text
stringlengths 6
976k
| token_count
float64 677
677
| cluster_id
int64 1
1
|
|---|---|---|
Why Are We Talking About Spheres?
The vertices of regular and semiregular polyhedra lie on the surface of an imaginary sphere, which is to say that all vertices are equidistant from a polyhedron's center. Given this fact, we can picture spherical versions of each polyhedron, in which the polyhedral edges have stretched outward to become great-circle arcs and the faces have expanded into curved surfaces, as if each shape had been blown up like a balloon. Figure 14-2 shows a spherical tetrahedron, octahedron, and icosahedron as examples. Comparing these systems with their planar counterparts, it is clear that polyhedral edges are actually chords of great-circle arcs. We can conclude that the shortest distance between two events of a system always involves a great circle.
Fig. 14-2. Spherical polyhedra. Click on thumbnail for larger image.
The concept (rather than the reality) of a sphere—i.e., an omnisymmetrical container-acts as a frame of reference for polyhedral systems. Spherical polyhedra thus introduce new versions of familiar characters. The topological information (that is, the numbers and valencies of vertices, edges, and faces) of any polyhedron are displayed on a spherical canvas. An obvious consequence of this type of display is that shape is no longer a variable. Shape similarities, which are so rigorously accounted for by A and B modules, are thus ignored; our investigation now focuses in on topological, or surface, characteristics. Transforming polyhedra into balloons temporarily equalizes shape and size, providing a "common denominator" for other comparisons. The process develops a somewhat unorthodox chart.
New Classification System
However, the chart is not yet complete. Simply projecting edges and faces onto a spherical surface does not teach us anything new. We have yet to exploit the nature of the sphere.
Spheres suggest spin. That's how synergetics initially arrives at the omnidirectional form. Spin any system in all directions, and ultimately the action will have defined a circumscribing spherical envelope. Fuller places considerable emphasis on the "spinnability" of systems, arguing that as everything in Universe is in motion, the different axes of spin inherent in systems are worthy of investigation.
Question: What is the relationship between the vertices of regular and semiregular polyhedra and an imaginary sphere? Answer: The vertices of regular and semiregular polyhedra lie on the surface of an imaginary sphere, and all vertices are equidistant from the polyhedron's center.
Question: What does the process of transforming polyhedra into spherical polyhedra develop? Answer: The process develops a somewhat unorthodox chart.
Question: What does Buckminster Fuller emphasize about systems in relation to spin? Answer: Buckminster Fuller places considerable emphasis on the "spinnability" of systems, arguing that the different axes of spin inherent in systems are worthy of investigation. | 677.169 | 1 |
Submit your word problem on hypotenuse:
Are you looking for hypotenuse word problems?
TuLyn is the right place. We have tens of word problems on hypotenuse and hundreds on other math topics.
Below is the list of all word problems we have on hypotenuse.
Hypotenuse Word Problems
The hypotenuse of a right triangle(#169)
The hypotenuse of a right triangle is 2 centimeters more than the longer side of the triangle. The shorter side of the triangle is 7 centimeters less that the longer ...
Question: If the longer side of the triangle in the first problem is x centimeters, what would be the length of the hypotenuse? Answer: The hypotenuse would be x + 2 centimeters. | 677.169 | 1 |
Tetrakis hexahedron square pyramid is a pyramid having a square base. If the apex is perpendicularly above the center of the square, it will have C4v symmetry.- Johnson solid :...
to each face of the central cube. Thus, the compound can be seen as a stellation
Stellationof the tetrakis hexahedron. A different form of the tetrakis hexahedron, formed by using taller pyramids on each face of the cube, is non-convex but has equilateral triangle faces that again lie on the same planes as the faces of the three octahedra; it is another of the known isohedral deltahedra. A third isohedral deltahedron sharing the same face planes, the compound of six tetrahedra
Compound of six tetrahedra
This uniform polyhedron compound is a symmetric arrangement of 6 tetrahedra. It can be constructed by inscribing a stella octangula within each cube in the compound of three cubes, or by stellating each octahedron in the compound of three octahedra....The stellated octahedron, or stella octangula, is the only stellation of the octahedron. It was named by Johannes Kepler in 1609, though it was known to earlier geometers...
. A fourth isohedral deltahedron with the same face planes, also a stellation of the compound of three octahedra, has the same combinatorial structure as the tetrakis hexahedron but with the cube faces dented inwards into intersecting pyramids rather than attaching the pyramids to the exterior of the cube.
The cube around which the three octahedra can be circumscribed has nine planes of reflection symmetry
Reflection symmetry Three of these reflection panes pass parallel to the sides of the cube, halfway between two opposite sides; the other six pass diagonally across the cube, through four of its vertices. These nine planes coincide with the nine equatorial planes of the three octahedra.
History, della Francesca already includes a drawing of an octahedron circumscribed around a cube, with eight of the cube edges lying in the octahedron's eight faces. Three octahedra circumscribed in this way around a single cube would form the compound of three octahedra, but della Francesca does not depict the compound.
The next appearance of the compound of three octahedra in the mathematical literature appears to be a 1900 work by Max Brückner, which mentions it and includes a photograph of a model of it.
Stars is a wood engraving print by the Dutch artist M. C. Escher which was first printed in October 1948, depicting two chameleons in a polyhedral cage floating through space.-Description:...
, used as the central figure of the woodcut a cage in this shape, containing two chameleons and floating through space. Escher would not have been familiar with Brückner's work and H. S. M. Coxeter
Harold Scott MacDonald Coxeter
Question: How many planes of reflection symmetry does the cube in the compound of three octahedra have? Answer: Nine
Question: Which mathematician is credited with naming the stellated octahedron? Answer: Johannes Kepler
Question: What shape is the base of a Tetrakis hexahedron square pyramid? Answer: Square | 677.169 | 1 |
e = 1
, so the conic is a parabola, and it has a horizontal directrix above
the pole. Because its directrix is horizontal, its axis must be vertical.
So the vertex will occur on the line
θ =
.
(r,) = (2,)
is the vertex of the parabola. Note:
Another way to find the vertex is to use the fact that
p
, the distance from
the focus to the directrix, is known to be
4
in this problem.
e = 2
, so the conic is a hyperbola. The directrix is vertical and to the left
of the pole. The transverse axis is horizontal. The vertices are at
(- 6, 0)
and
(2, Π)
. So the transverse axis is
8
units long, so
a = 4
.
Therefore
c = 8
, and
b = 4
.
Question: What is the orientation of the directrix when e = 1? Answer: Horizontal | 677.169 | 1 |
Major and minor Diagonals are identified by DG and DP ("Grandis" and "Parvus", in Latin).
I used Latin names because I wrote this program in Italian first, due to the necessity to test it with real homeworks of the local kids.
Anyway names of the variables shouldn't be a problem, since when you know what they means, you can write whatever you want on your notebook or exercise book.
I wish to thank my friend Alessandro Rudellat for helping me during the beta testing,
letting me try this program with his homeworks... I suspect he enjoyed it too... :^)
Question: Are the variable names important? Answer: No, the variable names are not important as long as you understand their meaning. | 677.169 | 1 |
We are now wondering when builders are building houses how do they measure the angle on the house that they are building. Do they have to carry a huge protractor to measure the house? When we were building the houses we had trouble keeping them straight and stable. I think that builders must have an easier way because they use metal in the middle and it is going straight up. Building with chopsticks it is hard to make it stay straight. Sometimes the way you put the elastic bands on is wrong and it pushes the chopstick to go at an angle that you don't want it to be.
