."
(i.e. The Orthocentre theorem)
Following on was exercise 47, and dipping in again can be found:
4.Draw a triangle ABC in which angle A is obtuse. Construct a point P on BA produced such that PB - PC = AB
15. A, B, C, D are four points on the circumference of a circle. Prove that the perpendicular bisectors of AB, AC, AD, BC, BD, CD are concurrent.
24. If H is the orthocentre of triangle ABC, prove that the angles BHC, BAC are equal or supplementary. [ Triangle ABC may be acute angled or obtuse angled.]
Question: Who is the author of the book mentioned in the text? Answer: Durell
Question: What were the successive stages Durell used to develop each group of geometrical facts? Answer: Examples for oral discussion, an exercise of (graded) numerical examples, formal proofs of the corresponding theorems, an exercise of readers.
Question: Which theorem is the topic of exercise 15? Answer: The perpendicular bisectors of the three sides of a triangle are concurrent.
Question: What is the topic of exercise 46 in the book? Answer: Loci | 677.169 | 1 |
Quadrilateral Properties
'Master the properties of shapes and take a step closer toward mastering space itself'
Works of art, architectural and industrial designs, packaging etc. show a real mastery of forms and their properties. The triangle is the building block of many two dimensional forms and the properties that result. This activity offers the first steps towards mastery: experiment to see what polygons can be made with different triangles, using the Powerpoint or paper print-outs in the resources section. Then create a flowchart to isolate the defining characteristic of each quadrilateral, before attempting the puzzle. The final challenge will be to use all the pieces to re-create an artistic masterpiece!
(b) On the second slide of the previous Powerpoint, or second side of the word document, can you create your own flowchart to correctly categorise the seven quadrilaterals? This activity can be done just as well on paper: cut out the quadrilaterals to place on a flowchart you have drawn on paper.
Description
Use the first Powerpoint "triangular building blocks" to examine what polygons can be made using equilateral, right-angled and isosceles triangles. How many different 2D shapes can you make for each type of triangle? Are some more limiting than others?
Using the second powerpoint, "classifying quadrilaterals and triangles", follow the flowchart to work out the unique properties of each of the four triangles.
You are now ready to use the textboxes and flowchart connectors on slide 2, to copy and paste as many additional textboxes and connectors as you need to design your own questions. Use the "properties help sheet" where necessary from the resource section above. Can you isolate each quadrilaterals uniqueproperty? Can you do so using as few questions as possible?
Open one of your classmates flowcharts and see if you can categorise all the quadrilaterals correctly. If you can't, is it your mistake or is it your classmates flowchart?
Your teacher will give you a set of a number of different triangles. Use them to construct the quadrilaterals on the sheets your teacher has given you. If you have understood the properties of the quadrilaterals, this should help you decide a lot faster which triangles can be used for which quadrilaterals, and which can't.
Now you've re-constructed the quadrilaterals, see if you can use all the pieces together to rebuild the original picture, created by Wassily Kandinsky
Question: Which type of triangles are mentioned in the text as being used to create polygons? Answer: Equilateral, right-angled, and isosceles triangles.
Question: What is the purpose of the flowchart mentioned in the text? Answer: The flowchart is used to isolate and categorize the unique properties of each of the four triangles.
Question: What should you do if you cannot categorize all the quadrilaterals correctly using your classmate's flowchart? Answer: Check if the mistake is yours or if there's an issue with your classmate's flowchart. | 677.169 | 1 |
Geometry & Shapes
Geometry (from the Greek geo = earth, metria = measure) arose as the field of knowledge dealing with spatial relationships. Geometry was one of the two fields of pre-modern mathematics, the other being the study of numbers.
In modern times, geometric concepts have been extended. They sometimes show a high level of abstraction and complexity. Geometry now uses methods of calculus and abstract algebra, so that many modern branches of the field are not easily recognizable as the descendants of early geometry.
Question: What were the two main fields of pre-modern mathematics? Answer: The study of geometry and the study of numbers | 677.169 | 1 |
The last definition is a bit confusing, since we don't have a very well-agreed upon name for this figure. But notice that ALL of Euclid's definitions are exclusive. A square is never an oblong. A square is never a rhombus. Each of the above quadrilaterals, according to Euclid, is mutually exclusive. These exclusive definitions persisted into the 19th century.
But sorry Euclid, no one likes your definitions anymore. I hate to say it, because everyone loves Euclid.
In his defense, he wasn't using these names for the same purpose we do. Nothing about his language is very technical and he doesn't say ANYTHING else substantial about these definitions. He doesn't use them to make categorical statements about quadrilaterals or to give properties that might be inherited. The names he uses are of little consequence to the rest of his work.
Can we lay this issue to rest yet? A parallelogram is always a trapezoid. Say it with me,
A parallelogram is a trapezoid.
A parallelogram is a trapezoid.
A parallelogram is a trapezoid.
Anything you can say about a trapezoid will be true about a parallelogram (area formulas, cyclic properties, properties about the diagonals). A parallelogram is a trapezoid.
19 thoughts on "Why I hate the definition of trapezoids (again)"
Thanks for your thoughts on the importance of definitions. As further support of your premise, I offer up the area formulas for the trapezoid and the parallelogram. Specifically, you can calculate the area for a parallelogram using the trapezoid formula: A = 1/2 * (b_1 + b_2) * h. Of course it will work since the "bases" of a parallelogram are equal and the formula reduces to A = b * h, where b = b_1 = b_2.
If the formula for the trapezoid's area also calculates the parallelograms area, the parallelogram is a trapezoid.
I've been told that the inclusive definition of trapezoid is the usual one in Canada (whereas the exclusive definition is the usual one in the US–which isn't to say you can't find perfectly good US sources for the inclusive definition). It's almost enough to persuade one to move north.
John – I admire your passion and tenacity on this issue. It takes a certain David v. Goliath form of courage to stand for what you believe against Euclid and the current majority consensus of folks who think about polygons. Plus it makes for entertaining debates in the math office! I think I'm starting to see the light about your preferred hierarchy and the logic of viewing parallelograms as specific cases of trapezoids. It makes sense to go from zero parallel sides to (at least) 1 pair of parallel sides to 2 pairs of parallel sides. Keep up the good fight!
Question: What is the area formula for a parallelogram, given in the text? Answer: A = b * h
Question: Which country typically uses the inclusive definition of a trapezoid? Answer: Canada
Question: Did Euclid use these definitions for the same purpose as we do today? Answer: No, he didn't. | 677.169 | 1 |
There are people who think that he wanted to use a word as table (or 'little table') for irregular (scalene) quadrilaterals. That would mean that the current trapezium with one pair of parallel lines never crossed Euclid's mind or he did not think it was special enough to give it a name.
You seem hung up on the fact that Euclid's definitions are exclusive. Why is this a bad thing. Of course, a square is a special kind of rectangle. But when I give a test to my pupils and draw a square and ask the pupils what the name of that figure is, I do not want them to write down rectangle…
Because of the properties of the square it is a special kind of rectangle, rhombus, etc. but in the first place, it is a square. Not a rectangle.
Lloyd :Best solution I can think of is this, using a geo triangle and a pair of compassess.
Start with the longest side (328). Set the distance of you pair of compasses to 220 and draw a circle with a centre that is the left tip of side 328 and a circle with a radius of 215 that has as a centre the right tip of side 328. Now you use the parallel lines on your geo triangle to draw a line parallel to side 328 with a length of 223 between the two circles.
I completely agree that we need to resolve the problem with the definitions of trapezoid. If you consult the premier mathematics sites (Dr. Math, Wolfram, etc) they agree with "at least one pair of parallel sides". Unfortunately most textbooks and standardized exams posit the "2 pair of parallel sides"
Question: What is the special kind of rectangle that is also a square? Answer: A square.
Question: Which mathematical sites agree with the "at least one pair of parallel sides" definition of a trapezoid? Answer: Dr. Math, Wolfram, etc. | 677.169 | 1 |
that as a starting point, we now tinker a bit to show that 90=100:
Draw the perpendicular bisectors to BE and AD;
call the point where they meet "C".
Wait -- does C really exist?:
Actually, we must prove that those two perpendicular bisectors really
do meet at all (i.e., that the point C even exists).
In this case, it turns out to be pretty clear —
it's not hard to argue that lines AD and BE aren't parallel,
and therefore their perpendicular bisectors aren't parallel,
and so they must intersect (in Euclidean geometry).
Still, be alert for people making glib assertions in proofs.
Figure 1: A construction to help prove that 90=100
Looking at this figure, some warning flags should be going up:
How do we know C lies below BD?
Might it lie above BD? Or exactly on BD?
It turns out that the argument below is the same in all of these cases,
though you'll certainly want to verify this to yourself later.
Exercise 1
If you feel this result is incorrect,
then the challenge for you is to find the first line which is false.
Solution
The flaw is extremely hard to find.
We won't actually give the solution, but here's a hint on
how to go about attacking the puzzle:
Note that finding the bug in the proof is the
same skill as debugging a program.
A good approach is to try various degenerate inputs.
In this case, there are a couple of "inputs" to
the construction—the length of CD is arbitrary; no matter
how long or short the proof should apply equally well.
Similarly, the angle 100° seems arbitrary;
fiddling with inputs like these (making them very small or very large)
might give you some clues as to where the bug is.
A very careful drawing will clear things up.
You may have noticed that the proof given here has some very minuscule
steps—e.g. "Congruent angles have equal measure."
Usually such simple steps can be omitted, since they are obvious to any reader.
We include them for a few reasons:
As a careful thinker, you should recognize that such small steps
really are part of the complete reasoning,
even if they're not worth mentioning continually.
If a computer is checking a proof, it needs to actually include
those steps.
Programmers do need to be concerned with distinctions about (abstract)
types—the difference
between angles and their measures, in this case.
Sometimes a line's justification is glibly given
as "by construction", when
that may not even be correct !-).
In this course, we'll spend a few weeks working with proofs which
do include all the small, pedantic steps,
to instill a mental framework for what a rigorous proof is.
But after that, you can relax your proofs to leave out
such low-level
steps, once you appreciate that they are being omitted
Question: What is a hint given to approach finding the bug in the proof? Answer: To try various degenerate inputs, such as changing the length of CD or the angle measure.
Question: Where does point C lie in relation to BD? Answer: C can lie below, above, or exactly on BD.
Question: What is the challenge for the reader if they believe the result is incorrect? Answer: To find the first false line in the proof.
Question: What is the difference between angles and their measures? Answer: Angles are abstract entities, while their measures are numerical values. | 677.169 | 1 |
Trig/Algebra Question - Sailboat
Trig/Algebra Question - Sailboat
So not being a math whiz I have a problem that I was hoping you all could help me with. I'll start with referring you to the following website which shows the parts of the sailboat that I'm trying to figure out.
My rig is a little less complex than this. I lost my mast (the big pole in the centre) last year and have found a replacement mast. The problem is I don't have the spreaders or rigging (specifically the upper/outer lower shrouds). My problem is how long should my spreaders be?
Constraints:
- The spreaders have to be 5 degrees from the horizontal (so they tilt slighly upwards and are not perpendicular to the mast as the picture shows)
- The angle from where the upper shroud and lower outershroud connect at the spreader must bisect (the angle below and above where the upper and outer lower shroud connect to the spreader must be the same)
Measurements
- Height of the mast (30 feet)
- Height from the bottom of the mast to where the spreaders connect (15 feet)
- Horizontal distance from the bottom of the mast to where the outer lower shroud connects (4 feet)
- Vertical distance from the bottom of the mast to where the outer lower shroud connects (1 foot)
The reason why the distance from the bottom of the mast to where the lower outer shroud connects is become my mast sits on top of a cabin which is higher than the deck of the boat (by 1 foot).
Solve: Solve for the length of the spreader
Length of the upper shroud and outer lower shroud
Now these are basic measurements so if you could give me a formula/process on how to solve if these dimensions were to change that would be awesome.
If I am understanding correctly, dropping a perpendicular line from the point where the shrouds meet the spreader to the deck forms a right triangle with the outer shrould being the hypotenuse. If we call the angle at the spreader "[itex]\alpha[/itex]" and the distance from the foot of the perpendicular "x", we have [itex]tan(\alpha)= x/15[/itex] or [itex]x= 15 tan(\alpha)[/itex]. Further, drawing a perpendicular from the point where the inner shroud is attached to the deck to the first perpendicular gives a second right triangle where the angle at the spreader is also [itex]\alpha[/itex], the height is 14, and the base is 4- x. That gives [itex]tan(\alpha)= (4-x)/14[/itex] or [itex]4- x= 14 tan(\alpha)[/itex]. Dividing one equation by the other eliminates "[itex]tan(\alpha)[/itex] leaving
[tex]\frac{4- x}{x}= \frac{14}{15}[/tex].
Question: What is the height from the bottom of the mast to where the spreaders connect? Answer: 15 feet
Question: What is the horizontal distance from the bottom of the mast to where the outer lower shroud connects? Answer: 4 feet | 677.169 | 1 |
How would I actually apply these maths to programming?
I have finally learned the Pythagorean theorem, Sine/Cosine, still working on Tangent/Cotangent(They just don't seem to click, I mean memorizing the formula), and am working on rise/run. I was wondering how do I actually apply these. For instance, with the Pythagorean theorem there seems to be a lot of things that have to be just right to find the distance between two objects. Why would I want to know the angle between to points, how would that be applicable, and why would I want to know the slope of a flat surface?
For instance, with the Pythagorean theorem there seems to be a lot of things that have to be just right to find the distance between two objects.
Nope, unless you're talking about finding ways around walls, it measure the exact distance under any circumstance and any angle (it works in 3D too, just add Z)
Quote
Why would I want to know the angle between to points?
I always used this (sin/cos) in platform and overhead games. If your character has a gun (or wide variety of other scenerios), you can point the mouse anywhere and the character will always look at it. Very useful for any shooting game. Sin/Cos is also the way you get things to move easily in a circular motion (in 2D of course, maybe 3D), and it's suprisingly difficult to do without it.
Quote
and why would I want to know the slope of a flat surface?
I never had much reason to use this, but I started building a 2D physics engine (which I failed at) and this topic came up quite a bit. As a warning though using that for physics is a bit hackish.
If you need any of these formulas, I have them sitting here on my hard drive collecting dust. Need to do anything specific in your game? :)
(\__/)
(='.'=) This is Bunny. Copy and paste bunny into (")_(") your signature to help him gain world domination. bunny also wants to fight spam: Click Here Bots!
No, I just recently got into the parts of math that teach about the Pythagorean theorem, I already knew how to Sine and Cosine. Just wondering how they were applicable. I need to know the formula for Sine to degrees, so that I don't have to have built in look up table.
Define how you get the two sides needed for the Pythagorean theorem, what I mean is give an example please.
From the pythagorean theorem, the distance between any pair of points (x1, y1) and (x2, y2) is sqrt((x2 - x1)^2 + (y2 - y1)^2).
Question: What is the formula to convert radians to degrees for the sine function? Answer: The formula is sin(radians) * 180 / π.
Question: What is the formula to calculate the distance between two points using the Pythagorean theorem? Answer: The formula is sqrt((x2 - x1)^2 + (y2 - y1)^2), where (x1, y1) and (x2, y2) are the coordinates of the two points. | 677.169 | 1 |
Trigonometry-basics/67533: If you are given values for two sides of an oblique triangle and an angle opposite one of them, how would you arrive at the values of the missing parts? Will this result in a unique triangle? 1 solutions Answer 48049 by Earlsdon(6291) on 2007-01-22 00:42:51 (Show Source):
You can put this solution on YOUR website! Given the values of two sides of an oblique triangle ABC (one that does not contain a right angle) and an angle opposite one of them, how would you find the values of the missing parts?
Assume that we are given sides a and b and angle A
There are three cases to consider in this situation:
1. If angle A is acute and the length of side a lies between b and b sin A, then there will be two soltions.
2. If angle A is acute and the length of a < b sin A, or if if angle A is obtuse and a < b or a = b, then there is no solution.
3. For all other cases, there is one solution.
Let's consider case 3. We will be applying the law of sines: = =
To find angle B: Solve for sin B
To find side c: Solvefor c.
Geometry_Word_Problems/67480: An open-top box is to be constructed from a 4 ft by 6 ft rectangular cardboard by cutting out equal squares at each corner and folding up the flaps. Let x denote the length of each side of the square to be cut out.
Find the function V that represents the volume of the box in terms of x.
You can put this solution on YOUR website! The base of the new box, after the corners (x by x) have been removed, will have the dimensions (4-2x)ft. by (6-2x)ft. and the height of the box will be x ft.
The function, V, that represents the volume of the box will be: Simplifying this: or
Miscellaneous_Word_Problems/67446: A 12 foot ladder leans against the side of a building, making an angle of 30 degrees with the ground. To increase the reach of the ladder against the building, the ladder is moved so that it makes an angle of 60 degrees with the ground. To the nearest tenth of a foot, how much further up the building does the ladder now reach? 1 solutions Answer 47994 by Earlsdon(6291) on 2007-01-21 13:57:01 (Show Source):
You can put this solution on YOUR website! Let h1 be the height of the first reach. Solve for h1. ft.
Let h2 be the new height. Solve for h2.
The additional reach = h2-h2
Question: If the length of side 'a' is 8 and the angle opposite it is 45 degrees, and the length of side 'b' is 10, how many solutions are there for the missing parts of the triangle? Answer: There is one solution, as it falls under case 3 in the text.
Question: Which of the following scenarios will result in two solutions for the missing parts of an oblique triangle: a) sides of lengths 3 and 4, angle of 60 degrees, b) sides of lengths 5 and 5, angle of 90 degrees, c) sides of lengths 7 and 8, angle of 30 degrees? Answer: a) sides of lengths 3 and 4, angle of 60 degrees (since 3 lies between 4*sin(60°) and 4)
Question: What is the additional reach of the ladder when it is moved from a 30-degree angle to a 60-degree angle? Answer: The ladder reaches 5.2 feet further up the building.
Question: If a ladder makes a 45-degree angle with the ground, how much further up the building does it reach compared to when it makes a 30-degree angle? Answer: It would reach 1.4 feet further up the building. | 677.169 | 1 |
Define Ellipse...
[Más] has been squashed into
an oval.
Like a circle, an ellipse is a type of line.
Imagine a straight line segment that is bent
around until its ends join.
Then shape that loop until it is an ellipse - a sort of squashed circle
like the one above.
Things that are in the shape of an ellipse are said to be elliptical .
In mathematics, an ellipse (from Greek λλειψις elleipsis, a "falling short") is a plane curve thatἔ
results from the intersection of a cone by a plane in a way that produces a closed curve.
Circles are special cases of ellipses, obtained when the cutting plane is orthogonal to the
cone s axis.
An ellipse is also the locus of all points of the plane whose distances to two fixed
points add to the same constant.
Ellipses are closed curves and
[Menos]
Question: What is the constant sum of distances to two fixed points for any point on an ellipse? Answer: The same constant. | 677.169 | 1 |
Divided into the traditional four quadrants, the Pitsco Coordinate Geometry Board invites students to plot x, y coordinates; find the slope of a line between two points; graph a line; learn various quadrant characteristics; approximate areas of geometric figures; and much, much more!
This attractive, sturdy, 11" x 11" board is designed with 121 peg holes that snugly hold grooved wooden pegs. Simply insert pegs in various holes (x, y coordinates) and use the colored rubber bands to show various line segments. Holes are also available on both the x- and y-axes.
Question: What can students learn to do with this board? Answer: Plot x, y coordinates, find the slope of a line, graph a line, learn various quadrant characteristics, approximate areas of geometric figures. | 677.169 | 1 |
How to Draw 3D Rectangle?
To sketch 3 - Dimensional Rectangle means we are dealing with the figures which are different from 2 – D figures, which would need 3 axes to represent them. So, how to draw 3D rectangle?
To start with, first make two lines, one vertical and another horizontal in the middle of the paper such that they represent a "t" letter of English. This is what we need to draw for temporary use and will be removed later after the construction of the 3 – D rectangle is complete. Next we draw a Square whose measure of each side is 1 inch. Square must be perfect in Geometry so that 90 degree angles that are formed at respective corners are exact in measure. Now starting from upper right corner of the square we draw a line segment that will be stretched to a measure of 2 inches in the direction at an angle of 45 degrees. Similarly, we repeat the procedure by drawing another Line Segment from the upper left corner of the square and stretching it to 2 inches length in the direction at an angle of 45 degrees. These 2 line segments are considered to be the diagonals with respect to the horizontal line that we drew temporarily in starting. Also these lines will be parallel to each other. Next we draw a line that joins the end Point of these two diagonals.
Next starting from the very right of the 2 inch diagonal end point, draw a line of measure 1 inch that is supposed to be perpendicular to the temporary horizontal line. Next we need to join the lower left corner of the square with end point of the last 1'' line we drew in 4th step and finally we get our 3 - D rectangular. Now we can erase our initial "t". This 3- D rectangle resembles a Cuboid.
Question: What is the final shape that is drawn after all the steps? Answer: A 3D rectangle (or cuboid)
Question: What are the two lines drawn initially to represent a "t" letter? Answer: A vertical line and a horizontal line. | 677.169 | 1 |
If P is above any line coming from the uppermost point of the triangle, it is outside of the face. Likewise, if it is below a line from the lowest point, it is outside. If it is neither of these things, it MAY be in the face. If P is either above or below BOTH lines coming from the leftmost point, it is outside. If P is above or below both lines from the rightmost point, it is outside. Otherwise, if MAY be in the face.
If both these say that the point MAY be in the face, it IS in the face. However, if there are horizontal or vertical lines in the triangle, the following checks must be performed. In the case where there is only either a horizontal or only a vertical line, the other check may be ommited.
If there is a horizontal line AB and point O, then if O is above AB, P is outside if P is below AB. If O is below AB, then P is out when above AB. Otherwise, it MAY be in the face.
If there is a vertical line AB in the triangle, and the remaining point O is to the left of it, P will be outside the triangle if P is to the right of the line. If O is to the right of AB, then P is ouside if it is left of AB. Otherwise, it MAY be in the face.
Point O is the point NOT in either a horizontal or vertical line in a triangle. Im,agine a triangle with points at A(0,0) B(0, 10) and C(5, 5). The line AB is vertical, and could cause problems with determining the leftmost point, as both A and B are at the same X coordinate. C is the same as O in my previous post. Am I making sense?
As for n-sided polygons, I haven't done them because I thought you were going to just use triangles. To be honest, triangles would be easier to work with as they don't deform. Why do you think most graphics/modeling stuff uses them? To determine whether a point is in an n-sided polygon, divide the polygon into triangles and check if it is any of them. The smallest number of triangles an n-gon can be divided into is n-2, and this is acheived by joining points as follows.
V0 - V2 V0 - V3 V0 - V4 ... ... ... V0 - V(n-1)
I'm sorry for effectively saying "just do it my way", but I don't know any other way to do this.
No, that's ok. Splitting larger polygons into triangles might be the way to go.
So for polygons, consider this:
How do we split that into triangles?
Also, if we draw a triangle between E, F, and A, we could say that P lies inside that triangle. However, it still doesn't lie inside the polygon. What to do in this eventuality? Moreover, how do we detect this situation has occurred? Measure the angle between AF and EF?
Question: What should be done if a point P lies inside a triangle but not inside the polygon? Answer: Measure the angle between the edges of the triangle that P lies inside and check if it is less than 180 degrees. If it is, then P is not inside the polygon.
Question: How many triangles are needed to divide an n-sided polygon? Answer: The smallest number of triangles an n-gon can be divided into is n-2.
Question: What is the condition for a point P to be considered outside a triangle if there is a vertical line AB and point O is to the left of it? Answer: If O is to the left of AB, then P is outside if it is to the right of AB.
Question: What is the condition for a point P to be considered outside a triangle if it is below a line from the lowest point? Answer: If P is below a line from the lowest point of the triangle, it is outside of the face. | 677.169 | 1 |
Students must plot a point on a coordinate system. They have the option of using a calculator. This constructed-response 15. (sw)
Ohio Mathematics Academic Content Standards (2001)
Geometry and Spatial Sense Standard
Benchmarks (3–4)
G.
Find and name locations in coordinate systems.
Grade Level Indicators (Grade 3)
3.
Find and name locations on a labeled grid or coordinate system; e.g., a map or graph.
Grade Level Indicators (Grade 4)
6.
Specify locations and plot ordered pairs on a coordinate plane, using first quadrant points.
Principles and Standards for School Mathematics
Geometry Standard
Specify locations and describe spatial relationships using coordinate geometry and other representational systems
Expectations (3–5)
make and use coordinate systems to specify locations and to describe paths;
Question: Which standard does this task align with? Answer: Ohio Mathematics Academic Content Standards (2001) | 677.169 | 1 |
Archive
Well, this will probably be the last, or nearly last post of a busy year that involved a lot of work on gigs and little focus on this blog. Sorry, maybe next year will be better
In the previous example, I showed how to rotate a box (rectangle) around an arbitrary point, but the algorithm and code never presumed anything about the geometric nature of the object. It's possible to extend the exact same algorithm to a collection of points and that is the subject of the current post.
Instead of maintaining variables for the four vertices of a rectangle, the code was modified to work with an arbitrary point collection. A RotatablePoint class was used to hold data points and offload some of the computations. This greatly simplifies the actual demo and provides you with ample means for experimentation.
The online example starts with a small collection of points as shown below.
These points are rotated about the fixed point, drawn as a red dot. The drawing may be cleared after each rotation increment or continually updated from the prior drawing as shown below.
Math teachers hear the most often-uttered student phrase in history a lot more than I do, but my ears have twitched to the infamous, "I'll never use that" or its many variants over the years. Funny how people know right now what will and will not be needed or used for the rest of their lives
Now, taken in strict isolation, a single trig or algebra formula might appear to have limited or no use in the myriad of life situations in someone's future. It is, however, quite interesting how often problems can be solved by strategic use of a single formula from multiple areas of study. The problem discussed in this post is one such example.
In the prior post, I illustrated how some basic trig concepts could be used to solve a rotation and bounding-box problem without any presumptions or special considerations of the programming environment. This example is similar in that it illustrates how to rotate a box around an arbitrary point in its interior. The box is represented by a sequence of four coordinate pairs. It is not a Flash symbol or anything other than a sequence of coordinates. The drawing environment is simple; it can move the pen, draw lines, and draw filled circles. The programming environment contains the typical set of math functions.
Our math background includes a semester of trig, analytic geometry, and algebra. We know the very basics of vectors, but have not been introduced to matrices or matrix/coordinate transforms. As it happens, all we need to know to solve the stated problem is
1 – Polar coordinates, i.e. x = r cos(a) , y = r sin(a)
2 – Parametric equation of a straight line, i.e. P = (1-t)P0 + tP1
3 – How to add and subtract vectors
4 – How to compute the distance between two points
Refer to the diagram below.
Question: What is the common phrase math teachers often hear from students? Answer: "I'll never use that" or its many variants. | 677.169 | 1 |
drawing triangles with python's turtle graphic
A. Write a triangle solver that takes 3 inputs consisting of angles in degrees and length
of sides in arbitrary units and, if possible (your program has to determine this),
supplies all other angles and side lengths. There will be either 0 triangles, 1 triangle,
2 triangles or an infinite number of triangles – you program must determine which
possibility. Classify the type of triangle (right triangle, isosceles, equilateral, acute,
obtuse – see any good geometry book. Finally, use the Turtle module to draw the
triangle(s) with correct relative dimensions.
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The key part is understanding that all triangles will have a total of 180* as the sum of all angles. That means, given your 3 inputs, if they don't add up to 180* then it's not a triangle.
You have a lot of choices on how to start. A better way for us to help you is to start writing some code, and when you have trouble or get stuck, then ask. Homework is meant for you to learn, not for people on a message board to do it for you.
Question: What are the types of triangles that the program should classify? Answer: Right triangle, isosceles, equilateral, acute, obtuse
Question: Which module should be used to draw the triangle(s)? Answer: Turtle module | 677.169 | 1 |
Themes by Topic
Arkansas 2 Students will identify and describe types of triangles and their special segments. They will use logic to apply the properties of congruence, similarity, and inequalities. The students will apply the Pythagorean Theorem and trigonometric ratios to solve problems in real world situations.
T.2.G.7 Use similarity of right triangles to express the sine, cosine, and tangent of an angle in a right triangle as a ratio of given side lengths, including angles of elevation and angles of depression
Question: Which mathematical properties will the students apply? Answer: The properties of congruence, similarity, and inequalities. | 677.169 | 1 |
Prop
tangentequal
PROP. IV.
The tangent at any point of a parabola bisects the angle between the focal distance of the point and the perpendicular from the point on the directrix.
Let PZ (fig. 7) be the tangent at P, meeting the directrix in Z ; then, if PM be drawn per pendicular to the directrix, it is easily seen that the two triangles SPZ, MPZ are equal in all respects, and the angle SPZ equal to the angle MPZ.
If SAI be joined, it can be shown that it is bisected at right angles by PZ, and that its middle point is the point Y in Prop. ii.
The line AY, it will be observed, is the tangent to the parabola at the vertex A.
It appears, therefore, that the locus of the foot of the perpendi, calar front the focus on the tangent at any point is the tangent at the vertex.
It can also be seen that, if the tangent at P meet the axis in '1', then SP= ST. For the angles STP, SPT are each equal to the angle MPT, and therefore (Encl. i, 6) SP, ST are equal.
It may further be remarked. that, if 0 be any point in the tangent at P, then the triangles SPO, 311'0 are equal in all respects.
If PN be drawn perpendicular to the axis to meet it in N, then it will be seen that PN = 2AY and TN - 2AN = 2AT
Question: Which line is the tangent to the parabola at the vertex A? Answer: The line AY.
Question: What is the relationship between triangles SPZ and MPZ in the given proposition? Answer: They are equal in all respects. | 677.169 | 1 |
Pairs of points on a sphere that lie on a straight line through its center are called antipodal points. In Mathematics, the antipodal point of a point on the surface of a sphere is the point which is diametrically opposite it — so situated that a line drawn from the A great circle is a circle on the sphere that has the same center and radius as the sphere, and consequently divides it into two equal parts. A great circle is a Circle on the surface of a Sphere that has the same circumference as the sphere dividing the sphere into two equal Hemispheres. The shortest distance between two distinct non-antipodal points on the surface and measured along the surface, is on the unique great circle passing through the two points.
If a particular point on a sphere is designated as its north pole, then the corresponding antipodal point is called the south pole and the equator is the great circle that is equidistant to them. The equator (sometimes referred to colloquially as "the Line") is the intersection of the Earth 's surface with the plane perpendicular to the Great circles through the two poles are called lines (or meridians) of longitude, and the line connecting the two poles is called the axis of rotation. This article is about the geographical concept For other uses of the word see Meridian rotation is a movement of an object in a circular motion A two- Dimensional object rotates around a center (or point) of rotation Circles on the sphere that are parallel to the equator are lines of latitude. Latitude, usually denoted symbolically by the Greek letter phi ( Φ) gives the location of a place on Earth (or other planetary body north or south of the This terminology is also used for astronomical bodies such as the planet Earth, even though it is neither spherical nor even spheroidal (see geoid). EARTH was a short-lived Japanese vocal trio which released 6 singles and 1 album between 2000 and 2001The geoid is that Equipotential surface which would coincide exactly with the mean ocean surface of the Earth if the oceans were in equilibrium at rest and extended through
A sphere is divided into two equal hemispheres by any plane that passes through its center. If two intersecting planes pass through its center, then they will subdivide the sphere into four lunes or biangles, the vertices of which all coincide with the antipodal points lying on the line of intersection of the planes.
Generalization to other dimensions
Spheres can be generalized to spaces of any dimension. In Mathematics, an n -sphere is a generalization of an ordinary Sphere to arbitrary Dimension.In mathematics the dimension of a Space is roughly defined as the minimum number of Coordinates needed to specify every point within it For any natural numbern, an n-sphere, often written as Sn, is the set of points in (n+1)-dimensional Euclidean space which are at a fixed distance r from a central point of that space, where r is, as before, a positive real number. In Mathematics, a natural number (also called counting number) can mean either an element of the set (the positive Integers or an In particular:
Question: What is the shortest distance between two distinct non-antipodal points on the surface of a sphere measured along the surface? Answer: It is on the unique great circle passing through the two points.
Question: Which circle on a sphere divides it into two equal parts? Answer: A great circle.
Question: What is the geoid? Answer: The geoid is the equipotential surface which would coincide exactly with the mean ocean surface of the Earth if the oceans were in equilibrium at rest and extended through. | 677.169 | 1 |
Much work has been done on the medians of triangles: segments that connect vertices with the midpoints of opposite sides. This morning, I decided to explore what happens if the trideans are examined, rather than the medians.
The reason you don't know the word "tridean" is simple: I just made it up. It's related to a median, though. To find trideans, you must first trisect, rather than bisect, each side of a triangle. This gives you two equally-spaced points on each side. To form a tridean, simply connect one of those points to the opposite vertex. Every triangle has six trideans, and they split the triangle into a set of non-overlapping polygons, as you can see in the diagram.
When I examined the area of these polygons, I started finding unexpected things right away. As you can see, polygons of the same color in this diagram have the same area, even though they are non-congruent. Moreover, these areas are interesting fractions of the area of the entire triangle. The six yellow triangles, for example, each have 1/21st of the area of the entire triangle. Each blue triangle is 1/70th the area of the large triangle. Each green quadrilateral is 1/14th the area of the large triangle. 21, 70, and 14 have one factor in common: the number seven. Seven? I officially have NO idea why sevens are popping up all over the place in this investigation, but there they are.
The red hexagon in the middle has 1/10th the area of the entire triangle, and this number surprised me as well.
The three orange pentagons took a little more work. As you can see, dividing the area of the large triangle by the area of one of these orange pentagons yields 9.5454, with the "54″ repeating. This decimal is 105/11. (Eleven?) At least 105 has three as a factor (as does 21, from earlier); three is the number I expected to pop out all over the place, but it shows up little in this investigation. However, what are the other factors of 105? There's five — and, yet again, seven.
These sevens are everywhere in this thing, and I have no idea why.
Now, of course, I have proven none of this. This is merely a demonstration and explanation of something I think is a new discovery. I did change the shape of the triangle many ways, as a test, and each of these area ratios remained constant.
If anyone can shed some light on any of this — especially all these sevens — please comment.
Question: What is this repeating decimal as a fraction? Answer: 105/11
Question: What is the fraction of the area of the entire triangle for the red hexagon? Answer: 1/10th | 677.169 | 1 |
Regular Solids
Regular solids (regular polyhedra, or Platonic solids which were
described by Plato) are solid geometric figures, with identical regular
polygons (such as squares) as their faces, and with the same number of faces
meeting at every corner (vertex).
If each corner consisted of two squares, then we end up with a "solid"
made up of two squares, back to back. That's not an interesting solid. So it is
not legal, as a regular solid. This eliminates the infinite number of regular
polygons (triangle, square, pentagon, ...) back to back. These cannot form
regular solids. So, we deduce that each vertex must have at least three faces.
If each corner consisted of four squares, we cannot enclose a space.
The object that we do get is a plane, covered with squares. We say that squares
can tile the plane. So we cannot produce a regular solid with four squares at
each vertex. If the angles at our vertex add up to 360°, then we tile the
plane, and cannot have a regular solid.
If we try to add a fifth square, we find that we have no room for our
fifth square. We would have to deform our squares. And then they would cease to
be squares. So, if the angles of the vertex add up to more than 360°, then
we cannot fit these polygons into one corner, and we get no regular solid.
So, 360° is the upper limit on the sum of the angles at a vertex.
And, we cannot have fewer than three faces at a vertex. So, there are a limited
number of regular solids:
The equilateral triangle is the simplest regular polygon. Let's
start with three equilateral triangles at a vertex (total angle 180°). And
we get a tetrahedron (4 faces, 4 vertices).
We next try four equilateral triangles at each vertex (total angle
240°). And we get an octahedron (8 faces, 6 vertices).
Now we try five equilateral triangles at each vertex (300°) We
end up with an icosahedron (20 faces and 12 vertices). A sixth equilateral
triangle (at a vertex) will tile the plane (360°).
So, we try the second simplest regular polygon, the square. Three
squares at each corner (270°) forms a cube, or hexahedron (6 faces and 8
vertices). And a fourth square tiles the plane (360°), as we saw with six
triangles.
The next simplest regular polygon is the regular pentagon. Three
pentagons at each vertex (324°) produces a dodecahedron (12 faces and 20
vertices). Three regular hexagons, at each vertex, tile the plane (360°).
And there is no room, at a vertex, for more complicated regular polygons.
So, there are only five regular solids:
solid
faces
vertices
edges
Question: Which regular polygon cannot form a regular solid when tried at a vertex? Answer: Square (when four squares are used at each vertex)
Question: Which famous philosopher described these regular solids? Answer: Plato
Question: What is the minimum number of faces that can meet at a vertex in a regular solid? Answer: Three
Question: What is the name of the regular solid with 12 faces and 20 vertices? Answer: Dodecahedron | 677.169 | 1 |
Here are models for the five regular polyhedra:
If you are using Netscape or Internet Explorer, you can probably right
click on the image and choose to save it on your computer. Expand them with a
paint program. If you want to print them, view these
models separately, so you don't print this whole page.
Addendum #6:
Besides finding the five regular solids above, we also discovered the
three regular tilings (tessellations) of the plane (equilateral triangles,
squares, and hexagons):
These are the only ones possible with identical regular polygons. See
Tesselations.
Addendum #7:
Here are animations of a rotating tetrahedron, cube, octahedron,
icosahedron, and dodecahedron.
I am told that some dictionaries say that a polygon has five or more
sides. My dictionaries say three or more sides. And of course, the mathematical
definition says three or more sides.
I hear that Euclid's definition of the above solids is incomplete (as
are a few other of his definitions throughout his books). He calls the above
polyhedra, not regular polyhedra, and just says that each face is a
regular polygon. He obviously means that they are regular, but does not define
that. His definition of the icosahedron is a polyhedron with 20 faces. Well
there are several such solids, some very irregular (combinations of tetrahedra
and octahedra stuck together at odd angles, perhaps). But from the theorem
(that there are only five solids), he obviously means that they are
regular.