If we were using the app on the IPad during the building of our models, we would be able to make it straighter. We could measure it and then adjust it as we are doing the building. If our building bent in the wrong places it would fall down or the roof would fall through. The only reason we are trying to build a house is because it is part of our Maths inquiry. Our central idea is "Geometric tools and methods can be used to solve problems relating to shape and space". Actually the main reason we are building a house is because we are hoping to build a 5×4 meter community house for the people at the Serpong landfill. Using geometric tools to build a house can help when you want to make the columns or the walls straight and so that you know what the angle of the roof is. Then you can work out how long the roof is and how many roof tiles or iron roofing you will need."
Here is the video that Ms Jane took of her lesson. Please watch it to learn about how we used the iPad app to measure the angles.
4 comments:
Jane, You do such an amazing job at showing the learning and curiosity in your classroom - it inspires me to up the anti with my own classroom blog. Thank you for sharing. Kirsten Moss in Nanaimo, Canada
We have a 1:1 too, but the students have MacBook Pros instead of iPads. I'm wondering how the angle measures might go. We have one class set of iTouches in the building. I'll see if we can get A+ Measuring finger onto the iTouches.
I also like your integration of bamboo. Hong Kong uses bamboo when they build and repair skyscrapers
Question: What is the name of the iPad app used to measure angles? Answer: The name of the app is not mentioned in the text.
Question: What tool do builders use to measure angles while constructing houses? Answer: They use geometric tools like rulers, squares, and levels. | 677.169 | 1 |
9.3. Representing Vectors in Terms of Their Components in a Coordinate System Using the Unit Vectors i, j, and k
• Vectors can be represented in terms of their components in a coordinate system. The vector components can be defined by their directions along the X, Y, and Z axes of a coordinate system using unit vectors denoted by i, j, and k. The i, j, andk unit vectors have magnitudes of one and directions pointing parallel to the X, Y, and Z axes, respectively, in a rectangular coordinate system.
Question: Can a vector be represented without using a coordinate system? Answer: No. | 677.169 | 1 |
Question 524491
They always say "practice makes perfect."
If I were you, I would look at some examples of proofs and ask myself, "Why must this always be true?" Then I would try some (or many) on my own. Note that proofs are not found exclusively in geometry courses; they come up in every branch of mathematics. Therefore you could try proving theorems from other areas of mathematics as well.
Above all, don't get scared! I see many students (including myself, sometimes) attempt to prove a question or theorem that they have never seen before, and shudder in fear, having absolutely no idea what to do (as a note, this happened to me when I had to prove that, in triangle ABC, tan B tan C = 3 if and only if the Euler line is parallel to BC).
In this case, you want to really sink your teeth into the problem: read word by word, draw pictures or diagrams, use whatever techniques are necessary. This enabled me to prove the Euler line problem in about half an hour.
Question: What is the author's personal experience with proving a theorem? Answer: They once struggled to prove that in triangle ABC, tan B tan C = 3 if and only if the Euler line is parallel to BC. | 677.169 | 1 |
The Distance Formula (Between Two Points)
This is nice and easy to use tool, that can show you the distance between any two points on a plane (Cartesian coodinate system), if you know the coordinates of these points.
How To Use It
For example if you have point A (2;5) and point B (3;10), you simply need to type in these coordinates into the appropriate fields and click the button to see, that the ditance - 5.0990195135927845 .
Distance Calculator
Calculate the distance from point A to point B
Point
A
Point
B
(xA,
yA)
(xB,
yB)
By the way, the calculation this widget works with is relatively simple and it's worth knowing it. If you are qurious about the theory used for it, you can visit this site: and read about this matter.
If you like this widget or/and find it useful, you can help us by sharing it with friends...
If you like this, please Share it:
COPY and PASTE this code to Share the The Distance Formula (Between Two Points) on:
Question: Can this tool calculate the distance between points on a 3D plane? Answer: No | 677.169 | 1 |
Prisms, Anti-prisms, Pyramids, and related Polyhedra
These sets consist of infinite series, and are generally generated using a
standard formula from two polygons of the same type. These polygons do not
even have to be regular or even convex polygons, though the anti-prism formula
would require some regularity to the starting polygon.
Prisms
Take two polygons and connect together with vertical edges of the same length
to form squares. Each vertex has 3 edges, the new edge being at right angles
to the edges of the original polygonal interface.
The cylinder is a special 'infinite' case of this class.
This method is also known as elongation of the original polygons (See Johnson Solids).
ASIDE: the "Pentagonal Prism Dual" and "Pentagonal Antiprism Dual" are the
only polyhedra that made a good ten-sided dice. The later is usually used as
the points are better distributed, and are common in the paper and dice
fantasy games such as "Advanced Dungeons and Dragons".
Pyramids and Di-Pyramids
Basically forming a tent on a polygon. Note however that only three of
these can be formed using edges of equal length, built using a triangle
(forming a tetrahedron), a square (regular or square pyramid, like in
Egypt), and a pentagon (or "pentagonal pyramid").
If you try this with a hexagon, the object is flat, with six triangles on
top of the hexagon, and no volume. After that it's impossible, unless you
use longer edges to form the pyramid, but then the object will not conform to
any .
Bicupolas are like pyramids, just two of the objects joined back to back.
In each case however they can be joined together in two different ways,
ortho-bi-cupola, are mirrored across the join, and gryo-bi-cupola have a small
twist so two triangles are not connected together by an edge.
Question: Can all polygons be used to create a pyramid with edges of equal length? Answer: No, only a triangle, square, and pentagon can be used to create a pyramid with edges of equal length. | 677.169 | 1 |
You are here
Sin, Cos and Tan
A right-angled triangle is a triangle in which one of the angles is a right-angle. The hypotenuse of a right angled triangle is the longest side, which is the one opposite the right angle. The adjacent side is the side which is between the angle in question and the right angle. The opposite side is opposite the angle in question.
In any right angled triangle, for any angle:
The sine of the angle = the length of the opposite side
  ...
To continue enjoying this article for free you need to login or register with Maths Revision. Registration is free and only takes a minute. Once you have registered you will have full access to all the Maths Revision content in full.
Question: What is the condition for a triangle to be considered right-angled? Answer: One of the angles must be a right angle (90 degrees) | 677.169 | 1 |
In Geometry, a point is defined as being where 2 or more lines intersect, or where a single line changes direction. For a chord and a circle, the intersection points are where the line passes through the circle, the intersecting line being that which forms the circumference of the circle. Even the diameter, or the Great Chord, holds true to this definition. Whilst it is true that the diameter passes through the centre of the circle, it still has only 2 points because, apart from its intersections with the circumference, there are no other intersections which will define a point.
Question: What is the definition of a point in Geometry according to the text? Answer: A point is defined as where 2 or more lines intersect, or where a single line changes direction. | 677.169 | 1 |
So let's start the explanation from the end.. The small pieces are exactly the same in the 2 images. If the 2 figures formed by them would have been the same (have the exact surface) it would have been impossible for one of them to have a missing part (square). So the full figures must be different. And the difference comes from the slope ("hypotenuse"), which is not actually a straight line. In figure #1 the slope is bent inwards, and in figure #2 in bent outwards. The surface created by the difference between the 2 inclinations is the exact size of the surface of the missing square. It was explained a couple of times before I did, but I wanted to add mine. Hope it makes sense
The triangles are not the same slope. The big triangle is 3/8 and the small is 2/5, so the "hypotenuse" of the whole diagram isn't a straight line. Compare where the two triangles touch in the lower diagram to the same spot in the upper: The upper one doesn't quite touch the grid corner.
how so? the red one is 8 squares along the bottom and 3 up the side while the blue one is only 5 squares long and two up its side. What colours do you see? I see red and blue triangles and the other two are kind of orange and green.
I'm gonna go with..... In the above picture, you have to take into account any square the color touches, you have to add the white portions as used space as well, as part of the equation. By moving the colored shapes, you are also moving the white space used in the equation. Best guess......??