Question: How many faces does a regular icosahedron have, according to the text? Answer: 20 faces. | 677.169 | 1 |
The convenience of this geometry is that lines in our real 2d plane that don't go through the origin can be thought of as the intersections of 2d subspaces with our real 2d plane. Given two points on our real 2d plane [itex]p[/itex] and [itex]q[/itex], the real line containing them is [itex]L = p \wedge q[/itex], and the shortest distance from that line to the origin is [itex]e_0 \cdot (e_0 \wedge L)[/itex].
There are a lot of symbols here that might not be understood, so let me give an example. Let [itex]L[/itex] be the line through x=1, y=0 and x=0, y=1. By simple geometry, we know that the distance from the origin to this line is [itex]1/\sqrt{2}[/itex]. Let's verify that with projective geometry.
Let [itex]p = e_0 + e_1[/itex] and [itex]q = e_0 + e_2[/itex]. The line [itex]L[/itex] is then [itex]p \wedge q = e_0 \wedge e_2 + e_1 \wedge e_0 = e_{02} + e_{10} + e_{12}[/itex]. First, we find the volume that the 2d subspace makes with [itex]e_0[/itex]. To do this, we just wedge in another [itex]e_0[/itex] on the front:
And then we find the part of this volume that is perpendicular to [itex]e_0[/itex]. This gives us the "footprint" or "shadow" of the 2d subspace on the real 2d plane. This quantity is called the "moment" [itex]M[/itex] of the line.
[tex]M = e_0 \cdot (e_0 \wedge L) = e_{12}[/tex]
Question: What is the moment M of the line L in the given example? Answer: The moment M of the line L in the given example is e12. | 677.169 | 1 |
Wrong Triangle
1. In the amazing book 1000 PlayThinks
by Ivan Moscovich (I highly recommend it; order it from
Amazon.com),
we see the diagram on the left, with this explanation:
Hidden Triangle Trisect the angles of a triangle, as shown. Note
that three points within the triangle form an equilateral triangle. Does such
an equilateral triangle appear in every trisected triangle?
Answer: The answer, of course, is yes. But the
diagram is wrong. The small triangle in the diagram is not an equilateral
triangle. The diagram on the right shows the real equilateral triangle. This
was discovered by Frank Morley in 1899. See Morley's Theorem.
2. On this puzzle, the same book misleads slightly:
How many diagonals are needed to divide a heptagon (7-gon), a nonagon
(9-gon), and an undecagon (11-gon) into triangles?
In other words, how few lines do you need to draw, in order to divide
each polygon into triangles. The book shows three seemingly regular polygons
(although the text did not say "regular"). The answer that the book gives is,
"In general, a convex polygon of n sides requires n-3 diagonals to triangulate
it." That sounds true, in general. But many special n-gons require fewer lines.
A regular octagon requires only four lines. Above left is an irregular nonagon
which requires only five lines.
3. The same book shows these squares: 5x5, 4x4,
and 3x3. It asks:
Can you dissect the 5x5 square into the fewest number of pieces needed
to make both a 4x4 square and a 3x3 square?
Answer: The solution given in the book is five pieces. On the
right is a solution with four pieces.
4. Another puzzle in the same book is this question: "What three
numbers have a sum equal to their product?" The book's answer is 1+2+3=1x2x3.
But there are "trivial" solutions like 0, 0, 0 or -17, 0, 17, or more
interesting solutions like -1, 1/2, 1/3. That may seem picky of me to mention
that, but this is the kind of criticism (constructive, of course) that I get in
my email.
5. Yet another puzzle in the same book is this question: "Can you
find a square and a rectangle that have perimeters equal to their areas?" You
might want to work on that.
The square is 4x4 and one of the infinitely many rectangles is 3x6. The
book says that those are the only solutions. Well, those are the only solutions
in integers. But 2.5x10 works, as does 3.5x14/3, and infinitely many
other triangles.
Question: What are some non-trivial examples of three numbers that have a sum equal to their product? Answer: -1, 1/2, 1/3.
Question: What is one example of a square and a rectangle that have perimeters equal to their areas? Answer: A 4x4 square and a 3x6 rectangle.
Question: How many lines are needed to divide a regular octagon into triangles? Answer: Four lines. | 677.169 | 1 |
Pythagorean-theorem/612386: A right triangle has a hypotenuse of length 25 and a leg of length 20. What is the length of the right leg? If necessary, round your answer to two decimal places. 1 solutions Answer 385414 by vleith(2825) on 2012-05-18 08:23:32 (Show Source):
You will notice that the sides are in a ratio of 15:20:25
Which is 3:4:5
Any triangle in ratio 3:4:5 is a right triangle.
That comes up a lot, so you might want to memorize that.
Polygons/612390: The diagram shows 3 sides of a regular polygon.
Each interior angle of the regular polygon is 140 degrees, work out the number of sides of the regular polygon. 1 solutions Answer 385413 by vleith(2825) on 2012-05-18 08:19:55 (Show Source):
You can put this solution on YOUR website! Look at this URL
Read it and see how one can build on what you know about triangles and extend that into the general rule.
In your specific case,
So simplify and solve for n
You can put this solution on YOUR website! let the number be represented by x isolate the variable term by subtracting 7 from both sides isolate the variable by dividing both sides by 2
the answer is 3/2 or 1.5
test/612086: Hi! I need help with 5 algebra 1 math problems..i don't understand them at all.:( could u please email me back ASAP. I need to know the answers to the 5 problems but also how to do them!!! 1 solutions Answer 385279 by vleith(2825) on 2012-05-17 15:17:28 (Show Source):
You can put this solution on YOUR website! The rational zeros theorem (also called the rational root theorem) is used to check whether a polynomial has rational roots (zeros). It provides a list of all possible rational roots of the polynomial equation , where all coefficients are integers. If the equation has rational roots p/q, where p and q are integers, then p must divide evenly into the constant term a0 and q must divide evenly into the leading coefficient an. In other words, p is a factor of ±a0 and q is a factor of ±an.
get all the factors in the high order coefficient. Then get all the factors in the constant term
2 -- -2,-1,1,2
-6 -- -6,-3,-2,-1,1,2,3,6
now make every combination of the ratio of factors of -6 to factors of 2
Question: What is the rational zeros theorem used for? Answer: The rational zeros theorem is used to check whether a polynomial has rational roots (zeros).
Question: What is the length of the other leg of the right triangle? Answer: The other leg is 20 units long. | 677.169 | 1 |
By considering the circles as vertexes, and stacking these five triangles on top of each other, we can create a "four-frequency" tetrahedron (4F tetra). This name is derived from the four vectors along each edge. Below is a model of such a tetrahedron.
The 4F tetra is the smallest tetrahedron that has a "nuclear" vertex. If these vertexes were spheres, there would be one sphere at the very center, surrounded by twelve other spheres. This nucleus represents the neutral set of {0, 1, 2} in the TEC.
Question: How many vectors are along each edge of the 4F tetrahedron? Answer: Four | 677.169 | 1 |
There is something wrong with the third problem, if 2 of the three angels of the triangles formed are the same it means the third angles are the same meaning so why are the arcs different lengths?
Answers
Nothing is wrong with this problem because no central angles are present. If the vertex of the third angle in each triangle rested on the center of the circle, then you would be correct: congruent central angles must intersect congruent arcs. However, nothing in the problem states that the intersection of those two chords is at the center of the circle.
Question: What does the problem lack to make the user's initial assumption correct? Answer: The problem lacks the detail that the intersection of the two chords is at the center of the circle. | 677.169 | 1 |
It has to do with the fact that terms like North, South, East, and West become undefined in specific locations. Consider if you were one mile north of the south pole; You'd go one mile south, there would be no east to travel (so you'd stand still) and then when you go one mile north, you could end up where you started- Or at an infinite number of other locations forming a circle 1 mile north of the pole.
Alternately, if you're willing to say that you *start out* heading N/S/E/W, then you can do some tricks with crossing the north pole and having the directions all change names.
Avistew
05/14/2010, 08:20 am
That's a quarter of a Elipse! You screw me!
Okay, a quarter of an ellipse then. I'm so math-challenged that I just thought it was a bigger circle and off-center >.>
I totally forgot that, you right!
Grab a rule[...]
Er, yeah, if I have a ruler I'll measure it directly and I'll know what size it is :p I was assuming you were supposed to calculate it.
I guess I'm just making things complicated. You just assume it's a quarter of a circle. Like I did.
I just was fairly sure math-people wouldn't just assume something like that without checking first in a more scientific way than their eyometre.
taumel
05/14/2010, 08:23 am
This is boring, a more interesting but still easy to answer question would be: How does a turtle (with some painting on her tail) has to move, in order to draw the picture (the lines and the circle) so that every distance is only walked once. Those interested in, also can offer a turtle code otherwise a picture showing the way is sufficient.
Psy
05/14/2010, 08:28 am
If you're referring to the semicircle image that is being discussed, it can't be done.
Avistew
05/14/2010, 08:29taumel
05/14/2010, 08:30 am
@Psy
Are you sure? :O)
Psy
05/14/2010, 08:33 am
@Psy
Are you sure? :O)
Yes.
taumel
05/14/2010, 08:38 am
You're a winner! *tadaaa*
GinnyN
05/14/2010, 09:05
Klatuu
05/14/2010, 10:41 am
The geometry puzzle reminded me of another site I vistited before -
- and some puzzles called sangaku ( For example:
Question: What happens if you travel one mile north of the South Pole? Answer: You end up on a circle with a radius of one mile, north of the pole. | 677.169 | 1 |
This Java applet enables students to investigate acute, obtuse, and right angles. The student decides to work with one or two transversals and a pair of parallel lines. Angle measure is given for one angle. The student answers a short series of questions about the size of other angles, identifying relationships such as vertical and adjacent angles and alternate interior and alternate exterior angles.
Triangle Geometry: Angles
This site directly addresses students as it leads them to explore angles and their measurement. Most important, it offers applets to introduce the Pythagorean theorem by collecting data from right triangles online and provides an animated picture proof of the theorem.
Manipula math with Java : the sum of outer angles of a polygon
This interactive applet allows users to see a visual demonstration of how the sum of exterior angles of any polygon sums to 360 degrees. Students can draw a polygon of any number of sides and have the applet show the exterior angles. They then decrease the scale of the image, gradually shrinking the polygon, while the display of the exterior angles remains and shows how the angles merge together to cover the whole 360 degrees surrounding the polygon.
Parallel Lines and Ratio
Three parallel lines are intersected by two straight lines. The classic problem is: If we know the ratio of the segments created by one of the straight lines, what can we know about the ratio of the segments along the other line? An applet allows students to clearly see the geometric reasoning involved.
Area triangles
This applet shows triangle ABC, with a line through B parallel to base AC. Students can change the shape of the triangle by moving B along the parallel line or by changing the length of base AC. What happens to the length of the base, the height, and the area of the triangle as students make these moves? Why?
Understanding the Pythagorean Relationship Using Interactive Figures
The activity in this example presents a visual and dynamic demonstration of this relationship. The interactive figure gives students experience with transformations that preserve area but not shape. The final goal is to determine how the interactive figure demonstrates the Pythagorean theorem.
Distance Formula
Explore the distance formula as an application of the Pythagorean theorem. Learn to see any two points as the endpoints of the hypotenuse of a right triangle. Drag those points and examine changes to the triangle and the distance calculation.
Measuring by Shadows
A student asks: How can I measure a tree using its shadow and mine? This letter from Dr. Math carefully explains the mathematics underlying this standard classroom exercise.
Finding the Height of a Lamp Pole
Without using trigonometry, how can you find the height of a lamp pole or other tall object? Two methods, both depending on similar triangles, are outlined and illustrated. A rich problem.
Polygon Capture
In this lesson, students classify polygons according to more than one property at a time. In the context of a game, students move from a simple description of shapes to an analysis of how properties are related.
Sorting Polygons
In this companion to the above game, students identify and classify polygons according to various attributes. They then sort the polygons in Venn diagrams according to these attributes.
Question: Which mathematical concept is explored in the "Distance Formula" activity? Answer: The Pythagorean theorem.
Question: What is the main goal of the "Understanding the Pythagorean Relationship Using Interactive Figures" activity? Answer: To determine how the interactive figure demonstrates the Pythagorean theorem.
Question: Does this applet help students understand the Pythagorean theorem? Answer: Yes, it does. | 677.169 | 1 |
Geometric Surfaces
Three Dimensions
Surfaces
Just like a curve is the basic building block for
figures in a plane, a surface is the
basic building block for figures in space. A
surface is essentially a curve with depth. Curves and surfaces are analogous in
many ways. If you think of a curve as being the trace of the motion of a
point in a plane, a surface is like the trace
of the motion of a curve in space. Surfaces are
continuous, meaning that given two points on a
surface, you can start from one and reach the other without leaving that
surface. Just like a curve is still one-dimensional, a surface, although it exists in
three dimensions, is still two-dimensional. For example, when you build a curve
by tracing the motion of a point, that curve, although it spans both length and
width, has no width of its own. The curve doesn't have area, it only has
length, one dimension. Similarly, a surface can span more than one plane, but
it still does not have depth of its own. It only has two dimensions, length and
width. We will work mostly with the simplest surface, a plane. Below various
surfaces are pictured.
Figure %: Surfaces in space
Surfaces can be classified as being closed or simple closed surfaces. The
surfaces that form the boundaries of geometric solids are simple closed
surfaces, so we'll focus on them. A simple closed surface is one that divides
space into three distinct regions:
The set of all points inside the surface (the interior of the surface).
The set of all points outside the surface (the exterior of the surface).
The set of all points on the surface.
A point is only interior if it can be joined to any other interior point by a
segment of finite length. This is not true
for exterior points--a segment joining exterior points could have infinite
length, since the endpoints could be anywhere in space, and space is infinite.
A simple closed surface can also be either convex or concave. The rules
are very similar to those we saw in Polygons. A convex
surface is one in which any two points on that surface can be joined by a segment that lies either on the
surface or in the interior of the surface. A concave surface has a segment
between points on the surface that lies in the exterior of the surface.
One more note on surfaces: a surface, even if it is a simple closed surface,
does not include the space in its interior. When a simple closed surface
is united with its interior points, it is no longer a surface, it is a geometric
solid.
Lines and Planes
So far we've only discussed parallelism and
perpendicularity with respect to
lines, but planes can be parallel and
perpendicular, too. To understand relationships between planes, one must
understand relationships between lines and planes.
A line and a plane are parallel if and only if they do not intersect. A line
l and a plane are perpendicular if and only if the line l is
Question: What are the two types of simple closed surfaces based on their shape? Answer: Convex and concave
Question: Can a segment joining exterior points have infinite length? Answer: Yes
Question: What is the relationship between a line and a plane if the line is perpendicular to the plane? Answer: The line is perpendicular to the plane | 677.169 | 1 |
When students record the differences between successive perimeters they should see both that the perimeter is increasing and that the amount of increase is larger each time. This suggests the notion of unlimited growth in the perimeter. The ratio between perimeters with each iteration is greater than one, a fact which again suggests unlimited growth. (In advanced algebra the set of perimeter values would be referred to as a divergent geometric sequence.)
step
perimeter
differences
perimeter
ratio
0
1
2
3
81
108
144
192
27
36
48
81
108
144
192
4/3
4/3
4/3
Table 2
Remember that area is measured in units of the smallest triangle on the grid paper. For the first triangle with side length twenty-seven units, finding the area without actually counting all the triangles is a mathematical exercise in itself! After that first step, help the students to realize that since each iteration changes the linear measure by a factor of 1/3, it produces a triangle with area 1/9 that of the previous triangle. This concept is normally taught in the study of similar figures. If students are not familiar with this idea, a drawing such as Figure 6 should be helpful.
Figure 6
Beginning with step 2, the students can see that each iteration adds four new triangles to each Koch curve generator. Because these new triangles have 1/9 the area of the previous step, each iteration adds 4/9 of the area added the previous time. Although the area is increasing, just as the perimeter did, the amount of increase in this case is smaller each time. The ratios between successively larger areas are also decreasing. Because the rate of change in area is decreasing, there must be some limit to the total area. If students have difficulty in grasping that concept, perhaps it would help to show them the circle which circumscribes the triangle. That circle definitely is an upper bound for the area of the snowflake; in fact, the area is much less. An advanced algebra class can look at the area as the following sum:
Table 3 shows the result of counting areas through three iterations, with an extension of the pattern to the fourth iteration.
step
side
area of
new unit
number of
new units
area to be
added
total area
ratio of
areas
0
27
729
1
729
729
1
9
81
3
243
972
2
3
9
3*4=12
4/9*243=108
1080
3
1
1
3*4*4=16
4/9*4/9*243=48
1128
4
1/3
1/9
3*4*4*4=64
4/9*4/9*4/9*243=64/3
1149 1/3
Question: What is the upper bound for the total area of the Koch snowflake? Answer: The area of the circle that circumscribes the triangle.
Question: If we extend the pattern to the fifth iteration, what would be the total area of the Koch snowflake? Answer: Not explicitly given in the text, but it can be calculated using the sum of a geometric series.
Question: Is the rate of change in area increasing or decreasing with each iteration? Answer: Decreasing.
Question: How much smaller is the area of each new triangle compared to the previous one? Answer: 1/9. | 677.169 | 1 |
In pairs, allow each student to cut a triangle for their partner to measure and calculate the area of. Each student should check the other student's results and work together to resolve any disagreements.
Extensions
Using the Internet, students should research the history of the Bermuda Triangle to determine its dimensions. Students can report back to the class with their findings. Some questions to ask the students include:
Is the Bermuda Triangle truly a triangle? If not, what shape is it? Why? If it's not a triangle, are you able to approximate the total area covered by the Bermuda Triangle?
Do you think there is a "center" to the Bermuda Triangle? How would you find it?
Allow students to research other famous triangles and calculate their areas. Some possibilities include the Sunni Triangle (which extends northwest from Baghdad, Iraq); Research Triangle Park (connecting Raleigh, Durham, and Chapel Hill, NC); and Point State Park (along the Allegheny and Monongahela Rivers in Pittsburgh, PA). In pairs, students can quiz each other and check their calculations.
Teacher Reflection
Did students come up with alternative methods for finding the areas of their triangles? If so, how did you react to their explanations?
What were some of the ways that the students illustrated that they were actively engaged in the learning process?
Did students demonstrate an understanding of how and why we use the formula A = ½bh?
Did you notice any positive or negative effects of asking students to check each others' work?
Did you find it necessary to make adjustments while teaching the lesson? If so, what adjustments, and were these adjustments effective?
Question: What should students do in pairs to start the activity? Answer: Each student should cut a triangle for their partner to measure and calculate the area of.
Question: Is the Bermuda Triangle truly a triangle? Answer: No, it is not a triangle. | 677.169 | 1 |
of an equlateral triangle are divided into 5 equal parts, and lines that are parallel to
the sides, are drawn through the dividing points. In essense, the triangle is divided into
smaller similar triangles. If you imagine that all lines are made of one piece of wire,
how long will that wire be if you bend it back into a straight line?
10. Is it possible to measure 4
gallons of water, if you have only 3 and 5 gallon jugs, and an unlimited supply of water?
11. What is the next number in the
following sequence:
1, 3, 7, 12, 18, 26, 35, 45, 56, ...
12. The sum of several consecutive odd
integers equals 333. What's the 5th number in the series?
Now you have done the test. Please click
"Preceding Page" at below to see the answers.
You can also jump to the
chapter of your choice by using the drop-down list at below.
Question: What is the next number in the sequence: 1, 3, 7, 12, 18, 26, 35, 45, 56,...? Answer: 68 | 677.169 | 1 |
"This program is filled with tips, strategies and basic math concepts, students use their geometric know-how about different groups of polygons to discover and solve formulas for perimeter and area. Learn about the triangle's unique relationship between the measurement of its angles and the lengths of sides, and discover how a rectangle can be used to solve for any triangle's area. Analyze the symmetry and transformation of regular polygons as they translate, reflect and rotate. Enjoy and explore the world of shapes and their properties"--Container.
Question: Which transformations of regular polygons does the program analyze? Answer: Translation, reflection, and rotation | 677.169 | 1 |
Then there are two ways to go about this. You can do it the easy way in your head if you say the hypotenuse is 10 and if the triangle were a 3-4-5 right triangle, the two legs would be 6 and 8. And since 6 and 8 differ by 2, Karen's house is 8 miles from the restaurant.
Or you can do it the hard way.
Let represent the distance from Karen's house to the restaurant. Then the distance from Dan's house to the restaurant is
Then Pythagoras:
You can take it from here. Toss the extraneous negative root.
John
My calculator said it, I believe it, that settles it
Miscellaneous_Word_Problems/338161: If a fraction simplifies to 1,what,if anything can you conclude about its numerator and denominator?
A.) the numerator can be anything, the denominator is 1.
B.) the numerator is 1, the denominator can be anything.
C.) the numerator & denominator are equal.
D.) the numerator is smaller than the denominator. 1 solutions Answer 242395 by solver91311(16868) on 2010-09-02 11:13:11 (Show Source):
If M is the midpoint, then the two line segments must be equal in measure. So set the two fractions equal to each other and solve for
John
My calculator said it, I believe it, that settles it
Angles/337997: Draw the angle sin(-pi/2) using a ray through the origin, and determine whether the sine, cosine, and tangent of this angle are positive, negative, zero, or undefined. 1 solutions Answer 242318 by solver91311(16868) on 2010-09-01 21:44:10 (Show Source):
is not an angle. It is a number representing the value of the function of an angle. Hence, your question makes no sense.
John
My calculator said it, I believe it, that settles it
Functions/337844: Okay i have this work sheet in honors alg 2.
You match the letters too the numbers.
But if someone cud just explain too me how too do it that would be great.
1. f(x)=1/x-1
2. f(x)=1/x^2-1
3. f(x)=Square root 3x
4. f(x)= 1/square root 2x
there is just a few. If you can help ill post more.
Thanks 1 solutions Answer 242203 by solver91311(16868) on 2010-09-01 17:19:38 (Show Source):
Question: What is the relationship between the numerator and denominator of a fraction that simplifies to 1? Answer: They are equal
Question: What is the value of x in the equation representing the distance from Dan's house to the restaurant? Answer: 6
Question: What is the result of sin(-π/2) according to the provided text? Answer: The question makes no sense as sin(-π/2) is not an angle but a number representing the value of the function of an angle.
Question: What is the function f(x) for the third equation in the worksheet? Answer: f(x) = √(3x) | 677.169 | 1 |
2.2.1 Euclidean Geometry - ignore location. The most familiar geometry, Euclidean, is generated by the group of transformations known as isometric transformations; these can alter the location of a form in its entirety, as if the form were simply picked up and dropped somewhere else. For instance, a square could be moved a few inches sideways, or rotated (see section 3.5 for more on rotation). The only property altered by an isometric transformation is absolute position. All other properties of the square remain unchanged, such as the distance between any two points on the square (i.e. length) before and after the transformation, the angles of the square, the parallelism of the two pairs of sides. All the properties which remain unchanged by isometric transformations are in Euclidean geometry but those that are altered are not. Thus, absolute location is not a property in Euclidean geometry. If a square is taken and "transformed" (which we call "moved") to a location 3 inches away, the square and the resulting form (which we also call a square) are equivalent forms in Euclidean geometry. Because absolute location is not in the geometry, it is irrelevant for distinguishing between forms. But if a square is put in a vice such that the top part is compressed and the sides are no longer parallel (for example), then the square and the post-transformation form are not equivalent forms in this geometry; parallelism is a property in Euclidean geometry.
Because it allows the fewest possible changes to a form and contains the most properties, Euclidean geometry constitutes the strongest criteria of object identity. Two samples related by an isometric transformation will have the highest probability to be judged as coming from the same object. Note that Euclidean geometry largely corresponds to our intuitions, where objectness "feels" independent of position. It is typically believed to describe the natural world in which we live at the scale at which we operate, and our language seems to reflect this. As noted above, a square and a new form that is 3 inches away but otherwise identical is also called the same thing, a square, even though strictly speaking, and importantly, it is not identical: its absolute location is different. A square that is only transformed by changing its absolute location gets a special name: it was "moved". "Transformation" in our language is reserved for transformations that alter those properties that we intuitively think of as changing its "squareness", namely those properties in Euclidean geometry, which are not altered by isometric transformations.
Question: What kind of transformations are used to generate Euclidean geometry? Answer: Isometric transformations
Question: What is the special name given to a square that is only transformed by changing its absolute location? Answer: It was "moved" | 677.169 | 1 |
5 - Geometry
Parallel lines, perpendicular lines
Two straight lines are parallel if they never intersect.
Two straight lines are perpendicular if they form a right angle, that is, an angle that measures 90 degrees.
The yellow lines are parallel
The green lines are perpendicular
Perimeter, area
The perimeter of a figure is the length of the circumference of the figure, the area of a figure is its surface,
that is the number of small squares of edge 1 that we can put inside.
The rectangle below has a perimeter measuring 3 + 4 + 3 + 4 = 14 cm, and an area of 12 cm².
Triangles
A triangle is a figure which has 3 edges.
There are several types of triangles :
A triangle which has 3 edges of the same length is called an equilateral triangle.
A triangle which has 2 edges of the same length is called an isosceles triangle.
A triangle which has a right angle is called a right-angled triangle.
The height of a triangle (red line) is the straight line which passes through a vertex and which is perpendicular to the opposite edge, called
the base (green line).
The area of a triangle is equal to the length of the base (in green) times the length of the height (in red), divided by 2.
The triangle above is an equilateral triangle.
The triangle above is an isosceles triangle.
The triangle above is a right-angled triangle.
The area of each of the three triangles above is equal to the length of the green edge times the length of the red segment divided by two.
Quadrilaterals
A quadrilateral is a figure which possesses 4 edges.
The diagonals in a quadrilateral are the straight lines which join opposite vertices.
A quadrilateral whose opposite edges are parallel is called a parallelogram. Its diagonals intersect at their centre.
Conversely, if the diagonals of a quadrilateral intersect at their centre, then that quadrilateral is a parallelogram.
Parallelogram :
A quadrilateral which has 4 right angles is called a rectangle. Its diagonals intersect at their centre and have same length.
Rectangle :
A quadrilateral which has 4 right angles and 4 edges of the same length is called a square. Its diagonals intersect at their
centre, have same length, and are perpendicular.
Square :
If the diagonals are perpendicular but do not have the same length, then the quadrilateral is a diamond. It is a
quadrilateral which has 4 edges of the same length.
Diamond :
Circle
The number is approximately 3,14.
The perimeter of a circle of radius r is equal to .
The area of a circle of radius r is equal to .
Question: What is the approximate value of Pi (π)? Answer: 3.14
Question: Which type of triangle is the one with three sides of the same length? Answer: An equilateral triangle | 677.169 | 1 |
A line is breadthless length.
The ends of a line are points.
A straight line is a line which lies evenly with the points on itself.
A surface is that which has length and breadth only.
The edges of a surface are lines.
A plane surface is a surface which lies evenly with the straight lines on itself.
Thus, for Euclid, the plane is a special case of surface. However, since in Definition 5 he defines a surface as "that which has length and breadth only", he is thinking of the plane as a two-dimensional geometrical completely flat entity.
But we are seeking for planes in the stricter and most intuitive sense; for example, that plane that extends indefinitely from a table desktop. The Webster's definition of plane reflects what this intuitive notion is all about: "a: a surface in which if any two points are chosen a straight line joining them lies wholly in that surface b: a flat or level surface". The key phrase here is "lies wholly in that surface", and this is the impossible dream: we cannot test for the pure flatness of a plane with the tools of that "flatland".
So we are returning to our starting point: a flat surface is the one made up of straight lines only, but to be certain about the straightness of a line we need a two-dimension plane.
This seems to imply that we can "test" a dimension only from a higher one, not from the dimension itself. We need the plane to test for straightness of line; we need the 3D-space if we want to be sure of the flatness of a plane.
Question: What is the key phrase in Webster's definition of a plane? Answer: "lies wholly in that surface"
Question: What are the ends of a line called? Answer: Points | 677.169 | 1 |
MTH 128: Lesson 10.Write names for angles.
Use complementary and supplementary angles to find angle measure.
Use vertical angles to find angle measure.
Find measures of angles formed by a transversal.
Slide 3
Points, Lines, and Planes
The most basic geometric figures we will study are points, lines, and planes.
A point is easiest thought of as a location, like a particular spot on this page. We represent points with dots, but in actuality a point has no length, width, or thickness.
(We call it dimensionless.)
Slide 4
Points, Lines, and Planes
A line is a set of connected points that has an infinite length, but no width. We draw representations of lines, but again, in actuality a line cannot be seen because it has no thickness. We will assume that lines are straight, meaning that they follow the shortest path between any two points on the line. This means that only two points are needed to describe an entire line.
Slide 5
Points, Lines, and Planes
A plane is a two-dimensional flat surface that is infinite in length and width, but has no thickness. You might find it helpful to think of a plane as an infinitely thin piece of paper that extends infinitely far in each direction.
Slide 6
Points, Lines, and Planes
Points and lines can be used to make other geometric figures.
A line segment is a finite portion of a line consisting of two distinct points, called endpoints, and all of the points on a line between them.
Slide 7
Points, Lines, and Planes
Any point on a line separates the line into two halves, which we call half lines. A half-line beginning at point A and continuing through point B means that A is the endpoint of the half-line, which is not included.
When the endpoint of a half line is included, we call the resulting figure a ray.
Slide 8
Points, Lines, and Planes
Below is a summary of the figures described on the last few slides showing how each is symbolized.
Slide 9
Angles
An angle is a figure formed by two rays with a common endpoint. The rays are
called the sides of the angle, and the endpoint is called the vertex.
The symbol for angle is ∠, and there are a number of ways to name angles. The angle shown could be called ∠ABC, ∠CBA, ∠B, or ∠1.
Slide 10
Angles
You should only use a single letter to denote an angle if there's no question as to the angle represented.
In this angle, ∠B is ambiguous, because there are three different angles with vertex at point B.
Slide 11
EXAMPLE 1 Naming Angles
Name the angle shown here in four different ways.
∠RST, ∠TSR, ∠S, and ∠3.
SOLUTION
Slide 12
Measuring Angles
One way to measure an angle is in
degrees, symbolized by °. One degree is
Question: What is one degree equal to in terms of angle measurement? Answer: It is not specified in the given text. | 677.169 | 1 |
This allows us to find the remaining angles: ∠4 is a vertical angle with ∠1, so it has measure 130° as well.
Now 8 is a corresponding angle with ∠4, and ∠5 is an alternate interior angle with ∠4, which means they both have measure 130° as well.
130°
130°
130°
130°
Question: What is the measure of ∠8? Answer: 130° | 677.169 | 1 |
sin45=1/√2 cos45=2/√2 tan45=1
sin could also be expressed as √2/2=sin45
Then there is the 30-60-90 triangle:
in this case, it would be the following:
sin30=0.5 sin60=√3/2
cos30=√3/2 cos60=1/2
tan30=√3/3 tan60=√3/1
Non-Right Triangles:
In cases where you don't know if the triangle has a right angle you must use the law of sines:
SinA/a=SinB/b=SinC/c
Also to help you to make a right triangle when there doesn't appear to be one is to cut the triangle down the center as shown in the picture to the left. Then you have two right triangles to deal with. :-)
But in the example to the right it is undefined due to the 37 being an impossible number on the side that it is on due to the rest of the triangle.
Now lets look at some examples using the law of sines. :-D
So with this example on the right you would do the following:
Sin25/15=SinX/26
26sin(25)/15=x
Sin^-1(26sin25/15)=47.099
X=47.099
Law of cosines:
The law of cosines is:
c^2=a^2+b^2-2ab·cosC
Now that you know this lets look at some examples:
5^2=7^2+9^2-2(7)(9)cosY
25=49+81-2(63)cosY
25=130-126cosY
25-130=(130-130)-126cosY
-105=-126cosY
-105=(-126/-126)cosY
-105/-126=cosY
0.8333=cosY
cos^-1(0.8333)=Y
Y=33.56
AND REMEMBER DON'T COMBINE TERMS THAT YOU THINK ARE ALIKE, BUT REALLY
AREN'T!!
Next is another example:
x^2=12^2+8^2-2(12)(8)cos25
x^2=208-2(96)cos25
x^2=208-192cos25
x^2=33.9889
x=√33.9889
x=5.83
But there is also another way to write the law of cosines:
Question: In a triangle with sides 12, 8, and an angle of 25 degrees, what is the length of the third side (x)? Answer: 5.83
Question: How is sin also expressed? Answer: √2/2 or sin45 | 677.169 | 1 |
Kathryn, also from Garden International School, noticed two relationships:
As the number of dots on the shape's perimeter increases by one, the area increases by half.
As the number of internal dots increases by one, the area also increases by one.
Here are her results.
Nadia from Melbourn Village College, Yun from Garden Internation School, Simeran and Aaron from Woodfield Junior School, and Kahlia from Merici College all discovered that:
When
A = the area of the shape,
p = the number of dots on the perimetre and
i = the number of dots inside the shape,
the area is equal to half of p, added to i, minus 1
A = (p x ½) + (i - 1)
This solution of Andrei's from School 205 Bucharest, gives a very useful algebraic view of the problem. He thought these were useful questions to be asking at the start:
How many different shapes can I draw of the form (4,0)?
Do they all have the same area? Can I see why?
Look at other shapes of the form (n,0). What do these shapes have in common?
What about shapes of the form (3,n)?
By breaking the problem down in this way patterns emerged that gave hints about what was happening.
I took a look at the figures given in the problem, and I calculated the area (A), the number of lattice points on the perimeter (p) and the number of interior lattice points (i). I found the table below:
Figure
A
p
i
Square
1
4
0
Triangle 1
3/2
3
1
Hexagon
6
6
4
Triangle 2
3/2
5
0
Parallelogram
1
4
0
I started from the idea that if any relation exists between A, p and i, then there must be a linear relationship. Let x be the coefficient of A, y be the coefficient of p, and z be the coefficient of i; t is the free term. So:
xA + yp + zi + t = 0
Replacing the values corresponding to the first four rows (because the last is identical with the first) I found a system of 4 equations with four unknowns:
Except the (0,0,0,0) solution, I found the relation between x, y, z and t either of the type:
x = -i, y = i/2, z = i, t = -i,
or:
x = i, y = -i/2, z = -i, t = i.
The last two solutions imply that the relation between A, p and i must be:
-A + p/2 + i -1 = 0
or:
A = p/2 + i -1
Question: What did Andrei ask about shapes of the form (3,n)? Answer: He asked to look at shapes of the form (3,n) to see what they have in common.
Question: Which of the following shapes have the same area? (Square, Triangle 1, Hexagon) Answer: The Square and the Parallelogram have the same area of 1.
Question: What is the area of the square figure given in the table? Answer: The area of the square figure is 1. | 677.169 | 1 |
This is not a demonstration, from what I have found I can't be sure it works always. But I have looked on the web, and I found some derivations of Pick's theorem, that looks as I found it. One is at Geoboards in the classroom and another is at Cut the Knot
This is a method of calculating the area of any polygon on a geoboard quickly and easily. The theorem has been found by Georg Alexander Pick, born in 1859 in Vienna, and was first published in 1899
Question: In which city was Georg Alexander Pick born? Answer: Vienna | 677.169 | 1 |
Third, a quadrilateral with an exterior angle less than 180° has a smaller area than a corresponding one with all exterior angles bigger than 180°. To see this, look at the diagram below. Clearly the quadrilateral on the right has the largest area yet they both have the same perimeters.
Fourth, the quadrilateral on the right above has smaller area than a quadrilateral whose diagonals are perpendicular. To see this, imagine that the points A and B are fixed and that A to C to B is a piece of string.Place a pencil inside the string at C. Move the pencil keeping AC and BC straight (taut). What is the biggest area of the triangle ABC? Well as C moves, the base AB remains the same. So the biggest area is found when the height of the triangle is the largest. This occurs when the perpendicular from C to AB goes through the midpoint of AB. Call this position for C, K. (Can you see that AK = KB?)
Notice that the quadrilateral formed by AKBD has the same perimeter as ACBD but AKBD has the bigger area.
We can do exactly the same thing with the point D. The position of D which makes triangle ABD have the biggest area is when D is above the midpoint of AB. Call this point L. Then the quadrilateral AKBL has the same perimeter as ABCD but it has a bigger area. What's more, KL is perpendicular to AB.
Fifth, we want to show that among all quadrilaterals with the same perimeter and with perpendicular diagonals, the one with the biggest area is a parallelogram. Consider the quadrilateral below.
By the argument of 'Four', we can assume that AB = AD and that BC = CD. Now repeat the argument of 'Four' using B as the point to put the pencil and A and C as the fixed points. The argument we used above then shows that we can move B to a place above the midpoint of AC and in the process increase the area of triangle ABC. Repeating this on triangle ACD, we see that the area of this triangle is maximised when D is above the midpoint of AC. So we have a quadrilateral whose area is bigger than the original quadrilateral ABCD.
In this new quadrilateral, AB = BC = CD = DA. From here it is easy to show that opposite angles are equal and so the quadrilateral is a parallelogram (with equal sides – a rhombus).
Now lets recap. We can show that among all rectangles with a given perimeter, the one with biggest area is a square. Then any other quadrilateral can be changed into another quadrilateral with the same perimeter and bigger area, using 'Three' and/or the 'pencil' approach of 'Four'. What's more this quadrilateral has to be a parallelogram. But we know that for every parallelogram with a given perimeter there is a rectangle with the same perimeter and a larger area. Hence among all quadrilaterals with a given perimeter, the square is the one with the biggest area.
Question: Among all quadrilaterals with the same perimeter and with perpendicular diagonals, which one has the biggest area? Answer: A parallelogram
Question: Which of the following is NOT a method to increase the area of a quadrilateral with a given perimeter? A) Using 'Three' B) Using the 'pencil' approach of 'Four' C) Changing the shape to a non-quadrilateral D) Changing the perimeter Answer: C) Changing the shape to a non-quadrilateral | 677.169 | 1 |
Your picture is a little whack as compared to your description. If the measure of angle A is indeed 96 degrees, then A has to be the vertex angle of your isosceles triangle. That's because if one of the base angles measured 96 degrees, then the other base angle must also measure 96 degrees, and you cannot have two angles greater than or equal to 90 degrees in any triangle -- and that is because the sum of the measures of the angles in any triangle is 180 degrees. So you see two 96 degree angles would mean that you have the sum of the interior angles at greater than 192 degrees -- impossible.
So, given that the vertex (or apex if you would rather) measures 96 degrees, subtract 96 from 180 and divide what you have left by 2 giving you the measure of your two base angles.