I'm colorblind but my guess would be the blocks that overlap on the top one balance out to 5 over 5...when they are offset it becomes 8 over 7 and an extra space is created even though all the measurements are otherwise equal? ? Just a guess
Neither triangle is really a triangle. You can actually see the slope angle more upward when it passes from red to dark green. It becomes less steep when it passes from dark green to red in the 2nd triangle. Just look back and forth at the 2 points on the grid where those 2 pieces meet to see the difference. Nice find Cat. :)
Partitions refer to the small (different color) pieces. And these are the same. But arranged differently they create a different result, a different slope. So even if the pieces are the same, stacked in a different order they create a different slope. The slope is not created only by the 2 pieces but also but how the pieces underneath them are ordered.. Shorter: Say you take out the orange and light green pieces. Then, you can arrange the 2 other pieces in an infinite way to create an infinite type of slopes because they only touch in one point..
Question: What causes the difference between the two figures? Answer: The difference comes from the slope, which is not a straight line and is bent inwards in figure #1 and outwards in figure #2.
Question: If the full figures were identical, would it be possible for one to have a missing part? Answer: No, it would not be possible.
Question: What is the ratio of the slope of the big triangle to the small triangle? Answer: The big triangle is 3/8 and the small is 2/5. | 677.169 | 1 |
Transformation Of Coordinates
posted on: 21 May, 2012 | updated on: 23 Aug, 2012
Before discussing the transformation of coordinates we will study what transformation is? Moving the shape to a different Position but the shape, size and area, line length and angle of that shape is same than this process is known as transformation, so when we move this shape then its coordinate also gets changed and this change is called transformation.
Now we will see coordinate transformation process.
We will study coordinate Transformations by Cartesian method.
A Cartesian coordinates system is totally defined by its origin and vectors along x-axis and y-axis.
Suppose we take first origin 'O' and unit vector OA1 and OA2 and we take a Point 'M' which has coordinates (r, s) relative to that coordinate system.
Then we have new coordinates with the origin O' and unit vectors OA1 and OA2 and point 'M' has coordinates (r', s') which is related to the new coordinate system.
Then the transformation formula between (r, s) and (r', s').
OM= OO' + O'M
=> M = O' + r' O'M1' + s' O'M2'
=> M = O' + r'(M1' - O') + s' (P2' - O')
=> With coordinates this becomes
(r, s) = (ro, so) + r'((a1, b1) - (ro, so)) + s'((a2, b2) - (ro, so))
=> r = so + (a1 - ro) r' + (a2 - ro) s'
s = so + (b1 - so) r' + (b2 - so) s'
With matrix notation this becomes
[r] [(a1 - ro) (a2 - ro)] [r'] [ro]
= +[s] [(b1 - so) (b2 - so)] [s'] [so]
Now we will see properties of transformation.
Some of the properties of transformation are given below.
1. Projective properties;
2. Affine properties;
3. Metric properties;
4. Affine Geometry;
5. Projective geometry.
Some theorem is also defined for the transformation which are:
Theorem of ceva for concurrent lines;
Theorem of Menelaus for Collinear Points;
Theorem of pappus – pascal;
And theorem of Desargues;
These all are the theorem of transformations.
Topics Covered in Transformation Of Coordinates
The term affine transformation is generally used in the Geometry. An affine transformation is a kind of transformation that preserves the straight lines by which we Mean to say that all those points which lie on the line initially now also lie on the line after the transformation has been done and also preserves the ratios of the distances that are between the poin...Read More
Question: What is the definition of transformation in the context of shapes? Answer: Transformation is the process of moving a shape to a different position while keeping its shape, size, area, line length, and angles the same. | 677.169 | 1 |
Definitions
GNU Webster's 1913
n.(Geom.) a straight line which bisects a system of parallel chords of a curve; called a principal axis, when cutting them at right angles, in which case it divides the curve into two symmetrical portions, as in the parabola, which has one such axis, the ellipse, which has two, or the circle, which has an infinite number. The two axes of the ellipse are the major axis and the minor axis, and the two axes of the hyperbola are the transverse axis and the conjugate axis.
Question: Which of the following is NOT an axis of an ellipse? A) Major axis B) Minor axis C) Long axis D) Short axis Answer: C) Long axis. | 677.169 | 1 |
It's perhaps easier to first consider |z-u^2|=|z-u|, i.e. the distance of z from u^2 is equal to the distance from you. This will be the perpendicular bisector between the two points representing u and U^2.
Since you actually want the distance to u^2 to be less, then the points satisfing this inequality will be the half plane closer to u^2.
Your first inequality is just points inside the circle radius 2, centred at the origin.
The first inequality is all complex numbers whose modulus is less than 2, i.e. all complex numbers that are within 2 of the origin on an argand diagram. So, this will be a circle (shaded in) about the origin, with radius 2.
The second inequality is saying that the distance in between z and u^2 is less than the distance between z and u - i.e. all points closer to u^2 than u. So, if you imagine a perpendicular bisector in between u and u^2, and then shade everything on the side that is closer to u^2.
Points that satisfy both inequalities will be the overlap between the two shaded regions - i.e. the part of the circle that is on the u^2 side of the perpendicular bisector.
(Original post by Azland)Yeah, and I would say normally when it starts talking about shading regions of an argand diagram, it's about looking at the inequalities and recognising the form, rather than doing fiddly algebra.
Question: What does the inequality |z-u^2|=|z-u| represent? Answer: It represents the perpendicular bisector between the two points representing u and u^2.
Question: Which of the following is NOT a requirement for a point to satisfy both inequalities: A) Being inside a circle with radius 2 centered at the origin B) Being closer to u^2 than u C) Having a modulus greater than 2 Answer: C) Having a modulus greater than 2 | 677.169 | 1 |
" The activity of the child has always been looked upon as an expression of his vitality. But his activity is really the work he performs in building up the man he is to become. It is the incarnation of the human spirit."
- Maria Montessori
Geometry has been defined as an awareness of the relationship between man and the objects in his environment. Montessori saw the study of geometry as practical, based on the physical reality of our world. Exploration of fundamental shapes, and their names and structure, provide the ground work for later studies ranging outside the confines of the classroom. As in most areas of study, the etymologies for geometry nomenclature both interest and add to the students' knowledge of concepts.
Reviewing Montessori sensorial geometry materials is the starting point for the curriculum at the Lower Elementary level. Children go on to study all aspects of lines: their types, parts and their relationships. This leads to the study of the different kinds of angles and their parts, and the introduction of congruency, similarity, and equivalence. A continuing study of triangles teaches their classification by kinds of sides and angles. Other plane figures are next, with classification of polygons, from 3-sided to dodecagons (12-sided figures) in both regular and irregular forms. Measurement of plane figures is then introduced.
Question: What is the last topic introduced in the Montessori geometry curriculum before measurement? Answer: Measurement of plane figures. | 677.169 | 1 |
Hey there guys, I'm learning processing.js, and I've come across a mathematical problem, which I can't seem to solve with my limited geometry and trigonometry knowledge or by help of Wikipedia.
I need to draw a rectangle. To draw this rectangle, I need to know the coordinate points of each corner. All I know is x and y for the midpoints of the top and bottom of the box, and the length of all four sides.
There is no guarantee on the orientation of the box.
Any help? This seems like it should be easy, but it is really stumping me.
8 Answers
If this quadrilateral is a rectangle (all four angles are 90 degrees), then it can be solved. (if it could be any quadrilateral, then it is not solvable)
if the points are (x1,y1), and (x2, y2), and if the two points are not perfectly vertical (x1 = x2) or horizontal (y1 = y2), then the slope of one edge of the rectangle is
m1 = (y2-y1) / (x2-x1)
and the slope of the other edge is:
m2 = - 1 / m1
If you know the lengths of the sides, and the midpoints of two opposite sides, then the corrner points are easily determined by adding dx, dy to the midpoints: (if L is length of the sides that the midpoints are on)
dx = Sqrt( L^2 / (1 + m2^2) ) / 2
and
dy = m2 * dx
NOTE: if the points are vertically or horizontally aligned, this technique will not work, although the obvious solution for those degenerative cases is much simpler.