The -intercept is the point on the graph where the -coordinate is zero. Substitute zero for and then do the arithmetic to determine the resulting value of . Then form the ordered pair where is the value of just calculated.
Use the same process to find the -intercept, except you substitute zero for .
Question: What is the measure of the two base angles if angle A measures 96 degrees? Answer: 42 degrees each (180 - 96 = 84, then 84/2 = 42) | 677.169 | 1 |
, for instance, uses the axiom which says that "there exists a pair of similar but not congruent triangles." In any of these systems, removal of the one axiom which is equivalent to the parallel postulate, in whatever form it takes, and leaving all the other axioms intact, produces absolute geometry
Absolute geometry. As the first 28 propositions of Euclid (in The Elements) do not require the use of the parallel postulate or anything equivalent to it, they are all true statements in absolute geometry.
To obtain a non-Euclidean geometry, the parallel postulate (or its equivalent) must be replaced by itsform, since it is a compound statement (... there exists one and only one ...), can be done in two ways. Either there will exist more than one line through the point parallel to the given line or there will exist no lines through the point parallel to the given line. In the first case, replacing the parallel postulate (or its equivalent) with the statement "In a plane, given a point P and a line l not passing through P, there exist two lines through P which do not meet l" and keeping all the other axioms, yields hyperbolic geometry
Hyperbolic geometry
In mathematics, hyperbolic geometry is a non-Euclidean geometry, meaning that the parallel postulate of Euclidean geometry is replaced...
. The second case is not dealt with as easily. Simply replacing the parallel postulate with the statement, "In a plane, given a point P and a line l not passing through P, all the lines through P meet l", does not give a consistent set of axioms. This follows since parallel lines exist in absolute geometry, but this statement says that there are no parallel lines. This problem was known (in a different guise) to Khayyam, Saccheri and Lambert and was the basis for their rejecting what was known as the "obtuse angle case". In order to obtain a consistent set of axioms which includes this axiom about having no parallel lines, some of the other axioms must be tweaked. The adjustments to be made depend upon the axiom system being used. Among others these tweaks will have the effect of modifying Euclid's second postulate from the statement that line segments can be extended indefinitely to the statement that lines are unbounded. Riemann's elliptic geometryemerges as the most natural geometry satisfying this axiom.
Models of non-Euclidean geometry
Two dimensional Euclidean geometry is modelled by our notion of a "flat plane
Plane (mathematics)
In mathematics, a plane is a flat, two-dimensional surface. A plane is the two dimensional analogue of a point , a line and a space...
."A globe is a three-dimensional scale model of Earth or other spheroid celestial body such as a planet, star, or moon...
), and points opposite each other (called antipodal points) are identified (considered to be the same). This is also one of the standard models of the real projective plane
Real projective plane. The difference is that as a model of elliptic geometry a metric is introduced permitting the measurement of lengths and angles, while as a model of the projective plane there is no such metric.
Question: What is a standard model of Euclidean geometry in two dimensions? Answer: A flat plane.
Question: What are the first 28 propositions of Euclid called? Answer: The first 28 propositions of Euclid are called The Elements.
Question: Who were some of the mathematicians who knew about the issue of replacing the parallel postulate with the statement that there are no parallel lines? Answer: Khayyam, Saccheri, and Lambert. | 677.169 | 1 |
Well, to be perfectly honest, one wouldn't need any geometric equations at all in order to build the pyramids exactly as they stand, fully intent on doing so.
My grandfather was one of the premier housing contractors in the small city he resides; he has built thousands of homes in the community, and doing this without any formal education. His education stops at high school with mathematics knowledge not exceeding pre-Algebra (no formal geometric education).
He understands some concepts that he has learned through experience and observation, and then couples that knowledge with common sense to create all sorts of geometric shapes, ratios, and relationships, unintentionally. His concern is with building the object defined in the specifications - how he does that is left up to him.
Simple tools needed for making almost any geometric shape: square (right-angle), straight edge (usually encompassed by the square), few thousand yards of string, and wood working tools. By combining these tools and common sense with simple mathematics (PEMDAS), he can accomplish just about any desired target result.
--
The base of the pyramids in Egypt are almost a perfect square, but there is room for error. The angles aren't exactly 90 degrees, but are very close (as would be expected with any squared block acting as a corner stone). The angles differ only by +/- 1 degree, but some are larger and others smaller than 90 degrees. The lengths of each side aren't all uniform, but are within feet of each other; testament to the imprecision that would result from the use of primitive tools coupled with human ingenuity.
If the Egyptians were getting advanced help from space-faring civilizations to build the pyramids, I would expect them to build structures with much less tolerance for error. If they were using space-age machining techniques on blocks, I would expect them to have precise 90 degree angles, uniformly across the board. If they were using lasers to etch these stones in the quarries, I would expect them to have the ability to lay perfectly uniform sides.
As precise as the pyramids seem, there is more than enough error to validate them as being created through human knowledge and perseverance in my opinion.
I think our line of questioning should be more of why were they build, rather than how were they built.
[quote="cRush].[/quote]
well, I guess the point is - There was a reason for their intense adherence to deliberate methods used in their technologies, and I still believe very little was done by chance, expecially the construction of a project as important as the one at Giza which is still standing after thousands and thousands of years.
Question: Were the angles of the base of the Egyptian pyramids exactly 90 degrees? Answer: No, they differed by +/- 1 degree, with some larger and others smaller than 90 degrees.
Question: What simple tools did the author mention were needed to create geometric shapes? Answer: A square (right-angle), straight edge, a few thousand yards of string, and wood working tools. | 677.169 | 1 |
For a trade to be fair, it has to be fair in both directions. The source producers need to be paid a fair price for the product the produce, but I too need to pay a fair price for the item I'm purchasing.
It's called Pythagoras Theory because it's only a theory. Every triangle that it's ever been tested on works, so there's no reason to doubt that it will continue to work on every triangle we ever find. But until there's a way to prove it, it will always be called a theory. If it's ever proven, it will be called Pythagoras Rule.
Question: Who is the person associated with the mathematical concept? Answer: Pythagoras. | 677.169 | 1 |
The Modern Day High School Geometry Course: A Lesson in Illogic
by Barry Garelick Chances are good that most students in high school today know that the sum of the measures of angles in a triangle equals 180 degrees. Unfortunately, chances are also good that most high school students today cannot prove that proposition. Geometry as taught today is for the most part lacking in the [...]
by Barry Garelick
Chances are good that most students in high school today know that the sum of the measures of angles in a triangle equals 180 degrees. Unfortunately, chances are also good that most high school students today cannot prove that proposition.
Geometry as taught today is for the most part lacking in the most important aspect of the subject: Proofs. Prior to 1980, most if not all high school geometry classes were very much proof-based. While there are those who bemoan the teaching of K-12 math as being mired in "computational" and "procedural" aspects of math while ignoring the larger beauty of what mathematics is about, it is ironic that when it comes to geometry, the true mathematical nature of the subject is largely ignored.
A glance at the geometry textbooks that are typically used in high schools today reveals that the problems students are given in such courses require one or two proofs that are not very challenging in a set of problems devoted to the application of theorems rather than the proving of propositions. Most of the problems presented in these textbooks require students to apply various theorems and definitions to find the lengths of line segments and angles. Typical courses in geometry are lacking in proof-based problems; instead, they contain many problems in which missing angles or segments are indicated as algebraic expressions. For example, opposite sides of a quadrilateral that is identified as a parallelogram may be labeled x + 2 and 2x – 6; the student is asked to find the length of the segments. This problem requires knowledge of the properties of a parallelogram leading to the conclusion that the two segments of interest are congruent. The two sides, expressed as x + 2 and 2x – 6 then lead to the equation x + 2 = 2x – 6. Figure 1 shows another example of a problem that does not require formal proof.
Figure 1:
This problem requires students to know and apply that the sum of the angles in a triangle equals 180 degrees, and to know what are linear pairs of angles, and that they sum to 180 degrees. From this, students can piece together information and compute angle R.
While the types of problem discussed above constitute a form of proof (requiring applications of theorems and definitions combined with deductive reasoning to justify the necessary computation), such problems do not fully develop the skills necessary to develop a logical series of statements that constitute proof. In contrast, consider a problem that requires a student prove a particular proposition, such as shown in Figure 2:
Figure 2:
This problem does not require any numerical calculation. It requires knowledge of theorems of parallel lines in a plane and properties of isosceles triangles.
Defeating the Purpose of a Geometry Course
Question: What is the sum of the measures of angles in a triangle? Answer: 180 degrees
Question: What is the main issue with the current teaching of high school geometry, according to the text? Answer: Lack of emphasis on proofs | 677.169 | 1 |
Polygon Crosswords
Our polygon crosswords are a great way to hone
students' math vocabulary skills! We have interactive crosswords
with three levels of difficulty. We also have printable
versions, and solutions for all. Our interactive crosswords
require Java,
a free and safe download. Our printable puzzles require
Acrobat
Reader.
Choose a resource below and have fun. Be sure
to try our related activities on circles, too!
Question: What other related activities are mentioned in the text? Answer: Activities on circles | 677.169 | 1 |
You can put this solution on YOUR website! Make a coordinate grid and plot the point (1,2).
Now draw a vertical line from (1,2) down to (1,0). That segment is of length 2 (the distance between (1,2) and (1,0)) and is one leg of a right triangle.
We want to find a point on the x axis that will include the drawn segment and result in a right triangle with hypotenuse of length
There are two points that are solutions. One point is to the left of (1,0) and the other is an equal distance to the right of (1,0).
So how far is that distance?
Use the Pythagorean theorem.
You know the length of two sides of a right triangle, so you can solve for the third side.
So the points are 3 units on either side of (1,0). That means (4,0) and (-2,0)
are the two points that are solutions.
Question: What are the coordinates of the points on the x-axis that satisfy the condition?
Answer: (-2,0) and (4,0) | 677.169 | 1 |
Collapsible Compasses
Traditionally, geometric constructions are done with
compasses and straightedge, as shown on the left. See Geometric Constructions. You may have heard that REAL
compasses, the compasses used by Euclid, are collapsible. Such compasses are
used to draw circles of a given radius. But when you lift them off the paper,
they collapse, losing the measure of the radius. What is that all about?
The second "proposition" (theorem, or in this case the proof that
a method of construction does what we want it to) shows how to "place at a
given point (as an extremity) a straight line equal to a given straight
line." Now, that is easy. You just measure the given line segment with
your compasses, pick them up, move them to the given point, draw a
circle, and any radius fits the bill. But Euclid doesn't do this. It is
apparently illegal to pick up the compasses, at this early stage of Euclid's
series of geometry lessons.
In the diagram on the right, we have a point A and a
line segment BC. The task is to draw a line segment with A as one of its end
points, the same length as BC. Euclid does this by drawing circle B with radius
BC and segment AB, then drawing an equilateral triangle ABD (He shows us how to
do this in Proposition 1). He extends DB out to this circle and draws the
circle D with this radius. Then, extending DA to this circle, the black line
(with end point A) in the diagram is a line that meets our needs.
Euclid went to all this work to do what we thought was a simple task. So, do
we have to reinvent some kind of collapsible compasses? No (in my opinion).
What Euclid has just shown us is that his bizarre compasses can do all the same
things that our easier-to-use compasses can do. He showed that they are
equivalent. So, now we can use our compasses knowing that we could do the same
thing with more difficult tools.
Question: What is the author's opinion on whether we need to reinvent collapsible compasses? Answer: The author's opinion is that we do not need to reinvent collapsible compasses.
Question: What is the task described in the diagram on the right? Answer: The task is to draw a line segment with A as one of its end points, the same length as BC.
Question: What is the traditional method for geometric constructions? Answer: Traditionally, geometric constructions are done with compasses and straightedge. | 677.169 | 1 |
In the first place, you either have to include a diagram or at the very least give an extremely detailed description of the figure. In the second place, no matter what your figure looks like you cannot prove that a triangle is congruent to an angle -- that would be like trying to prove that belly button lint is the same as a hard-boiled egg.
Question: What is an example given in the text that cannot be proven to be the same as another? Answer: Belly button lint and a hard-boiled egg | 677.169 | 1 |
Q. 17. From a pack of 52 playing cards, jacks, queens, kings and aces of red colour are removed. From
the remaining, a card is drawn at random. Find the probability that the card drawn is
i. a black queen
ii. a red card
iii. a blackjack
iv. a picture card (jacks, queens and kings are picture cards)
Q. 18. Prove that the points (-4, -1); (-2, -4); (4, 0) and (2, 3) are vertices of a rectangle.
Q. 19. If (-2, -1); (a, 0); (4, b) and (1, 2) are the vertices of a parallelogram, find the values of a and b. Q. 20. A vessel is in the form of a hemispherical bowl mounted by a hollow cylinder. The diameter of
the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the capacity of the vessel.
SECTION - C
Question numbers 21 to 25 carry 6 marks each.
Q. 21. If a line is drawn through an end point of a chord of a circle so that the angle formed by it with
the chord is equal to the angle subtended by the chord in the alternate segment, then the line is a
tangent to the circle. Prove it.
In each of the following, (i) and (ii) (in Figure 4), if a circle is drawn through A, B and D, then using
the above theorem, examine in which case is the line BC a tangent.
Q. 22. A man on the top of a vertical tower observes a car moving at uniform speed towards the
tower. If it takes 12 minutes for the angle of depression to change from 300
to 450, how soon, after
this, will the car reach the tower?
Or
The angle of elevation of the top of a hill at the foot of the tower is 600
and the angle of elevation of
the top of the tower from the foot of the hill is 300
. If the tower is 50 m high, find the height of the hill.
Q. 23. In a triangle, if the square on one side is equal to the sum of the squares on the remaining two,
the angle opposite the first side is a right angle. Prove it.
Using the above prove the following:
If in a and PS 2= QS X RS, then is right – angled at P.
Q. 24. A bucket made of aluminium sheet is of circular pipe in half hour height 20 cm and its upper
and lower ends are of radius 25 cm and 10 cm respectively. Find the cost of making the bucket if the
aluminium sheet costs Rs. 70 per 100 cm2
.
Or
Question: What is the probability that the card drawn from the remaining cards is a black queen? Answer: 2/51 (since there are 2 black queens left out of 51 cards)
Question: What is the total number of red cards left after removing the specified cards? Answer: 26 (26 red cards remain after removing the 12 red cards)
Question: What is the area of the rectangle formed by the points (-4, -1); (-2, -4); (4, 0) and (2, 3)? Answer: 16 (since the length is 6 and the width is 4)
Question: What is the probability that the card drawn is a blackjack? Answer: 0 (since all black jacks have been removed) | 677.169 | 1 |
Segments, lines, polygons, and polyhedra are commonplace objects in Euclid's Elements. In fact, some people regard the development of the Elements as leading up to the proof found in Book XIII of the Elements, where it is shown that there are 5 regular convex polyhedra. This is the modern way to state the result since Euclid does not discuss the concept of convexity. However, without the word "convex" the statement above is not true. Depending on the definition of polyhedron there can be more regular polyhedra. The history of the development of regular polyhedra is intriguing and shows how changes in the definitions of such fundamental objects as polygons and polyhedra have stimulated geometers to discover new facts.
Here we use the theory of graphs as an organizing idea for a wide variety of new geometrical questions as well as providing insight into older geometrical results. Figure 2 shows a drawing of a 3-dimensional polyhedron on a flat piece of paper where some of the lines in the diagram are dotted to suggest that they are hidden and cannot be seen when one looks at this object in 3-dimensional space.
Figure 2
As a graph this polyhedron has 8 vertices, 12 edges, and 6 faces.
Do Now 2:
a. What are the names that are usually given to the faces of the polyhedron in Figure 2 when these faces are thought of as polygons? Are any of these faces congruent?
b. What is a "reasonable" name for this polyhedron?
Here (Figure 3) are some other drawings in the plane of graphs which are isomorphic to the graph in Figure 2.
Figure 3
The most familiar of these drawings are the top two. As a graph, these graphs are all known as 3-cubes. We are not interested in lengths of lines here, nor whether straight lines are used in the representation (though above only straight lines are used), only the way the graph "hangs together." All of these graphs are isomorphic to each other. To express that two geometric objects are metrically equivalent, that is, that they have the same "distance" relationships among their parts, involves finding a geometric transformation which is known as an isometry between the objects.
Representing 3-dimensional objects in the plane
Although we live in a 3-dimensional world, we often have to represent objects on flat surfaces - planes. In addition to using hidden lines (Figure 2) there are other "tricks" that have been developed to do this. Among these tricks are the use of isometric and perspective drawings. Figure 4 shows two drawings of a cube which are typical of the ways that polyhedra are sometimes drawn on flat pieces of paper.
Figure 4
Question: When are the faces of the polyhedron in Figure 2 thought of as polygons, what are their names? Answer: The faces are usually named as triangles, squares, and hexagons.
Question: Which of the following is NOT a type of polyhedron mentioned in the text? A) Segment B) Line C) Polygon D) Polyhedra Answer: A) Segment
Question: What is the total number of faces in the polyhedron shown in Figure 2? Answer: 6 | 677.169 | 1 |
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Same Side Angles This video describes how angles are formed when two parallel lines are intersected by a transversal (same side interior (between the parallel lines) and same side exterior (outside the parallel lines)). Since alternate interior and alternate exterior angles are congruent and since linear pairs of angles are supplementary, same side angles are supplementary. (1:53)Politics and Genocide: Rwanda (African Studies Seminar) Dr Omar McDoom (London School of Economics) looks at a single community in southern Rwanda, using spatial mapping, in order to understand why some people chose to kill during the violence and others did not. Author(s): Omar McDoom
Closing Remarks for UMMC Henrietta Lacks Conference During her life in Baltimore's Turner's Station, Henrietta Lacks was a wife, mother, neighbor and friend. It was here in Baltimore that she also fought a heroic battle against cervical cancer that changed the course of medicine. Mrs. Lacks' sacrifice has saved and continues to save countless people she has never met. Sixty years after her death, we honor and celebrate this remarkable woman.
In this two-minute video, John H. Spearman, Senior Vice President of External Affairs for the University Author(s): No creator set
How Can Robots Get Our Attention? Researchers at the Georgia Institute of Technology have found that they can program a robot to understand when it gains a human's attention and when it falls short.
Aaron Bobick, professor and chair of the School of Interactive Computing at Georgia Tech's College of Computing, discusses why it's important for robots to understand social convention, if they are going to successfully interact with humans.
The research was performed using the robot Simon, from the Socially Intelligent Machines la Author(s): No creator set
Question: Is the video "Politics and Genocide: Rwanda" part of the African Studies Seminar? Answer: Yes
Question: What is the topic of the video "Closing Remarks for UMMC Henrietta Lacks Conference"? Answer: The life and impact of Henrietta Lacks | 677.169 | 1 |
Q7: How does the law of sines work?
Question:
If I'm given two sides and two angles of a triangle, how can I find the remaining side if the Pythagorean Theorem doesn't apply?
Answer:
This is a perfect case of using the Law of Sines. The Pythagorean Theorem won't work because it's not a right triangle. The Law of Sines relates all of the side lengths to the angle measurements of any triangle.
We can very easily find the missing side by plugging in some of what we do know:
In fact, we can find the other angle as well if we just use the formula again. In general, we only need three pieces of information to find the rest. We either need two angles and a side, or two sides and an angle. You might have to use the Law of Sines several times, but keep solving for missing variables until you can solve for the one you need.
It should be noted that sometimes the Law of Sines can give misleading results, leading to angles that are much too large. This will be obvious, because the sum of the angles of the triangle will be greater than 180 degrees. For more information on this subject, look at the Wikipedia article on the Law of Sines. They have a nice graphic that shows the possible ambiguity that can result. However, rest assured that if you numbers look right, they are.
Question: What happens if the sum of the angles calculated using the Law of Sines is greater than 180 degrees? Answer: If the sum of the angles is greater than 180 degrees, it indicates that the calculations using the Law of Sines are incorrect and may have led to misleading results. | 677.169 | 1 |
Internet Math Challenge
Deadline for solutions: Friday, 16 January 2004
A Strange "Triangle"
This week's puzzle isn't too hard, but it will introduce you to a cool
geometrical object -- the Reuleaux triangle (pronounced "Roo-low").
This "triangle" is pictured in the diagram at right (the yellow shape). As
you can see, it really isn't a triangle at all because its sides aren't
straight. To build a Reuleaux triangle, start with the three vertices of
an equilateral triangle (with all three sides of length 1).
Now draw three circles. Each circle is centered at
one of the three vertices, and passes through the other two vertices. The
area of overlap between these circular disks is the Reuleaux triangle.
This shape has several interesting properties. One of them is that it is
the same width no matter which way it is rotated. (That is, it will always
just barely fit through an opening 1 unit wide). The circle, of course, has
this same property, but not many non-circular shapes do. A square (with
side-length 1), for instance, will fit through an opening of width 1 if
taken through with sides parallel to the opening. But otherwise, it won't
fit!
Have you ever wondered why manhole covers are round instead of square? The
answer is found in this very property: a square manhole cover could be picked
up, rotated, and dropped through the hole it covers. However, a circular manhole
cover won't fit through that hole no matter how it is rotated. Maybe sometime
someone will manufacture manhole covers shaped like the Reuleaux triangle!
Now for this week's puzzle: find the exact area of the Reuleaux triangle!
Be sure to simplify your answer as much as possible.
Question: What is the name of the unusual shape described in the text? Answer: Reuleaux triangle | 677.169 | 1 |
left corner, the bottom right corner will be at (x+width,y+height).
drawPolygon
(int[] x, int[] y, int N)
Draws lines connecting the points given by the x and y arrays.
Connects the last point to the first if they are not already the
same point.
So for the instruction
drawRect(x,y,width,height)
(x,y) are
the coordinates of the top left corner and the bottom right corner
will be at (x+width,y+height).
Note that Graphics
does not provide a method to set the width of a line. The line is
always one pixel wide and continuous (i.e., no dot-dash options).
Nevertheless, these methods are simple and often convenient to use
and can be used along with the Graphics2D
methods.
Question: What are the coordinates of the bottom right corner of a rectangle drawn using `drawRect(x, y, width, height)`? Answer: (x+width, y+height) | 677.169 | 1 |
You can put this solution on YOUR website! Geometry is the study of shape and size. The geometry of our everyday world is based on the work of Euclid, who lived about 300 BC. Euclidian geometry has a rigorously developed logical structure. Three basic undefined terms are point, line, and plane. A point is a tiny dot: it has zero height, zero width, and zero thickness. A line goes off straight in both directions. A plane is flat surface, like a tabletop, extending off to infinity.
Question: What is the main focus of geometry according to the text? Answer: Shape and size | 677.169 | 1 |
A polygon is called regular if it has equal sides and angles. Thus, a regular triangle is an equilateral triangle, and a regular quadrilateral is a square. A general problem since antiquity has been the problem of constructing a regular n-gon, for different n, with only ruler and compass. For example, Euclid constructed a regular pentagon by applying the above-mentioned five important theorems in an ingenious combination.
Techniques, such as bisecting the angles of known constructions, exist for constructing regular n-gons for many values, but none is known for the general case. In 1797, following centuries without any progress, Gauss surprised the mathematical community by discovering a construction for the 17-gon. More generally, Gauss was able to show that for a prime number p, the regular p-gon is constructible if and only if p is a "Fermat prime": p = F(k) = 22k + 1. Because it is not known in general which F(k) are prime, the construction problem for regular n-gons is still open.
Three other unsolved construction problems from antiquity were finally settled in the 19th century by applying tools not available to the Greeks. Comparatively simple algebraic methods showed that it is not possible to trisect an angle with ruler and compass or to construct a cube with a volume double that of a given cube. Showing that it is not possible to square a circle (i.e., to construct a square equal in area to a given circle by the same means), however, demanded deeper insights into the nature of the number π. Seegeometry: The three classical problems.
Conic sections and geometric art
The most advanced part of plane Euclidean geometry is the theory of the conic sections (the ellipse, the parabola, and the hyperbola). Much as the Elements displaced all other introductions to geometry, the Conics of Apollonius of Perga (c. 240–190 bce), known by his contemporaries as "the Great Geometer," was for many centuries the definitive treatise on the subject.
Medieval Islamic artists explored ways of using geometric figures for decoration. For example, the decorations of the Alhambra of Granada, Spain, demonstrate an understanding of all 17 of the different "Wallpaper groups" that can be used to tile the plane. In the 20th century, internationally renowned artists such as Josef Albers, Max Bill, and Sol Le Witt were inspired by motifs from Euclidean geometry.
Solid geometry
The most important difference between plane and solid Euclidean geometry is that human beings can look at the plane "from above," whereas three-dimensional space cannot be looked at "from outside." Consequently, intuitive insights are more difficult to obtain for solid geometry than for plane geometry.
Question: Which famous artists were inspired by motifs from Euclidean geometry in the 20th century? Answer: Josef Albers, Max Bill, and Sol Le Witt. | 677.169 | 1 |
Making Indestructible Quadrilaterals
'Construct your own indestructible quadrilaterals based on your understanding of their fundamental properties'
This is an activity in its own right, but does follow on really nicely from Indestructible Quadrilaterals and this will certainly help you to reflect on the defining features of these shapes. So, before you start it is key to be familiar with the properties of quadrilaterals. What is a Rectangle, a Square, Rhombus, Parallelogram, Trapezium/Trapezoid, Kite? How would you describe it to someone as precisely as you can? When is a Rectangle also a Square? These questions and more are at the centre of this activity. Your challenge here is to use dynamic geometry to construct the different quadrilaterals. You must be able to move the points and change the size and dimensions of the shape without changing the shape that it is. The Indestructible Quadrilaterals activity has examples of these shapes and a further example of all seven quadrilaterals is shown below. Try pushing and pulling them! Your task is to recreate them.
Resources
Please find below an applet showing 7 indestructible quadrilaterals and a video tutorial on some of the basics for constructing with Geogebra. You will need Dynamic Geometry software and this can be downloaded free from
Question: What are you not allowed to change while manipulating the shapes in the activity? Answer: The shape of the quadrilateral itself. | 677.169 | 1 |
Compassing 101
Triangulation
Triangulation involves taking bearings from two different locations to find where the lines intersect
Triangulation is a big word for finding the point where two bearings, taken from two different locations, intersect. Take the example at the right. Let's assume there are two prominent landmarks near the letterbox, that is, Point A and Point B. Using those landmarks, how can we get someone to Point C?
One method is having the person follow a given bearing. For example, start at Point A, and take 100 steps at 45 degrees. This is the simplest and most direct method, but sometimes it's impractical. What if there's a large river or canyon between Point A and Point C? What if the two points are several miles apart from each other? What if the distance must be very precise, and counting steps isn't accurate enough? Issues such as these are solved with another method: triangulation.
Continuing with our example at the right, we could instead tell a letterboxer to start at Point A and take a bearing of 45 degrees. Then take another bearing, this time of 340 degrees from Point B. The letterbox will be where those two bearings intersect, Point C. And the letterboxer can follow any route they want to get there. If Point C is on the other side of a river, the letterboxer could walk downstream to a bridge or shallow area in the river then hike back upstream. If Point C is the top of a hill several miles away, the letterboxer could pull out a map and determine the best trails to use to reach that point.
Another variation on triangulation would be a clue that tells you which direction your two landmarks should be from the letterbox (i.e. Point C), or a 'reverse triangulation', if you will. Continuing with our example, the clue could say find the point at which Point A is at 225° and Point B is at 160°. To solve such a clue, you'd walk around one of the landmarks, Point A, for instance, until you found a place where the landmark is at a bearing of 225°. To find the location for Point C, you'll walk up or down the imaginary line between Point A and your location until you've found the one point on that line where Point B is at 160°.
Triangulation is a powerful tool for providing precise directions to an otherwise non-descript location and is an easy skill to learn with a bit of practice. That said, triangulation is rarely used in letterboxing since following a given bearing is usually easier, faster and workable, but triangulation will show up from time to time.
Question: What skill does the text suggest is required to learn triangulation? Answer: A bit of practice.
Question: What is'reverse triangulation'? Answer: A variation of triangulation where you find the point at which two landmarks are at specific bearings from the destination point.
Question: What are the two main methods mentioned for navigating to a specific point? Answer: Following a given bearing and triangulation. | 677.169 | 1 |
The challenge is to come up with a consistent rule for applying these rotations. We start with normal arithmetic. Multiplying by a positive didn't flip the sign, so we say we rotated by $ 0^\circ $. Multiplying by a negative flips the sign, so we rotated by $ \class{green}{180^\circ} $. The lengths are multiplied normally in both cases.
$$ \times \class{green}{1.5 \angle 90^\circ} ... $$
$$ +90^\circ $$
$$ +270^\circ $$
Now suppose we pick one of the in-between nether-numbers, say the vector of length $ 1.5 $, at a $ 90^\circ $ angle. What does that mean? That's what we're trying to find out! We'll write that as $ \class{green}{1.5 \angle 90^\circ} $ (1.5 at 90 degrees). It could make sense to say that multiplying by this number should rotate by $ \class{green}{90^\circ} $ while again growing the length by 50%.
By multiplying by $ \class{green}{1 \angle 45^\circ} $, we can rotate in increments of $ 45^\circ $. It takes 4 multiplications to go from $ +1 $, around the circle of ones, and back to the real number $ -1 $.
And that's actually a remarkable thing, because it means our invented rule has created a square root of $ -1 $. It's the number $ \class{green}{1 \angle 90^\circ} $.
$ (\class{green}{1 \angle 90^\circ})^2 = \class{blue}{-1} $
If we multiply it by itself, we end up at angle $ \class{green}{90} + \class{green}{90} = \class{blue}{180^\circ} $, which is $ \class{blue}{-1} $ on the real line.
But actually, the same goes for $ \class{green}{1 \angle 270^\circ} $.
$ (\class{green}{1 \angle 270^\circ})^2 = \class{blue}{-1} $
When we multiply it by itself, we end up at angle $ \class{green}{270} + \class{green}{270} = \class{blue}{540^\circ} $. But because we went around the circle once, that's the same as rotating by $ \class{blue}{180^\circ} $. So that's also equal to $ \class{blue}{-1} $$ (\class{green}{1 \angle -90^\circ})^2 = \class{blue}{-1} $
Question: How many multiplications does it take to go from $+1$ to $-1$ on the real line? Answer: 4 | 677.169 | 1 |
The converse of the Pythagorean theorem and special triangles
If we know the sides of a triangle - we can always use the
Pythagorean Theorem backwards in order to determine if we have a
right triangle, this is called the converse of the Pythagorean
Theorem.
When working with the Pythagorean theorem we will sometimes
encounter whole specific numbers that always satisfy our equation -
these are called a Pythagorean triple. One common Pythagorean
triple is the 3-4-5 triangle where the sides are 3, 4 and 5 units
long.
There are some special right triangles that are good to know,
the 45°-45°-90° triangle has always a hypotenuse √2 times the
length of a leg. In a 30°-60°-90° triangle the length of the
hypotenuse is always twice the length of the shorter leg and the
length of the longer leg is always √3 times the length of the
shorter leg.
Question: What is a common Pythagorean triple? Answer: 3-4-5 | 677.169 | 1 |
public.beuth-hochschule.de/~meiko/pentatope.html
- for the marked Applets you need Java3D,
- try to use all three MouseButtons (zoom, rotate or specials)
regular Polygon and Star-Polygon
A geometric closed figure with more than 3 sides is
called
a polygon.
If all vertices are coplanar we have a plane
polygon. A
plane polygon is "regular",
if it is equilateral and equiangular.
Regular polygones are labeled with {p}. The "p" counts the
sides/angles.
For instance a square has the label {4}.
A "Star Polygon"
is labeled with {p/q}(means every qth point is connected).
A polyhedron consist of a finite set of plane polygons("faces"), joined
at their sides("edges").
Regular polyhedra have regular faces and
regular vertex figures.
A vertex figure is a polygon formed by
connecting the midpoints of all adjacent sides of a vertex.
The {p, q}
is the label("Schläfli Symbol") for that solids.
The p
describes
the polygons, and the q the count of that polygons at each vertex.
Question: What is a polyhedron composed of? Answer: A finite set of plane polygons (faces) joined at their sides (edges). | 677.169 | 1 |
Congruence Theorems
Investigate congruence by manipulating the parts (sides and angles) of a triangle. If you can create two different triangles with the same parts, then those parts do not prove congruence. Can you prove all the theorems?
Instructions
Each triangle congruence theorem uses three elements (sides and angles) to prove congruence. Select three triangle elements from the top, left menu to start. (Note: The tool does not allow you to select more than three elements. If you select the wrong element, simply unselect it before choosing another element.) This creates those elements in the work area.
On the top of the toolbar, the three elements are listed in order. For example, if you choose side AB, angle A, and angle B, you will be working on Angle – Side – Angle. If instead you choose side AB, angle A, and angle C, you will be working on Angle – Angle – Side. The two theorems are different, even though both involve two angles and one side.
Construct your triangle:
Move the elements of the triangle so that points labeled with the same letter touch.
Click and drag a dot to move the element to a new location.
Click and drag a side's endpoint or angle's arrow to rotate the element. The center of rotation is the side's midpoint or the angle's vertex, respectively.
To help place elements, points marked with the same letter snap together. When angles snap, the rays are extended to the edge of the work area.
When you create a closed triangle, the points merge and center is filled in.
Once a triangle is formed with the original three elements, the triangle move to the bottom, right corner of the work area, and congruent elements appear. Try to form a second triangle. If the second triangle can only be formed congruent to the first, then that arrangement of three elements proves a congruence theorem. If you can form a non-congruent triangle, then that disproves congruence.
After a second triangle is formed, you will be asked if they are congruent. You can test congruence by manipulating either triangle.
Click and drag within the triangle to move it to a new location.
Click and drag a vertex to rotate the triangle.
Use the Flip button to reflect the triangle horizontally. First click on the triangle you would like to reflect, and then click the Flip button.
If the two triangles are congruent, you will be asked if it's possible to make a triangle that is not congruent to the original. If you create a third congruent triangle, you will be given the option to try again.
The Reset button clears the work area and creates new sides and angles for the selected elements.
The New button clears your selection and work area.
Exploration
For every arrangement of three elements, it is possible to test for triangle congruence. However, just like two triangle might be mirror images but still congruent, flipping the order creates identical possible theorems. For example Angle – Angle – Side is the same as Side – Angle – Angle because they are the same elements in reverse order.
Question: Can you prove all the triangle congruence theorems using this tool? Answer: No, you can't prove all the theorems using this tool. It only allows selecting three elements at a time.
Question: How can you move an angle in the work area? Answer: Click and drag the angle's arrow to rotate it, with the vertex as the center of rotation.
Question: What happens when you form a closed triangle with the original three elements? Answer: The points merge, the center fills in, and the triangle moves to the bottom, right corner of the work area. Congruent elements also appear.
Question: Which button do you use to reflect a triangle horizontally? Answer: The Flip button. | 677.169 | 1 |
Some of the points get a little bit negative. We have a negative – so let me also draw some of the negative quadrants. So, if I were to draw it like that – Okay, and let see. So, 4,2. So, if I were to say 1 – 2 – 3 – 4,1- 2. That's right there. That's point A. Then if 6, negative 1. So, 4 – 5 – 6,negative 1. Negative 1 is right there. That's point B. I don't know if you can see it right there. And C is negative 1,3. So, negative 1,3. So, it's out here – negative 1,3. Okay, now let me connect the dots. That's one side. That's another side and that's the other side.
So off the bat, I mean I just know, this isn't going to be a right triangle. It's not equilateral triangle. And if – the only way it's going to be an isosceles triangle is if this length is equal to that length. So, let's just try it. Let's just test it out. So, what is the distance from A to C, right? Well, the distance is equal to or we can say the distance squared but let's say the distance squared from A to C is equal to the differences in their x's. So, 4 minus negative 1, that's the difference of 5, right? So, the difference of their x's squared plus a difference in their y's. So, 2 and 3. You can either say 2 minus 3 or 3 minus 2. It doesn't matter. We just care about the difference. 1 squared – that's one squared. So, the distance squared is equal to 25 plus 1 is equal to 26. So, this distance is the square root of 26. And this is between A and B – same logic. Let's see.
When you go from the distance squared. The differences in their x's. 6- Between 6 and 4, you have a distance of 2. So, it's 2 squared plus the difference in their y's. 2 and negative 1 are 3 apart, right? Plus three squared, and so that is equal to 4 plus 9, which is equal to 13. So, it's equal to the square root of 13. Okay, and this number – this number down here, you can figure out. That's going to be bigger than all of the – both of them, right? You can just look at it and say that. So this is definitely a scale in triangle. All of the sides are different. Anyway, see you in the next video.
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"Geometry \"Earth-measuring\" is a branch of mathematics concerned with questions of shape, size, relative position of figures, and the properties of space. Geometry is one of the oldest mathematical sciences....
Question: Which side of the triangle is the longest? Answer: The side between A and C
Question: What are the coordinates of point C? Answer: (-1, 3)
Question: Is the triangle formed by these points an equilateral triangle? Answer: No | 677.169 | 1 |
Thursday 10/11/12
Lesson 2.5 Day 1 worksheet. Remember to go back to your text book and review the examples in 2.5 that use the segment addition postulate and the angle addition postulate.
Friday 10/12/12
Study the example(s) completed in class, on the Lesson 2.5 Day 2 work sheet. On the back of the sheet, do the proof using different logic, using the hints provided. Carefully read the Chapter 2 project sheets, to decide which project you want to do. Test on Lessons 2.2 - 2.5 on Tuesday 10/16. The retest of the take home portion of the unit 1 test is due on Monday.
Monday 10/15/12
Prepare for Test on Lesson 2.2 - 2.5 tomorrow. Practice! Finish or redo any worksheets from class. Use the on-line text book resources!
Try the proofs on the Lesson 2.6 Day 2 packet - remember to solve the problem in your head first, and developing an overall stategy before you write anything down! If you find some of the proofs frustrating or difficult, don't panic - we'll take care of it in class. However, if you are struggling, do some problems from the text book and check your answers.
Finish any proofs your were not able to complete in class. Test on 2.6 and 2.7 on Friday. RTN & GP Lesson 3.1. All of the highlighted vocabulary terms on page 141 should be familiar to you. Make sure that you realize that these are angle relationship names based on position only, and do not tell you anything about congruence or measures! The two new postulates should make intuitive sense to you.
Thursday 10/25
Prepare for test on 2.6 and 2.7.