With this method you'll need to make sure you add special cases for when x1 == x2 or y1 == y2 otherwise you'll end up dividing by zero. Calculating with vectors instead of slopes avoids this (although you still need to check that your two points aren't exactly the same). – DaveFeb 8 '10 at 2:27
If you know your quadrilateral is a rectangle, then you can use some simple vector maths to find the coordinates of the corners. The knowns are:
(x1,y1) - the coordinate of the midpoint on the top line
(x2,y2) - the coordinate of the midpoint on the bottom line
l1 - the length of the top and bottom lines
l2 - the length of the other two lines
First, we find the vector between the two known points. This vector is parallel to the side lines:
(vx, vy) = (x2 - x1, y2 - y1)
We need to normalize this vector (i.e. make it length 1) so we can use it later as a basis to find our coordinates.
vlen = sqrt(vx*vx + vy*vy)
(v1x, v1y) = (vx / vlen, vy / vlen)
Question: What is the method to find the coordinates of the corners of a rectangle using vector maths? Answer: First, find the vector between the two known points, normalize it, then use it as a basis to find the coordinates.
Question: What are the coordinates of the midpoints of the top and bottom of the rectangle? Answer: (x1, y1) and (x2, y2)
Question: What is the length of all four sides of the rectangle? Answer: The length is not specified in the text.
Question: What is the slope of one edge of the rectangle if the points are (x1, y1) and (x2, y2) and they are not perfectly vertical or horizontal? Answer: m1 = (y2-y1) / (x2-x1) | 677.169 | 1 |
Pages
Friday, April 6, 2012
VFC: Intro to Trigonometry
Here is a Virtual Filing Cabinet for an introduction to unit circle trigonometry. The goal of this unit is roughly to get students conversant in the sine, cosine and tangent functions. Following this unit in my class is a full unit on graphing the trigonometric functions.
What am I missing here? Point me to your favorite trigonometry resources in the comments.
[Last Updated: 4/5/2012]
The Hard Parts
Part of what makes this unit challenging for me is that students may or may not come in knowing the right triangle trigonometry definitions of sine and cosine. Some teachers like to start with the right triangle definitions and expand them to all real numbers. Because I know that I can't rely on my kids remembering the old definitions, my solution is typically to offer a fresh start with the circular definitions of the trig functions, and then later tie them back in to right triangle definitions.
What I like about trigonometry is that so much of it can be connected to one central model -- the unit circle. This is empowering. But kids had better know their unit circle stuff, and so kids really need a firm grasp of the unit circle by the end of the unit.
Kate and Riley both have good starting points for the beginning of this unit, but my understanding is that they start with the right triangle definitions of sin, cos, and tan. I don't like to start there for two reasons. First, because my students usually don't remember those definitions. Second, because I'd rather give them a clean foundation and connect it to the right triangle definitions later.
In general, I like the worksheets and implied approach of the eMathematics textbook. I use their worksheets on angles and rotation terminology after I've reviewed the equations of circles with my classes. You can find them at the link below, though everything from their Lesson #6 and on I consider part of my next unit, on graphing the trig functions.
I don't know where I first saw that the tangent function is the slope of a radius in the unit circle. I never learned it that way, but it makes so much sense to me and my kids than just defining tan directly in terms of sin and cos.
Here's a full course through problem sets. I haven't looked through it carefully yet, but it looks good:
Oh yeah, radians. And if you're in NY, arc length. I spend a day converting unit with my kids, and radians <---> degrees is an application of converting units. Sometimes I use this not-so-great resource that I made that just contains some goofy units of measurement to shake kids out of their comfort zone.
Question: What is the author's preferred approach to teaching sine and cosine? Answer: The author prefers to start with the circular definitions of the sine and cosine functions and then tie them back to the right triangle definitions later. | 677.169 | 1 |
Today we went back to triangles, specifically right triangles. First we reviewed the types of triangles and introduced this type.
We talked about 2 special types of right triangles, just for the sake of knowing them, because until we start solving right triangles with trigonometric functions, we won't do anything with them.
Then, we moved onto the Pythagorean theorem, which is a²+b²=c².
Finally, we learned the area of a triangle, which is 1/2 x base x height.
Procedures
In this lesson, there were three pictures of right triangles. One picture of a generic right triangle, one is of a 45-45-90 degree triangle, and the other is of a 30-60-90 degree triangle. Again, as stated above, we won't be using these until later, but I thought I would point out the special types.
The animation was a construction of a right triangle.
And finally, there was a picture of a right triangle with its hypotenuse missing so could use the Pythagorean theorem and find its area.
This time around, we took a step back and learned about the basics of geometry. Hopefully you know what a point, line, ray, line segment, and planes are. If you don't, here's a refresher.
Point has no dimensions, and two points form a line. A line plus a point for the next dimension creates a plane. A ray is a line with one endpoint, whereas a line segment has two endpoints. It was a pretty easy lesson.
Procedures
Once again, in this lesson there are several pictures for each new term introduced.
Then, the constriction for a plane, explaining why you need three points to form a plane. This animation really helped me, so I hope it helps you!
There wasn't much else, because this lesson was basically just definitions. It was all still important though. Onto the examples, then!
]]>
Wed, 06 Mar 2013 23:45:50 +0000iamwill99 make lessons make more sense, and so you can follow along like this website is a classroom, I have provided lesson plans for each lesson. Teachers can use this to help teach their students, and the people who visit this blog can use it to find more example. This is Daily Lesson Plan #1
This lesson's objective is for you to be able to classify triangles by their sides and angles. You will also learn the definition of obtuse, acute, and right angles and scalene, isosceles, and equilateral triangles. After you fully understand the difference between triangles, you will also learn to find angle measures in triangles.
Procedures
To help students, there were many things in that lesson to help them understand. There were pictures of each kind of triangle. It labeled congruent sides and angles.
In addition, there was an animation showing the construction of an equilateral triangle.
Afterwards, there was a picture of a triangle with it's angle measures, showing they all added up to 180.
Question: What is the sum of the angle measures in any triangle? Answer: The sum of the angle measures in any triangle is 180 degrees.
Question: Which of the following is NOT a dimension of a point? A) Length B) Width C) Height Answer: C) Height
Question: What is the difference between a ray and a line segment? Answer: A ray is a line with one endpoint, whereas a line segment has two endpoints.
Question: Which type of triangle does the animation in the second text show? Answer: The animation shows the construction of a right triangle. | 677.169 | 1 |
Formula for the Curvative of a Curve
Date: 10/10/2002 at 21:10:15
From: Adele Champlin
Subject: Dimensions
I have an assignment in English class to convince someone who believes
a wall has only one side that there is another side to it. I was
wondering if there is a formula or theorem I can use for this?
Date: 11/17/2002 at 20:26:48
From: Doctor Nitrogen
Subject: Re: Dimensions
Hello, Adele:
Suppose both of you are in a very large room with a very, very long
wall.
1. Imagine both of you traveling along the long length of the wall. If
the wall is perfectly flat, you might never be able to convince the
other party there is another side to the wall.
2. If the very long wall is curved away from both of you as you travel
alongside it inside the room, you could convince the party there must
be another side to the wall, since it is curved and curving in some
direction away from both of you and in some direction normal
(perpendicular) to the wall.
3. If the very long wall is curved inward toward the room and toward
both of you, you could convince the other party there must be another
side to the wall, since it is curved and curving in some direction
toward both of you and in some direction normal (perpendicular) to the
wall.