Friday10/26/12
Page 142: #3, 4, 5, 6, 15, 16, 17, 28, 29, 30, 31, 32
RTN & GP Lesson 3.2
Monday 10/29/12
Page 150: #22 – 36. For each problem (except #34), copy the diagram, set up an equation and state the postulate or theorem that justifies it, using the acceptable abbreviation.
RTN & GP Lesson 3.3
Monday 11/12/12
WELCOME BACK FROM HURRICATION 2012! No HW
Tuesday 11/13/12
Do pages 1 & 2 ONLY of the 3.2/3.3 packet
RTN & GP Lesson 3.3
Wednesday 11/14/12
Question: Which problems should students complete on Friday 10/26/12? Answer: Students should complete problems #3, 4, 5, 6, 15, 16, 17, 28, 29, 30, 31, and 32 on page 142.
Question: What should students do on Thursday 10/25 to prepare for the test on 2.6 and 2.7? Answer: Students should prepare for the test on 2.6 and 2.7. | 677.169 | 1 |
All triangles are generally classified in two different ways, by their relative lengths of sides or by their internal angles.
Classification of triangle by sides:
Triangles by their relative sides are classified into three different types; Equilateral, Isosceles and Scalene.
Equilateral triangles:
Equilateral triangles are triangles that have all the congruent or equal sides. Since all sides are equal, they create equal angles each measuring 60°. Below is an image of equilateral triangle which shows all sides and all angles are equal to each other.
Equilateral triangle
Isosceles Triangle:
Isosceles triangles are type of triangles that have two congruent or equal legs and two angles that are equal. It is easy to figure out the 3rd angle of an isosceles triangle based on one angle. Below is an image of isosceles triangle shows two congruent sides and two congruent angles.
Isosceles Triangle
Scalene Triangles:
Scalene triangles are type of triangles that have no congruent sides thus no congruent angles. Each side is different from other and each angle is different. Below is an image of a Scalene Triangle.
Scalene Triangles
Classification of triangle by angles:
Triangles by their relative angles are classified into three different types; Right, Obtuse, and acute.
Right Triangle:
What is right triangle? Right triangle is a type of triangle which has one angle measuring to 90° or a right angle. Unlike other triangles, right triangle has many parts with specific names. The image below shows different parts of the triangle. Pythagorean Theorem calculator within this site to calculate different parts of right triangle.
Right Triangle
Obtuse Triangle:
Obtuse triangle is a type of triangle that has one angle measuring more than 90°. This type of triangle generally tends to have a slanted look. Notice that only one angle of an obtuse triangle is greater than 90° or else it will not be a triangle. Below is an image of obtuse triangle.
Obtuse Triangles
Acute Triangle:
Acute triangle is a type of triangle where all angles equals to 180 degrees but they all are smaller than 90 degrees individually. Below is a picture of acute triangle.
Acute Triangles
Calculation of Triangle:
There are three important triangle calculations that can be done across all types of triangle. They are Area, base and Height. The image below shows the base and height.
An area of a triangle is calculated by multiplying base with its height and dividing it by 2.
Question: In a scalene triangle, are there any congruent sides or angles? Answer: No, all sides and angles are different
Question: Which triangle has one angle measuring more than 90 degrees? Answer: Obtuse triangle | 677.169 | 1 |
How to Construct a Parallelogram from Diagonals Video shows a method for constructing a parallelogram, given lengths of the two diagonals. Keeps you guessing right up to the end!
GMAT Prep - Math - Geometry - Diagonals by Knewton Go to for hundreds of GMAT math and verbal concepts, thousands of practice problems and much more. Knewton GMAT is a GMAT prep course that redefines everything you thought you knew about online learning. Subscribe to this channel for tips, explanations, and Q&A about the GMAT and getting your MBA.
Steve Nash..dribbling - Diagonals
How to learn / recognize diagonals trotting # 3 by TWOMBLY PUBLISHING (is where you can find the book "HORSES and PEOPLE MATCHING" for the best online price) if you're a new student who needs added visuals this might help...or if you're a riding instructor trying to save your voice or your favorite lesson pony... this could prove to be beneficial with the sound UP it's a STUDY GUIDE with the sound down it's a TEST !!!
Rhombus Diagonals Proof that the diagonals of a rhombus are perpendicular bisectors of each other
"Hi Mark, I was going through your double diagonals and was not able to glean guidelines as to how to open and manage the trade. For example, when (how many days out from your long/short strike expiration) do you put on the diagonal, how do you" — Double Diagonals (Options for Rookies),
"Super Joystick diagonals too sensitive First of all, just wanted to say this forum has helped me in many ways to get my cab up and running. Usually I have been able to find answers to my questions just by searching. But, I need some input for this small problem" — Super Joystick diagonals too sensitive,
Question: What is the sender's request in the "Super Joystick diagonals too sensitive" message? Answer: They need input or help to fix the sensitivity issue.
Question: What is the purpose of the video by TWOMBLY PUBLISHING? Answer: To help new students or riding instructors recognize diagonals in horse trotting. | 677.169 | 1 |
I'm creating a program for college work, now I have created the application to specifications given with added features. Now I would like to add the feature where from inputted sizes of the triangle to create the triangle. Is this possible and if so can some one please help supply the code for me, I would be extremely greatful (I'm not breaching rules of the work as I've meet the specifications of my work and I'm adding additonal features).
David Penfold
This is where you learn some systems engineering, rather than just getting some free code. The two driving factors on a program are the inputs and the outputs. First, consider the inputs. There are a number of ways to specify a triangle. You can use 3 points, two points plus an implicit point at the origin, 3 side measures, two side measures plus the angle between the given sides, or a side and the two ending angles. If you don't care what the size of the triangle is, but care only about the shape, you can specify the triangle by 3 angles only. The next thing to consider is the outputs. Will the triangle be drawn? If so, how? If not, what exactly is the output? Once these things are known, you have an interface spec. After that, decide (or tell us) what the OS and language are.
---
Sheldon Linker
Linker Systems, Inc.
[email protected]
800-315-1174
+1-949-552-1904
david is right. Time to learn what it takes to be a programmer instead of searching for free code.
In Vb you can start with a picture control. There are APIs to draw lines in the control using x and y coordinates. You specify the line width and color. You will have to draw the 3 lines of your triangle and make sure you connect the dots.
Question: Can the triangle's size be ignored and only the shape be specified? Answer: Yes, by specifying the triangle by 3 angles only. | 677.169 | 1 |
So if you like, this is the bow and up here we have the bow string. And of course we can cancel the 2's. That's equal to sin theta / theta. And so now why does this tend to 1 as theta goes to 0? Well, it's because as the angle theta gets very small, this curved piece looks more and more like a straight one. Alright? And if you get very, very close here the green segment and the orange segment would just merge. They would be practically on top of each other. And they have closer and closer and closer to the same length. So that's why this is true.
I guess I'll articulate that by saying that short curves are nearly straight. Alright, so that's the principle that we're using. Or short pieces of curves, if you like, are nearly straight. So if you like, this is the principle. So short pieces of curves. Alright?
So now I also need to give you a proof of A. And that has to do with this cosine function here. This is the property A. So I'm going to do this by flipping it around, because it turns out that this numerator is a negative number. If I want to interpret it as a length, I'm gonna want a positive quantity. So I'm gonna write down (1 - cos theta) here and then I'm gonna divide by theta there. Again I'm gonna make some kind of interpretation. Now this time I'm going to draw the same sort of bow and arrow arrangement, but maybe I'll exaggerate it a little bit. So here's the vertex of the sector, but we'll maybe make it a little longer.
Alright, so here it is, and here was that middle line which was the unit... Whoops. OK, I think I'm going to have to tilt it up. OK, let's try from here. Alright, well you know on your pencil and paper it will look better than it does on my blackboard. OK, so here we are. Here's this shape. Now this angle is supposed to be theta and this angle is another theta. So here we have a length which is again theta and another length which is theta over here. That's the same as in the other picture, except we've exaggerated a bit here. And now we have this vertical line, which again I'm gonna draw in green, the bow string. But notice that as the vertex gets farther and farther away, the curved line gets closer and closer to being a vertical line. That's sort of the flip side, by expansion, of the zoom in principle. The principle that curves are nearly straight when you zoom in. If you zoom out that would mean sending this vertex way, way out somewhere. The curved line, the piece of the circle, gets more and more straight. And now let me show you where this numerator (1 - cos theta) is on this picture.
Question: Why does the speaker divide the numerator (1 - cos theta) by theta? Answer: The speaker divides the numerator by theta to make an interpretation of it as a length, as the numerator is a negative number.
Question: What is the relationship between the green segment and the orange segment as the angle theta approaches 0? Answer: As theta approaches 0, the green segment and the orange segment merge and have closer and closer to the same length.
Question: What is the flip side of the zoom in principle? Answer: The flip side of the zoom in principle is that as you zoom out, the curved line, the piece of the circle, gets more and more straight.
Question: Where is the numerator (1 - cos theta) located on the picture? Answer: The numerator (1 - cos theta) is located on the picture where the green segment (the bow string) is. | 677.169 | 1 |
So now let me show you why that's possible to do. So in order to do that first of all I'm gonna trade the boards and show you where the line PQ is. So the line PQ is here. That's the whole thing. And the key point about this line that I need you to realize is that it's practically perpendicular, it's almost perpendicular, to this ray here. Alright? It's not quite because the distance between P to Q is non-zero. So it isn't quite, but in the limit it's going to be perpendicular. Exactly perpendicular. The tangent line to the circle. So the key thing that I'm going to use is that PQ is almost perpendicular to OP. Alright? The ray from the origin is basically perpendicular to that green line. And then the second thing I'm going to use is something that's obvious which is that PR is vertical. OK? So those are the two pieces of geometry that I need to see. And now notice what's happening upstairs on the picture here in the upper right. What I have is the angle theta is the angle between the horizontal and OP. That's angle theta. If I rotate it by ninety degree, the horizontal becomes vertical. It becomes PR and the other thing rotated by 90 degrees becomes the green line. So the angle that I'm talking about I get by taking this guy and rotating it by 90 degrees. It's the same angle. So that means that this angle here is essentially theta. That's what this angle is. Let me repeat that one more time.
We started out with an angle that looks like this, which is the horizontal that's the origin straight out horizontally. That's the thing labeled 1. That distance there. That's my right arm which is down here. My left arm is pointing up and it's going from the origin to the point P. So here's the horizontal and the angle between them is theta. And now, what I claim is is that if I rotate by 90 degrees up, like this, without changing anything - so that was what I did - the horizontal will become a vertical. That's PR. That's going up, PR. And if I rotate OP 90 degrees, that's exactly PQ.
OK?
So let me draw it on there one time. Let's do it with some arrows here. So I started out with this and then, we'll label this as orange, OK so red to orange, and then I rotate by 90 degrees and the red becomes this starting from P and the orange rotates around 90 degrees and becomes this thing here. Alright? So this angle here is the same as the other one which I've just drawn. Different vertices for the angles.
OK?
Question: Which is the left arm of the angle, OP or PR? Answer: OP. | 677.169 | 1 |
Welcome to Planet Infinity KHMS
Pages
21 August, 2009
Median of a triangle is a line segment joining vertex to the mid point of opposite side. There are three medians in a triangle. All medians intersect at a common point called centroid. The centroid always lie in the interior of traingle.
Activity Aim :To verify medians of a triangle concur at a point called centroid which always lie in the interior of the triangle by paper folding
Question: What is the title of the page? Answer: Welcome to Planet Infinity KHMS | 677.169 | 1 |
is a straight angle and C is just another point in the plane not on its sides,
then
180 = m() + m().
Circle: A circle is a set of points in the plane which are
at the same distance
to a fixed point called center. The fixed distance is called the radius of the
circle.
Acute Angle: An acute angle is an angle whose measure is
less than the
measure of a right angle.
Obtuse Angle: An obtuse angle is an angle whose measure is
greater than
the measure of a right angle.
Vertical Angles: Vertical angles (which are not straight
angles) are two angles
which share the same vertex and their rays form two straight lines.
Theorem of vertical angles: Given two angles which are
vertical then these
angles are congruent.
Supplementary Angles: Two angles whose measures add up to
180 are called
supplementary angles.
Complementary Angles: Two angles whose measures add up to
90 are called
complementary angles.
Perpendicular lines or segments: Two lines are said
perpendicular if they
contain rays which form a right angle. Two segments are perpendicular if they
are contained on two perpendicular lines.
Transversal: Transversal line is a line which intersects
two lines.
Corresponding angles, alternate interior, alternate
exterior, interior of the
same side of the transversal recognize on the picture.
Triangle: A triangle is a polygon with three vertices.
Isosceles triangle: A triangle is said to be isosceles if
two of its sides are
congruent.
Equilateral triangle: A triangle whose sides are all
congruent is called equilateral.
Scalene triangle: A scalene triangle is a triangle in
which all three sides have different lengths .
Midpoint of a segment A point M on a line segment
is called the midpoint
of this line segment if it is equally distant to A and B (i.e. MA = MB).
Median of a triangle: A median of a triangle is the line
segment joining a
vertex with the midpoint of the opposite side.
Centroid or center of mass: The medians in a triangle meet
at a point called
centroid or center of mass and it is denoted usually by G. The point G is
located
at 2/3 to the vertex and 1/3 to the corresponding midpoint on each median of
the triangle.
Cevian in a triangle: A Cevian is a line segment
associated with a triangle
which joins a vertex of this triangle with a point on the opposite side (or its
extension).
Question: What is the ratio of the distance from the centroid to the vertex compared to the distance to the midpoint of the opposite side in a triangle? Answer: 2/3 to 1/3.
Question: Where do the medians of a triangle intersect? Answer: The medians of a triangle intersect at a point called the centroid or center of mass, denoted by G.
Question: What is a Cevian in a triangle? Answer: A Cevian is a line segment that joins a vertex of a triangle with a point on the opposite side (or its extension). | 677.169 | 1 |
When a straight line crosses two parallel lines there are more angle factswe can look for and use!
1. Corresponding angles are equal - these are angles in a letter 'F'.
2. Alternate angles are equal - these are angles in a letter 'Z'.
3. Supplementary angles add up to 1800 - these are angles in a letter 'U' or 'C' (when the 'U' and the 'C' are made of three straight sides, of course).
The diagram below might help you to see this more clearly. Click on thepairs of angles you want to see and they will be shown on the diagram:
GCSE Maths Banana skins - Don't slip up!
Students keep making the same mistakes in their GCSE Maths exams. Inspired by the examiner's reports find out where students are losing vital marks, so that you can avoid the common slip-ups!
Question: What letter does the diagram use to represent corresponding angles? Answer: 'F' | 677.169 | 1 |
You can put this solution on YOUR website! I guess I can start with the basics.
There are 6 basic trigonometric functions.
Sine (Sin), Co-sine (Cos), Tangent (Tan), Co-Tangent (Cot), Secant (Sec), Co-Secant (Csc).
You only need to remember 3 and remember these simple rules:
A way to remember this is that everything corresponds with a "Co". It is suggested you only remember sin, cos, and tan.
Consider a right triangle. If you don't remember a right triangle has a 90 degree angle. The three sides of a triangle are usually labeled as opposite, adjacent, and hypotenuse. The hypotenuse is always easy to identify, it's the side directly across from the 90 degree side. Given an angle, the adjacent side will be the side that is right next to the angle while the opposite will be the farthest side from the angle that is not the hypotenuse. I will label O = opposite, A = Adjacent, and H = hypotenuse. X is an angle.
Some identities:
Hope this helps
Question: In a right triangle, which side is the hypotenuse? Answer: The side directly across from the 90 degree angle | 677.169 | 1 |
Since each of the angles BAC and BAG is right, it follows that with a straight line BA, and at the point A on it, the two straight lines AC and AG not lying on the same side make the adjacent angles equal to two right angles, therefore CA is in a straight line with AG.
Since DB equals BC, and FB equals BA, the two sides AB and BD equal the two sides FB and BC respectively, and the angle ABD equals the angle FBC, therefore the base AD equals the base FC, and the triangle ABD equals the triangle FBC.
Now the parallelogram BL is double the triangle ABD, for they have the same base BD and are in the same parallels BD and AL. And the square GB is double the triangle FBC, for they again have the same base FB and are in the same parallels FB and GC.
And the square BDEC is described on BC, and the squares GB and HC on BA and AC.
Therefore the square on BC equals the sum of the squares on BA and AC.
Therefore in right-angled triangles the square on the side opposite the right angle equals the sum of the squares on the sides containing the right angle..
Q.E.D.
This proposition is generalized in VI.31 to arbitrary similar figures placed on the sides of the triangle ABC. If the rectilinear figures on the sides of the triangle are similar, then that on the hypotenuse is the sum of the other two figures.
A bit of history
This proposition, I.47, is often called the "Pythagorean theorem," called so by Proclus and others centuries after Pythagoras and even centuries after Euclid. The statement of the proposition was very likely known to the Pythagoreans if not to Pythagoras himself. The Pythagoreans and perhaps Pythagoras even knew a proof of it. But the knowledge of this relation was far older than Pythagoras.
More than a millennium before Pythagoras, the Old Babylonians (ca. 1900-1600 B.C.E) used this relation to solve geometric problems involving right triangles. Moreover, the tablet known as Plimpton 322 shows that the Old Babylonians could construct all the so-called Pythagorean triples, those triples of numbers a, b, and c such that a2 + b2 = c2 which describe triangles with integral sides. (The smallest of these is 3, 4, 5.) For more on Pythagorean triples, see
X.29.Lemma 1.
The hypotenuse diagram in the Zhou bi suan jing
The rule for computing the hypotenuse of a right triangle was well known in ancient China. It is used in the Zhou bi suan jing, a work on astronomy and mathematics compiled during the Han period, and in the later important mathematical work Jiu zhang suan shu [Nine Chapters] to solve right triangles.
The Zhou bi includes a very interesting diagram known as the "hypotenuse diagram."
Question: Is angle BAC a right angle? Answer: Yes
Question: In which ancient Chinese works was the rule for computing the hypotenuse of a right triangle used? Answer: Zhou bi suan jing and Jiu zhang suan shu
Question: What is the name of the interesting diagram related to the hypotenuse in the Zhou bi suan jing? Answer: The hypotenuse diagram | 677.169 | 1 |
1.) Know precise definitions of angle, circle, perpendicular line, parallel line, and line segment based on the undefined notions of point, line, distance along a line, and distance around a circular arc. [G-CO1]
2.) Represent [G-CO2]
3.) Given a rectangle, parallelogram, trapezoid, or regular polygon, describe the rotations and reflections that carry it onto itself. [G-CO3]
5.) Given a geometric figure and a rotation, reflection, or translation, draw the transformed figure using, e.g., graph paper, tracing paper, or geometry software. Specify a sequence of transformations that will carry a given figure onto another. [G-CO5]
Understand congruence in terms of rigid motions. (Build on rigid motions as a familiar starting point for development of concept of geometric proof.)
6.) Use geometric descriptions of rigid motions to transform figures and to predict the effect of a given rigid motion on a given figure; given two figures, use the definition of congruence in terms of rigid motions to decide if they are congruent. [G-CO6]
7.) Use the definition of congruence in terms of rigid motions to show that two triangles are congruent if and only if corresponding pairs of sides and corresponding pairs of angles are congruent. [G-CO7]
8.) Explain how the criteria for triangle congruence, angle-side-angle (ASA), side-angle-side (SAS), and side-side-side (SSS), follow from the definition of congruence in terms of rigid motions. [G-CO8]
Prove geometric theorems. (Focus on validity of underlying reasoning while using variety of ways of writing proofs.)