If you take a magic marker and draw a long curve on the curved wall
as both of you travel around the big room, you can calculate the
radius of curvature for that curve, and that perpendicular radius of
curvature would either point out from the big room (the wall curving
away from you both) or into the big room (the wall curving inward
toward you both).
The direction the radius of curvature would be along would be
perpendicular to the direction in which you both are traveling, and
perpendicular to the up or down direction of the room, so for a
curved wall you could hypothesize there lies another side to the
wall.
In Calculus the following formula computes such curvature for a
curve drawn on a curved surface:
Curvature K = y"/[1 + (y')^2]3/2.
y' is the derivative of y with respect to x, and y" is the second
derivative of y with respect to x.
I hope this helped answer the questions you had concerning your
mathematics problem. You are welcome to return to The Math
Forum/Doctor Math whenever you have any math-related questions.
- Doctor Nitrogen, The Math Forum
Question: Who provided the response to Adele's query? Answer: Doctor Nitrogen.
Question: What is the formula provided by Doctor Nitrogen to calculate the curvature of a curve drawn on a curved surface? Answer: Curvature K = y"/[1 + (y')^2]^(3/2). | 677.169 | 1 |
180°) whenever possible. To accomplish this, pre-survey reconnaissance is recommended. An oft-made mistake is to construct the traverse while collecting the observations. This
technique works in low-order surveys, but frequently results in poorly designed control traverses.
For long traverses, checks on the measured horizontal angles can be obtained by making periodic astronomical azimuth
observations. These should agree with the values computed from the direction of the starting line and the measured horizontal
angles. However, if a traverse extends an appreciable east-west direction, as illustrated in Figure 19-13, meridian convergence will
cause the two azimuths to disagree. In Figure 19-13, for example, azimuth FG obtained from direction AB and the measured
horizontal angles should equal astronomic azimuth FG + q, where q is the meridian convergence. A very good approximation for
meridian convergence between two points on a traverse is
(19-11)
where q" is meridian convergence, in seconds; d the east-west distance between the two points in meters; Re the mean radius of the
earth (6,371,000 m); f the mean latitude of the two points; and
r the number of seconds per radian (206,264.8"/rad). Because of meridian
convergence, forward and back azimuths of long east-west lines do not differ by exactly 180°, but rather by 180° ±
q. (A sketch of
the situation will clarify whether the sign should be plus or minus.) From Eq. (19-11) an east-west traverse of 1-mi length at latitude
of 30° produces a convergence angle of approximately 30". At a latitude of
45°, convergence is approximately 51"/mi east-west. These
calculations illustrate that the magnitude of convergence can be appreciable, and must be considered when astronomic observations
are made in connection with plane surveys that assume the y axis parallel throughout the survey area.
Procedures for precise traverse computation vary depending on whether a geodetic or a plane reference system is used. In either
case, it is necessary first to adjust angles and distances for observational errors. Closure conditions are enforced for (1) azimuths or
angles, (2) departures, (3) and latitudes. The most rigorous process, the least-squares method, should be used because it
simultaneously satisfies all three conditions and gives residuals having the highest probability.
Trilateration
Trilateration, a method for horizontal control surveys based exclusively on observed horizontal distances, has gained acceptance
because of electronic distance measuring capability. Both triangulation and traversing require time-consuming horizontal angle
measurement. Hence trilateration surveys often can be executed faster and produce equally acceptable accuracies.
The geometric figures used in trilateration, although not as standardized, are similar to those employed in triangulation. Stations
should be intervisible, and therefore placed on the highest peaks, perhaps with towers to elevate instruments and observers.
Question: What causes astronomic and computed azimuths to disagree in an east-west traverse? Answer: Meridian convergence causes astronomic and computed azimuths to disagree in an east-west traverse.
Question: What is the convergence angle for an east-west traverse of 1 mile at a latitude of 30°? Answer: The convergence angle for an east-west traverse of 1 mile at a latitude of 30° is approximately 30".
Question: What is the formula for calculating meridian convergence? Answer: The formula for calculating meridian convergence is q" = (d r) / (Re cos(f)), where q" is meridian convergence, d is the east-west distance, Re is the mean radius of the earth, f is the mean latitude, and r is the number of seconds per radian. | 677.169 | 1 |
I assume you're trying to find an angle between two points? I'm sure there are more efficient ways of doing this but I used some stuff I learned from Calculus to develop a Get_Angle() proc which uses inverse cosine and some vector math to determine the angle between point (x0,y0) and point (x1,y1).
atan2() (or arctan2() or arctangent2() or whatever you want to call it) is different from just the arctangent function. atan() takes just one argument, which is the ratio y/x. However, the issue here is that the domain and range is then limited to half of the unit circle: y/x = -y/-x, and -y/x = y/-x. The range of the function, therefore, is only -pi/2 to pi/2. The atan2() function solves this by taking two arguments - y and x - and then using that to give a full range.
Granted, it's only a small difference in terms of lines of code (as shown by Lummox's example) but it really makes things neater and less prone to error.
Also that's a really poor explanation in retrospect. Anybody who's actually a mathematician able to say things properly?
Question: Which function is better for calculating the angle between two points and why? Answer: atan2() is better for calculating the angle between two points because it provides a full range of results, making it less prone to errors. | 677.169 | 1 |
Chapter 6, Major Exercise 6
Above is an interactive version of figure 6.32. You can drag any of the red
points around using the mouse.
Recall, you are to show that if lines l and m (green)
are parallel but not limiting parallel,
we construct AA' and BB' (purple) to be perpendicular
to m. Then, assuming that AA' is longer than BB', we
can create ray EF (dark green)
so that EA' is congruent to BB', and so that
angles A'EF and B'BG are congruent.
You are to show that EF intersects line l at a point H.
In order to do this exercise, you may use Major exercises 2, 3, 4, and 5.
Indeed, you almost certainly must use them.
This exercise is needed to show that Hilbert's construction of the mutual
perpendicular to lines l and m exists. Below is that
construction, which works as follows. (see page 263 of the text).
First, we create ray EF as above, and from the exercise, we
know it intersects AB at a point H. Now we find the
unique point K on ray AB so that EH is congruent
to BK. Drop perpendiculars HH' and KK',
and observe that they are congruent (since Lambert
quadrilaterals A'H'HE and B'K'KB are congruent).
Since we have two congruent sides perpendicular to the base,
quadrilateral H'K'KH is a Saccheri quadrilateral.
We showed earlier that the line joining the midpoint M
of the summit HK to the midpoint M' of the base H'K'
will be perpendicular to both lines.
Question: Are lines l and m parallel? Answer: Yes, lines l and m are parallel but not limiting parallel.
Question: What is the length relationship between AA' and BB'? Answer: AA' is longer than BB'. | 677.169 | 1 |
Lets say there live two dudes on an island (or two gals, or one dude one gal, idc really!). AnywayQuestion is, how to divide the islands surface area into two equal pieces with just this equipment (the radius of the island is known and it is r).
calculo.png (3.01 KiB) Viewed 2757 times
image lifted from brenok's post -jr
So to clarify, A should equal B / the green surface should equal the orange surface. The thing we would like to find is a relation between r and q...Yeah I guess that is a solution... but I was aiming for a relation between r and q. The story about the island is just for... to make it less boring .
your initial work is right, but gets nowhere because you have 2 equations and 3 unknowns. (well actually the unknown q is what you are looking for, and it wasn't even included in the equations) anyways, all you need to solve it is an equation for the area of B in terms of r and q, something I don't want to do at 11pm, but I'll work on later.
Why put off till today what you could just as easily get done tomorrow?
For an analytic solution, write the equations of the circles in terms of q and r and find the points of intersection. Draw the line between those points (for reference, not for construction) and find the area of B on either side of the line. Set the sum of those areas equal to pi*r2/2. That's what my first approach would be, anyway.
I'm not going to worry about how they might come up with the correct distances right now, though. Or what they should do if the island is lumpy.
Let's call the centers of the two circles O and P, and the points where they intersect Q and R. Then the area of the region labeled B is easily given in terms of the angles s=<QOR and t=<QPR by r2(s-sin s)+q2(t-sin t), since we can think of it as a union of two wedges (the positive terms) minus the corresponding triangles (the negative terms).
Now s is twice one angle of a triangle with sides r, r, and q, so from the law of cosines we have q2=r2+r2-2r*r*cos (s/2), or s=2*cos-1 (1-q2/2r2).
Now that we know what s is, t is just pi-s/2, so t=pi-cos-1 (1-q2/2r2).
Substituting s and t and simplifying, we get the following for the area of B in terms of q and r: r2(2*cos-1 (1-q2/2r2)-sin 2*cos-1 (1-q2/2r2))+q2(pi-cos-1 (1-q2/2r2)-q/r * sqrt (1-q2/4r2))
Question: What is the radius of the island? Answer: The radius of the island is r.
Question: What are the centers of the two circles called? Answer: The centers of the two circles are called O and P.
Question: What are the two colors used to represent the two equal pieces of the island's surface area? Answer: The two colors used are green and orange.
Question: What is the user trying to find a relation between? Answer: The user is trying to find a relation between the radius of the island (r) and the distance from the center of the island to the point where the two circles intersect (q). | 677.169 | 1 |
Proposition 17
If two straight lines are cut by parallel planes, then they are cut in the same ratios.
Let the two straight lines AB and CD be cut by the parallel planes GH, KL, and MN at the points A, E, and B, and at the points C, F, and D, respectively.
I say that the straight line AE is to EB as CF is to FD.
Join AC, BD, and AD. Let AD meet the plane KL at the point O. Join EO and FO.
Now, since the two parallel planes KL and MN are cut by the plane EBDO, therefore their intersections EO and BD are parallel. For the same reason, since the two parallel planes GH and KL are cut by the plane AOFC, their intersections AC and OF are parallel.
And, since the straight line EO is parallel to BC, one of the sides of the triangle ABD, therefore proportionally AE is to EB as AO is to OD. Again, since the straight line FO is parallel to CA, one of the sides of the triangle ADC, therefore proportionally AO is to OD as CF is to FD.
Question: What is the ratio of AO to OD? Answer: The ratio of AO to OD is the same as the ratio of AE to EB, which is also given as CF to FD. | 677.169 | 1 |
GOING AROUND IN CIRCLES
The art element of PATTERN is richly embedded in line, texture, color and shape repetitions of unending configurations. I have always enjoyed the illusions of parabolic line in both linear line and 3-dimensional form . These sections deal exclusively with simple circles and some parabolic connections which create innumerable ways to extract and define patterns within them.
We began by marking circle "points" along its circumferance. We used both 36 and 72 points (finer lines) for our constructions. Since we did not have protractors, we used a compass and our best "gestimates" to divide the spaces as equally as possible. The diagram below shows our method.
Our practices were with 3" circles, 4 of them arranged on a 12"X18" drawing paper. The first step, marking off the 6 divisions, is defined with the original compass setting. From there, students used their best judgments (optical spacing) to subdivide the areas as equally as they could.
Now lets move on to simple constructions with Practice 1 and grow from there.
Question: What shapes and forms does the text mention as part of this pattern? Answer: Line, texture, color, and shape | 677.169 | 1 |
A line through three-dimensional space between points of interest on a spherical Earth is the chord of the great circle between the points. The central angle between the two points can be determined from the chord length. The great circle distance is proportional to the central angle.
The great circle chord length, , may be calculated as follows for the corresponding unit sphere, by means of Cartesian subtraction:[4]
The shape of the Earth closely resembles a flattened sphere (a spheroid) with equatorial radius of 6,378.137 km; distance from the center of the spheroid to each pole is 6356.752 km. When calculating the length of a short north-south line at the equator, the sphere that best approximates that part of the spheroid has a radius of (which equals the meridian's semi-latus rectum), or 6,335.439 km, while the spheroid at the poles is best approximated by a sphere of radius , or 6,399.594 km, a 1% difference. So as long as we're assuming a spherical Earth, any single formula for distance on the Earth is only guaranteed correct within 0.5% (though we can do better if our formula is only intended to apply to a limited area). There are different approximations used as a radius for a non-spherical body such as Earth:
the average geocentric radius (≈(a+b)/2) or (ab).5;
the equatorial radius (a);
the volume and total surface area radius (≈(2a+b)/3 or (a2b)1/3);[5][6];
comes out to be 25.958 degrees, or 0.45306 radians, and the great-circle distance is the assumed radius times that angle:
So assuming a spherical earth, distance between LAX and BNA is about 2887 km or 1794 statute miles (× 0.621371) or 1559 nautical miles (× 0.539957). Geodesic distance between the given coordinates on the GRS 80/WGS 84 spheroid is 2892.777 km.
Question: What is the formula to calculate the great circle chord length? Answer: The formula is derived from Cartesian subtraction.
Question: What is the great-circle distance between LAX and BNA assuming a spherical Earth? Answer: About 2887 km | 677.169 | 1 |
investigation
Grade 8: Investigation of geometry
Study Unit 1: Angles: the basics
Introducing the concept of the angle
The diagram below shows the angle
ABˆCABˆCA hat{B}C . AB and BC are called the arms of the angle.
ABˆCABˆCA hat{B}C is the angle. B is the vertex, which is the point where the angle is.
Figure 1
Naming angles
In the diagram below the angle at vertex B is called
BˆBˆhat Bor
ABˆCABˆCA hat{B} C or
CBˆACBˆAC hat{B} A.
If we use a capital letter to name an angle we must put a cap on it to distinguish it from a point.
Figure 2
Note that if there are two or more angles that share the same vertex, as in the diagram below, we cannot simply speak of
BˆBˆhat{B} . We must distinguish between
ABˆCABˆCA hat{B} C,
MBˆCMBˆCM hat{B} C and
ABˆMABˆMA hat{B} M.
Figure 3
We may also put a small letter near the vertex and name the angle using the small letter. We do not put caps on small letters because they are not referring to points. Also note that a small letter can either mean the angle or the size of the angle. So in the diagram below we have:
ABˆMABˆMA hat{B}M or x.
MBˆCMBˆCM hat{B}C or y.
Also:
ABˆC=x+yABˆC=x+yA hat{B}C = x + y.
Figure 4
Measuring angles (in degrees)
The ancient Babylonians believed that there were only 360 days in a year (the amount of time for the earth to round the sun) and therefore divided the circle up into 360 equal parts, which they called degrees, denoted by the symbol °.
Figure 5
It follows that half a circle (a semi-circle) may be divided into 180° and a quarter of a circle may be divided into 90°.
Figure 6
Angles on a straight line
If half a revolution is 180°, then angles on a straight line must add up to 180°. In the diagram below
x+y+z=180°x+y+z=180°x + y + z = 180^{circ}
Key concept: Angles on a straight line add up to 180°.
Test your knowledge 1:
1) Consider the diagram below. If x = 50°, what is the value of y?
Figure 7
2) Consider the diagram below. If x = 50° and y = 100°, what is the value of z?
Figure 8
Study unit 2: Polygons
Question: What are the two parts of an angle called? Answer: The arms of the angle.
Question: What does it mean when a small letter is used to name an angle? Answer: It can refer to either the angle itself or the size of the angle.
Question: In the diagram with angles at vertex B, what are the different ways to name the angle at B? Answer: ABˆCA hat{B}C, CBˆAC hat{B}A, or using a small letter like x or y.