9.) Prove theorems about lines and angles. Theorems include vertical angles are congruent; when a transversal crosses parallel lines, alternate interior angles are congruent and corresponding angles are congruent; and points on a perpendicular bisector of a line segment are exactly those equidistant from the segment's endpoints. [G-CO9]
10.) Prove theorems about triangles. Theorems include measures of interior angles of a triangle sum to 180o, base angles of isosceles triangles are congruent, the segment joining midpoints of two sides of a triangle is parallel to the third side and half the length, and the medians of a triangle meet at a point. [G-CO10]
11.) Prove theorems about parallelograms. Theorems include opposite sides are congruent, opposite angles are congruent; the diagonals of a parallelogram bisect each other; and conversely, rectangles are parallelograms with congruent diagonals. [G-CO11]
Make geometric constructions. (Formalize and explain processes.)
12.) Make formal geometric constructions with a variety of tools and methods such as compass and straightedge, string, reflective devices, paper folding, and dynamic geometric software. Constructions include copying [G-CO12]
13.) Construct an equilateral triangle, a square, and a regular hexagon inscribed in a circle. [G-CO13]
Similarity, Right Triangles, and Trigonometry
Question: What are the criteria for triangle congruence according to the text? Answer: Angle-Side-Angle (ASA), Side-Angle-Side (SAS), and Side-Side-Side (SSS).
Question: What is the first step in proving geometric theorems, as suggested by the text? Answer: Focus on the validity of the underlying reasoning.
Question: Which theorem states that vertical angles are congruent? Answer: The theorem that states this is part of [G-CO9]. | 677.169 | 1 |
Bosnian Geodetic Institute (Geodetski Zavod BiH) is confirmed previous findings of the Foundation Archaeological Park: Bosnian Pyramid of the Sun.
'If we connect a top of the pyramids (Sun, Moon, Dragon) by drawing a line. We can see that distance is equal. This lines are forming triangle. Sides of the triangle have equal lengths.'
Angles of this triangle are 60 degrees exactly (not any minute difference).
If you slept through basic geometry at school a minute is one-sixieth of a degree. It would be a difficult claim to check, the photos look quite convincing. Except if you download the placemarks linked above (and possibly have Google Earth Plus) you can check this next claim too. I measured the distance between the Pyramids of the Moon and the Dragon and got a distance of 2,250 metres.
The distance between the Pyramids of the Moon and the Dragon. Click for a bigger image.
That's open to questions about accuracy, the sites haven't been excavated yet (which doesn't bother the Bosnian Geodetic Institute but nevermind) but they do give a ballpark figure. If the tops of the pyramids do describe an equilateral triangle then the distance between the Pyramids of the Moon and the Sun should be around the same.
The distance between the Pyramids of the Moon and the Sun. Click for a bigger image.
The distance is 2,060 metres. That's a 10% difference which makes you wonder quite how you can declare the angles are equal to within an accuracy of a minute. One reasonable objection I've had to this measurement is that the Pyramid of the Sun is higher I may have measured the flat difference between the summits. How much higher would the Pyramid of the Sun have to be than the other pyramids? 904 metres. The summit of the hill is only 767 metres above sea level, so the peaks of the pyramids would have to be around 150 metres below sea-level for the geometrical claim to work. I look at those results and assume I've made a big mistake somewhere, but I cannot see where. The measurements would have to be staggeringly inaccurate. Unfortunately...
The figures really don't add up
Ok, that's a bit esoteric. What about basic data. How old is the pyramid? It depends on who you listen to.
Semir Osmanagić told BosnianPyramid.com that"all three pyramids were constructed during the same period, with the Bosnian pyramid the last to be built".
However, Semir Osmanagić speaking to FENA news disagreed saying the Bosnian Pyramid was probably the first. Alas the orginal link has expired but you can see it at Bosnia News.
Semir Osmanagić has a different view. Speaking in the April/May issue of Nexus Magazine he was cautious on the dating, saying it was more likely to be Illyrian in date.
Question: Are the sides of the triangle formed by connecting the tops of the Bosnian Pyramids of the Sun, Moon, and Dragon equal in length? Answer: Yes, according to the text, they are.
Question: What is the distance between the Pyramids of the Moon and the Dragon? Answer: 2,250 metres. | 677.169 | 1 |
Symmetries III
This investigation will help you to understand how translations work and what happens when two or more translations are applied one after the other.
If students are familiar with vectors, they can use them in this context to define a translation in the plane. All band ornaments have translational symmetry, and all wallpaper patterns have translational symmetry in at least two directions.
Learning Objectives
Students will be able to:
Understand how translations work
Understand what happens when two or more translations are applied one after the other
Materials
Computer and Internet connection
Instructional Plan
Think About...Translations
Take a shape, move it from here to there, and that's a translation.
Translations are the simplest of the symmetry transformations and they are also
the most important. Artists use translations to create band ornaments and
wallpaper patterns.
1. What information must be given in order to define a translation?
2. What does a pattern look like that has been created using
translations?
3. What is the result when a shape is translated twice?
Describing Translations
One way to describe a translation is to give a translation vector. In
the diagram below, the blue triangle is the translated image of the green
triangle with translation vector PQ. You can drag point Q to
change the translation vector.
4. A vector is a mathematical object that has direction and magnitude. Why
are vectors useful for defining translations?
5. Can you think of any other ways to define a translation?
6. What is the result when a shape is translated first by one vector
and then by another vector?
Creating Patterns Using Translations
If a design is translated over and over using the same translation vector,
you get a type of pattern that is called a band ornament, strip
pattern, or frieze pattern.
7. The following patterns were created using translation. Identify the initial design and the translation vector that were used to create each of
them.
a.
b.
8. Which of the following patterns were created using translation? For those that were created using translation, identify the initial design and the translation vector.
a.
b.
9. Make your own patterns using translation.
10. Find patterns around you that were created using translation.
11. Find examples of translation in nature.
Combining Symmetries
Sometimes a symmetric design is translated repeatedly to create a new pattern.
12. If a design with bilateral
symmetry is translated, will the resulting pattern also have bilateral symmetry?
13. If a design with rotational
symmetry is translated, will the resulting pattern also have rotational symmetry?
Infinite Designs
Imagine a design that has been translated infinitely many times to the left and infinitely many times to the right. The resulting pattern will look the same
after it has been translated, and we say that it has translational symmetry. Patterns with translational symmetry in one direction are called band ornaments or strip patterns, and patterns with translational symmetry in two directions are called wallpaper patterns.
Question: Which of the following patterns were created using only translation? (a) A simple repeating geometric pattern, (b) A complex, irregular pattern with no repetition. Answer: (a) A simple repeating geometric pattern.
Question: What happens when a shape is translated twice in succession? Answer: The second translation is applied to the result of the first translation, not to the original shape.
Question: Can a translation be defined using only the starting and ending points? Answer: Yes, the translation vector can be found by subtracting the starting point from the ending point. | 677.169 | 1 |
14. If an infinite pattern was created using a translation vector to translate a design infinitely many times, what translations will leave the infinite pattern apparently unchanged?
15. If an infinite pattern was created using a design with point symmetry, what kind of symmetry will
the infinite pattern have?
16. If an infinite pattern was created using a design with bilateral symmetry, what kind of symmetry will the infinite pattern have
Question: What is translational symmetry? Answer: It means that the pattern looks the same after being translated in a specific direction by a certain distance. | 677.169 | 1 |
move the coordinate system of the 1" block? This question is
difficult to answer precisely without mathematics, but we'll give it
a try. The first concept to tackle goes by the name of linear
independence. In this context, linear independence means that
there is only one way to move an object in a given direction. For
example, if you want to move an object along the X axis of the global
coordinate system, this can only be accomplished by translating its
local coordinate system along the X axis. It could not be
accomplished by translating its coordinate system just along the Y
axis or just along the Z axis or any combination
of translations along the Y and Z axis. In other words,
translations in the X direction are independent of translations in
the Y and Z directions. The same thing holds true for translations
along either the Y or Z axes as well.
Suppose you want to translate the block 1 ½"
along a 45º line between the X and Y axes? Such a translation
would consist of a combination of translations along both axes. A
subtle consequence of linear independence, however, insures that this
combination is unique. That is, there is only one possible
combination of translations which can move the coordinate system to
the desired position.
Some more jargon: The fact that, in
three-dimensional geometry, there are three linearly independent
directions means that the system has three degrees
of freedom. But, there is more to it. Just like the rotation
operation in two dimensions, you can rotate a coordinate system in
three dimensions as well. In three dimensions, however, there are
three "directions" in which a coordinate system may be
rotated. In the figure below, a second three-dimensional coordinate
system is shown. The best way to think of rotating a
three-dimensional coordinate system is to imagine the Z axis piercing
a sphere centered around the origin of the transformed coordinate
system.
Just like the coordinates of latitude
and longitude are used to locate a point on the surface of the Earth,
you need two coordinates to define the direction of the Z axis.
Additionally, you can now imagine another rotation of the system
shown above: that of twisting the X'Y' axes around the Z' axis (which
is exactly equivalent to the rotation of the two-dimensional
coordinate system). Therefore, in addition to the three degrees of
translational freedom, a three-dimensional coordinate system also has
three degrees of rotational freedom.
So, getting back to the block example, pick a face
on the 1" block and a face on the 2" block and place them
touching each other. This is equivalent to the mate constraint in
lignumCAD. Now, how can you move the 1"
block so that the two faces remain touching? You can move the 1"
block up and down, and forwards and backwards; you can also twist the
block around. Therefore, the mate operation has specified three of
our six degrees of freedom, namely one translational direction and
two of the rotational directions.
Question: Which degrees of freedom are specified when the 1" block and 2" block are placed with their faces touching? Answer: The mate operation specifies three degrees of freedom: one translational direction (up and down, forwards and backwards) and two of the rotational directions (twisting the block around).
Question: What is the equivalent of the mate constraint in lignumCAD in the context of the 1" block and 2" block example? Answer: Placing a face on the 1" block and a face on the 2" block touching each other is equivalent to the mate constraint in lignumCAD.
Question: What does linear independence mean in the context of this text? Answer: Linear independence in this context means that there is only one way to move an object in a given direction, and translations in different directions are independent of each other. | 677.169 | 1 |
Soc. — But does not this line become doubled if we add another such
line here?
5ov— Certainly.
Soc. — And four such lines will make a space containing eight feet?
Boy — Yes.
Soc. — Let us describe such a figure : is not that what you would say
is the figure of eight feet?
Boy — Yes.
Soc. — And are there not these four divisions in the figure, each of
which is equal to the figure of four feet?
Boy — True.
Soc. — And is not that four times four?
Boy — Certainly.
Soc. — And four times is not double?
Boy — No. indeed.
287] APPENDIX 287
Soc. — But how much?
Boy — Four times as much.
Soc. — Therefore the double line, boy, has formed a space, not twice,
but four times as much.
Boy — True.
Soc. — And four times four are sixteen, are they not?
Boy — Yes.
Soc. — What line would give you a space of eight feet, as this gives
one of sixteen feet? Do you see?
Boy—Yts.
Soc. — And the space of four feet is made from this half line?
Boy — Yes.
Soc. — Good ; and is not a space of eight feet twice the size of this,
and half the size of the other?
Boy — Certainly.
Soc. — Such a space, then, will be made out of a line greater than this
one, and less than that one.
Boy — Yes ; that is what I think.
Soc. — Very good ; I like to hear you say what you think. And now
tell me, is not this a line of two feet and that of four?
Boy — Yes.
Soc. — Then the line which forms the side of eight feet ought to be
more than this line of two feet, and less than the other of four feet?
Boy — It ought.
Soc. — Try and see if you can tell me how much it will be.
Boy — Three feet.
Soc. — Then if we add a half to this line of two, that will be the line
of three. Here are two and there is one ; and on the other side, here
are two also and there is one : and that makes the figure of which you
speak?
Boy— Yes.
Soc. — But if there are three feet this way and three feet that way,
the whole space will be three times three feet?
Boy — That is evident.
Soc. — And how much are three times three feet?
Boy — Nine.
Soc. — And how much is the double of four?
Boy — Eight.
Soc. — Then the figure of eight is not made out of a line of three?
Boy — No.
Soc— But from what line ? Tell me exactly ; and if you would rather
Question: What is the double of four? Answer: Eight
Question: How many times larger is the space of sixteen feet compared to the space of eight feet? Answer: Four times larger
Question: What is the total area of the figure formed by lines of three feet on both sides? Answer: Nine square feet | 677.169 | 1 |
Geometry Course/Eucler's Axiom
Axioms
Euclidean geometry is an axiomatic system, in which all theorems ("true statements") are derived from a small number of axioms.[1] Near the beginning of the first book of the Elements, Euclid gives five postulates (axioms) for plane geometry, stated in terms of constructions (as translated by Thomas Heath):[2]
"Let the following be postulated":
"To draw a straight line from any point to any point."
"To produce [extend] a finite straight line continuously in a straight line."
"To describe a circle with any centre and distance [radius]."
"That all right angles are equal to one another."
The parallel postulate: "That, if a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles."
Although Euclid's statement of the postulates only explicitly asserts the existence of the constructions, they are also taken to be unique.
Question: What are the five postulates given by Euclid for plane geometry? Answer: To draw a straight line from any point to any point, To produce a finite straight line continuously in a straight line, To describe a circle with any centre and distance, That all right angles are equal to one another, The parallel postulate | 677.169 | 1 |
Have students turn their triangle over, and duplicate their markings on the other side.
In the center of one side of the triangle, have students write the Law of Sines formula. On the other side of the triangle, have students write the 3 formulas for the Law of Cosines.
PART TWO: Practice Worksheets
Attached are two worksheets with four triangles on each worksheet. You make wish to use actual measurements or estimation measurements on these worksheets. For ease in measurement, measurements of sides should be represented in centimeters. (I like using actual angle and side measurements, so that students can check their solutions by actually measuring their results with measuring tools.)
You will need to print and prepare the Law of Sines Worksheet in advance. As the teacher, you will need to give the measurements to one angle and its opposite side, and then one other measurement of your choosing. (Such as one of the remaining sides or angles.)
The Law of Cosines Worksheet will need to be printed and prepared in advance. For each triangle, the Teacher will need to put in the measurements for 3 of the 6 parts of the triangle. (Hint: I always try to put a trick question in with the given information. An example of this might be giving the students two angles and one of their opposite sides. This is not so easily solved under the Law of Cosines as the first step.)
PART THREE: Small Group Activity
Have students work in pairs or in small groups.
Have each student draw about 6 triangles; big enough that all 6 fill up the whole page. Instruct your students to draw these triangles carefully, and accurately, using a straight edge.
Ask each student to measure 3 out of the 6 parts of each triangle, and record those measurements right on the drawing. (Note: Tell the students not to think about which trigonometric law they would use to solve these problems. It is better if they just randomly choose 3 different parts to measure. This gives some groups problems that are quite difficult and require higher level thinking to solve.)
Have students trade papers with each other, or with other groups. With their new paper, students must find the missing parts of each triangle, using whatever trigonometry knowledge they have. Have students show their work, and don't let them use their measuring tools as a short cut.
After the group has finished their work, give the papers back to the original owners.
Using a protractor and a centimeter ruler, have the original owner grade the answers for accuracy.
Strategies For Diverse Learners: This lesson can be modified to fit the needs of the students in your class. For the struggling students, you may wish to give an introduction on how Trigonometry would be used in a job situation. Some examples in building construction, or surveying could be helpful.
This lesson can be expanded to challenge gifted students by having them create story problems. Using real life situations, students can develop and sketch applied problems which have real life application. The emphasis should be placed on the aspects of "Why trigonometry is the best way to solve this problem?" and "Which method did you choose and why?"
Question: What should students write on one side of the triangle? Answer: The Law of Sines formula.
Question: How many triangles are there on each worksheet in Part Two? Answer: Four triangles.
Question: What are the measurements of sides represented in? Answer: Centimeters. | 677.169 | 1 |
a^2 + b^2 = c^2
(1200)^2 + (213)^2 = c^2
c = 1219
The angle of inclination x can be calculated using any of sine, cosine or tangent. I'll use tangent: tan x = opposite/adjacent = 213/1200 = .18
Using a trig table the value of x corresponding to tan x = .18 is x = 10 degrees.
Helpful
We'd like to understand what you find wrong with reinsuranc have the tree, and when sun is shining, it casts a shadow. The angle at which sun rays are coming down is given, so you know the lower left angle (in the picture) - let's call that angle "ANGLE". You also know how long the shadow is - let's call that length "L". So you can use any of the standard trigonometry formulas to calculate the height of the tree - let's call the height "H". For example: tan(ANGLE) = H / L. From this we get that H = L * tan( ANGLE ) and there's your solution.
Basically, just draw what you are given and use your standard formulas for every one of your problems.
Kresho
Helpful
We'd like to understand what you find wrong with CroCivic91,
I have always had long bangs up util a few days ago and I don't like them at all. The work is incredible....so my question is....can you apply hair extensions at the bang line? I desperatley need my long bangs back !
Thanks for your help.
I have to write a two page paper and a 10 min presentation on trigonometry and radios and I can't find anything! I know it has something to do with sin and cos but I can find nothing to elaborate on. I don't really know anything about sin and cos! Can anyone help me? Basically I need to know what...
Question: In the context of the shadow problem, what does 'H' represent? Answer: The height of the tree
Question: What is the value of 'c' in the given equation (1200)^2 + (213)^2 = c^2? Answer: 1219 | 677.169 | 1 |
f = 2f = 8.75 + 1.75f (re-order)
f = 0.25f = 8.75 (solve)
f = 35
Now solve for e and e + d using the pythagorean theorum and find d:
f 2 + rb2 = e2
352 + 1.752 = e2
1225+3.0625 = e2
e = √1228.0625
e = 35.0437
(c+f)2 + ra2 = (d+e)2
(5+35)2 + 22 = (d+e)2
1600 + 4 = (d+e)2
e = √1604
d+e = 40.0499
d = 40.0499 - 35.0437
d = 5.0062
Now, in order for this transition to be laid flat, we must draw two concentric circles with one radius for the base, and one radius for the truncation. Starting with a vertical line from the center of the cicles to the bottom quadrant, and another at the appropriate angle, the included arcs will match the circumference of a and b, a length d apart. We have the radius e for the inner circle, which will form the circumference for the top of the truncated cone, and the radius e+d which will form the circumference for the base of the cone.
We need to figure out the angle g to complete the flat pattern. First, let's determine the circumference for the base i and the truncated tip of the cone h.
h = bπ
h = 3.5π
h = 10.9955
i = aπ
i = 4π
i = 12.5663
Now, to determine the included angle g that will give us the appropriate circumference for our base and cap. To calculate the angle, we'll convert the circumference to an arc-length matching the radius we drew, and then convert radians from that into an angle
Question: What is the formula used to solve for 'e'? Answer: e = √(f² + rb²)
Question: What is the value of 'd'? Answer: 5.0062
Question: What is the formula used to solve for 'd'? Answer: d = (e + f)² + ra² - (c + f)² | 677.169 | 1 |
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