Question: In the diagram (Figure 8), if x = 50° and y = 100°, what is the value of z? Answer: z = 180° - (50° + 100°) = 30° | 677.169 | 1 |
Inscribed Angles
Popular Tutorials in Inscribed Angles
When you're given the measurement of the intercepted arc, you can find the measure of the inscribed angle with a few short steps! Follow along with this tutorial to learn how to find an inscribed angle when you know the intercepted arc!
Just about everything in math has a name! Did you know that a fraction of the area of a circle is known as a sector? This tutorial introduces you to the term sector and gives you examples of sectors. Take a look
Question: What does the phrase "Just about everything in math has a name!" imply? Answer: It implies that mathematical concepts often have specific names or terms associated with them | 677.169 | 1 |
Other notation, such as for rays, arcs, etc, is usually defined
in the text. Unfortunately, as old as geometry is, the notation does not seem, even today, to be
entirely standardized. So pay particular attention to how your book does things, so you can follow
along, but don't be surprised if your instructor does something else.
For trigonometric functions, powers are indicated directly on the
function names. For instance, "the square of the sine of beta" is written as sin2(β),
and this notation means [sin(β)]2. Multipliers on the variable go inside the argument: sin(2β) does not mean the same thing
as sin2(β). Some texts omit the function-notation parentheses, writing
sin2β and sin2β, which can lead to confusion, especially when these expressions are hand-written. Try
to remember to use the parentheses, so you can be clear in your own work. Also, try not to get
in the lazy habit of omitting the arguments of the functions, writing things like sin2 + cos2 = 1,
as this will lead to severe problems when the argument is not something simple like just
"x".
The final "convention" I'll mention is really an assumption
that you should remember not to make: Do not assume that pictures are "to scale"!
Question: What assumption should you not make when looking at pictures in geometry? Answer: You should not assume that pictures are "to scale". | 677.169 | 1 |
Just me and my thoughts on life.
Completely amazed..
I recently came across a revelation when learning trig identities sin and cos. Here it is:
Trig Reference Angle Cheat Hand
Observe...
Flip down the finger that corresponds to the angle whose sine and cosine you need.
The number of fingers to the left gives you the sine, and the number of fingers to the right gives you the cosine.
So if you flip down your index finger which corresponds to 30 degrees...
there is one finger to the left.
and there are three fingers to the right.
Try it for the fingers that correspond to the other reference angles. For example, if you flip down your pinky, there are four fingers to the left (sin(90) = radical 4 over 2 = 1) and zero fingers to the right (cos 90 = radical 0 over 2 = 0.) It works!
It's just another way of organizing the cofunction behavior of sine and cosine to remember the values of five reference angles, but adults and kids both flip out when I show them. Kids especially feel that they "don't have to memorize" if they know this method.
Question: What is the sine value of the 30-degree angle according to this method? Answer: The sine value is 1 (since there is one finger to the left). | 677.169 | 1 |
Scalene Triangle
posted on: 21 Apr, 2012 | updated on: 11 Sep, 2012
The figures, which have the same starting and the ending points are called the closed curves. Polygons are also the examples of the closed curves, which are formed by line segments. These are the figures formed by joining 3 or more line segments and so they form the closed figures.
We say that the polygons with three line segments are Triangles. And this way more different names are given to the polygons formed by joining different line segments. Here we will learn about scalene triangle in this unit. In Geometry we say that the triangles can be classified based on the length of their line segments. We observe that if line segments of the triangle are same, then we call it the Equilateral Triangle. On another hand we say that if the two sides of the triangles are equal, then the triangles are called Isosceles Triangle. Here we are learning about scalene triangles. The triangles which have the different length of the line segments, then the triangle so formed are called, scalene triangles. In scalene triangles, we say that the length measure of the triangles is different, so the length measure of the angles of the triangles are also of different measure. As the different sides of the triangle are not of same length in the scalene triangle are not same, but the sum of the angles sum of the triangle always remains same.
This sum of angles of the triangle is 180 degree. Thus we say that the property of sum of angles is equal is called angle sum property of the triangle. In scalene triangle, we say that if the two angles of the triangle are known, then we can find the third angle of the triangle by simply subtracting the sum of two angles of the triangle by 180 degrees.
Question: Which type of triangle has all sides of equal length? Answer: Equilateral Triangle. | 677.169 | 1 |
Question 612826: A disc jockey must play 14 commercial spots during 1 hour of a radio show. Each commercial is either 30 seconds or 60 seconds long. If the total commercial time during 1 hour is 11 min., how many 30-second commercials were played that hour? How many 60-second commercials? Click here to see answer by stanbon(57342)
Question 613324: Your are given a triangle ABC with a line drawn in the middle of the diagram that connects the midpoints of the two sides of the triangle. In the diagram, points D and E are midpoints of sides AB and AC, respectively. If the measure of angle B is 68 degrees, than what is the measure of angle ADE.
I tried to use the Midline Th., but the angles did not work out for me. Please help. Thank you. Click here to see answer by solver91311(16885)
Question 614421: Jane has cut out a triangle using cardboard.
The base of the triangle is from F to H and measures 1.1m in lenght.
The left side of the triange is F to G and measures 0.7 m in length.
The right side of the triangle is G to H and measures 0.9 m in length.
Which would be the largest angle? Could you please explain?
A. angle F
B. angle G
C. angle H
D not enough information Click here to see answer by Alan3354(30993 rfer(12657 [email protected](15651)
Question 615608: Twenty more than five times the measure of an angle is 60 degrees. What is the measure of the angle?
Question 616018: Can someone PLEASE help me with this problem? THANK YOU!
A 212 ft tower is located on the side of a mountain that is inclined 19° to the horizontal. A guy wire is to be attached to the top of the tower and anchored at a point 71 ft downhill from the base of the tower. Find the shortest length of wire needed.
Length of guy-wire = ? ft Click here to see answer by dragonwalker(72)
Question 616960: 3rd request. The angle of elevation from the tip of the shadow of a 12-ft flag pole to the top of the pole is 60 degrees. How far is it from the tip of the shadow to the top of the pole? Round your answer please.
12 ft pole with a 90 dgree angle. measures 60 degrees x should be 30. Not sure if correct. Please assist me. Thank you. Click here to see answer by vleith(2825)
Question: If the length of the guy wire were 100 ft, would it be sufficient? Answer: No, it would be too short.
Question: How many 30-second commercials were played in the hour? Answer: 7
Question: How far is it from the tip of the shadow to the top of the 12-ft flag pole with a 60-degree angle of elevation? Answer: 10 ft
Question: What is the measure of the angle if twenty more than five times the measure is 60 degrees? Answer: 10 degrees | 677.169 | 1 |
Question 617616: Use the table to identify the relationship of the SECOND sentence to the FIRST sentence.
1. A triangle is a right triangle if and only if it has a right angle.
2. A triangle has a right angle if and only if it is a right triangle.
CO = contrapositive
BI = biconditional
LE = logical equivalent
CV = converse
IN = inverse Click here to see answer by stanbon(57342 jim_thompson5910(28595 Alan3354(30993)
Question: What is the logical equivalent of the first sentence? Answer: The first sentence is its own logical equivalent. | 677.169 | 1 |
4. Give examples of how the same absolute error can be problematic in one situation but not in another; e.g., compare "accurate to the nearest foot" when measuring the height of a person versus when measuring the height of a mountain.
9. Show and describe the results of combinations of translations, reflections and rotations (compositions); e.g., perform compositions and specify the result of a composition as the outcome of a single motion, when applicable.
2. Describe and compare characteristics of the following families of functions: square root, cubic, absolute value and basic trigonometric functions; e.g., general shape, possible number of roots, domain and range.
2. Represent and analyze bivariate data using appropriate graphical displays (scatterplots, parallel box-and-whisker plots, histograms with more than one set of data, tables, charts, spreadsheets) with and without technology.
8. Differentiate and explain the relationship between the probability of an event and the odds of an event, and compute one given the other
Question: What is the general shape of a square root function? Answer: It is a rightward-opening parabola with a vertex at the origin | 677.169 | 1 |
axiom A: people are born by other people lemma 1.1 "for every person x there exists y such that y is x's parent" proof suppose x is a person that has no parents. by axiom A people are born by other people. contradiction.
lemma1.2 "if you go back in time sufficiently far and kill someones parent, he will cease to exist"
Guessing you've already learned what you need to know, but the only thing you need to remember is SOH CAH TOA. That is; Sin = O/H, Cos = A/H, and Tan = O/A, whereas O is the opposite side of the triangle to the given angle, A is the closest side that's not the hypotenus, and H is the hypotenus.
Like Priscilla said; cos(t) = h/a, you can invert the formula to find the angle; cos^-1(h/a) = t, ex. tan^-1(20/20) = 45 degrees, which makes sense since both the opposite and adjacent are the same length, so you get a right-angle triangle.
To find individual sides of the triangle, you can multiply the function by the denominator on the other side, ex. If I want to find the opposite of the aforementioned right angle triangle, knowing only the angle and adjacent, I go from... tan(45) = o/a tan(45)a = o
>>10016 I just rolled the dice on random thread and appeared here. I am sorry OP I cannot answer your question as I do not take or have ever taken calc. I am going to guess and say because that is what Berry said?...
Math, what is your favorite calculator and why? what do you use it for, what features make it stand apart from others? what don't you like about it?
my preference is a casio classpad 330. several years back in a calc2 class we were given them by a casio rep to test and keep in exchange for our feedback.
split screening in graphs/tables etc is very nice, also has auto window sizing methods and shit. im a very visual person, it makes seeing things quick and easy.
my absolute favorite part is the 2d representation. you can literally build things as they look on paper, quick and easy. it makes it super easy to check your work or evaluate complicated things without having to fight with a billion parentheses.
easily evaluates integrals both definite and indefinite (not that im at all dependent but it could honestly be defined as a cheating machine), including the nasty tables that i don't care to read though. so much easier to just type it it and get an answer.
cons: bit of a learning curve to figure functionality but its a very powerful device for everything ive seen. i don't know how it compares for much more complex mathematics
I was the kid who brought absolutely nothing to class, and obviously do not have a calculator. Ended up using my pure braib skills. Nowadays my ex schoolmates often get stuned by how a lazy person like me could handle crazy calculations.
Question: Is the statement "People are born by other people" a true statement according to the text? Answer: Yes, as stated in Axiom A.
Question: What is the relationship between 'x' and 'y' in Lemma 1.1? Answer: 'y' is 'x's parent.
Question: What does SOH CAH TOA stand for? Answer: SOH CAH TOA stands for Sin = O/H, Cos = A/H, and Tan = O/A.
Question: What is one feature the user does not like about their favorite calculator? Answer: The user mentioned a bit of a learning curve to figure out the functionality of the calculator. | 677.169 | 1 |
2008-Apr-27, 10:14 PM
The line of sight distance ( is approx sqrt(13h) kilometers, where h is in meters. 26 feet is about 9 meters, so 13*9 is about 121, and the square root is 11 kilometers. The radius of the Earth is about 6400 kilometers, so the tangent of the angle is 11/6400, or 0.00171875. Since that is approx. the same as theta, we can find the sine of that, which is 0.001718749. Multiply the difference 0.000000001 by the radius of the earth and I get half a centimeter. Hmmm, I should have got one centimeter, about half an inch.
It's more than 26 feet??
Delvo
2008-Apr-27, 11:07 PM
It's not a matter of choosing between mathematical symbols and language. Mathematical symbols ARE a part of the language. It might be a specialized part that only some people use, but other examples of that aren't separated from the language and treated as not a part of it; they're just called "jargon" or something like that.
grant hutchison
2008-Apr-27, 11:13 PM
It's more than 26 feet??I iterate it out to ~10.5m.
I'm iterating because I can't be bothered attempting to solve:
sqrt(2rh+h2) - r.arccos(r/[r+h]) = 1/2"
where r is the equatorial radius of the Earth (6378136m) and h is the height to be found.
(On the left of the equality is the difference between the tangent length of the "pinched up" cord and the length of the arc below it on the surface of the Earth, which should amount to half the extra length added to the cord.)
Grant Hutchison
mugaliens
2008-Apr-28, 12:46 AM
I'm all agrins, as they say in the South, from the responses to my OP. So I'll try to address them one on one.
mugaliens
2008-Apr-28, 12:47 AM
DatGUm it Mugaliens, just give us a Link to Wikipedia!!!
Eta: Ok, I agree with the OP:p on this one.
Thanks, Neverfly.
mugaliens
2008-Apr-28, 12:48 AMWell said, Henrik! A fine condensation, to be sure!
mugaliens
2008-Apr-28, 12:49 AMThank you, geonuc.
mugaliens
Question: What is the tangent of the angle in the given scenario? Answer: 0.00171875
Question: Is the line of sight distance more than 26 feet? Answer: Yes | 677.169 | 1 |
Scalene Triangle
posted on: 21 Apr, 2012 | updated on: 11 Sep, 2012
The figures, which have the same starting and the ending points are called the closed curves. Polygons are also the examples of the closed curves, which are formed by line segments. These are the figures formed by joining 3 or more line segments and so they form the closed figures.
We say that the polygons with three line segments are Triangles. And this way more different names are given to the polygons formed by joining different line segments. Here we will learn about scalene triangle in this unit. In Geometry we say that the triangles can be classified based on the length of their line segments. We observe that if line segments of the triangle are same, then we call it the Equilateral Triangle. On another hand we say that if the two sides of the triangles are equal, then the triangles are called Isosceles Triangle. Here we are learning about scalene triangles. The triangles which have the different length of the line segments, then the triangle so formed are called, scalene triangles. In scalene triangles, we say that the length measure of the triangles is different, so the length measure of the angles of the triangles are also of different measure. As the different sides of the triangle are not of same length in the scalene triangle are not same, but the sum of the angles sum of the triangle always remains same.
This sum of angles of the triangle is 180 degree. Thus we say that the property of sum of angles is equal is called angle sum property of the triangle. In scalene triangle, we say that if the two angles of the triangle are known, then we can find the third angle of the triangle by simply subtracting the sum of two angles of the triangle by 180 degrees.
Question: If two angles of a scalene triangle are known, what is the formula to find the third angle? Answer: Subtract the sum of the two known angles from 180 degrees.
Question: Which property of a triangle is always constant, regardless of its shape? Answer: The sum of its angles. | 677.169 | 1 |
Solutions 1. Suppose there are m women. Then the last woman knows 15+m men, so 15+2m = 47, so m = 16. Hence there are 31 men and 16 women.
2. Answer:m ≤ -3
For real roots we must have (m+3)2 ≥ 4m+12 or (m-1)(m+3) ≥ 0, so m ≥ 1 or m ≤ -3. If m ≥ 1, then -(2m+6) ≤ -8, so at least one of the roots is < -2. So we must have m ≤ -3.
The roots are -(m+3) ±√(m2+2m-3). Now -(m+3) ≥ 0, so -(m+3) + √(m2+2m-3) ≥ 0 > -2. So we need -(m+3) - √(m2+2m-3) > -2, or √(m2+2m-3) < -m-1 = √(m2+2m+1), which is always true.
3. ABCD is a tetrahedron. A' is the foot of the perpendicular from A to the opposite face, and B' is the foot of the perpendicular from B to the opposite face. Show that AA' and BB' intersect iff AB is perpendicular to CD. Do they intersect if AC = AD = BC = BD? 4. The tetrahedron ABCD has BCD equilateral and AB = AC = AD. The height is h and the angle between ABC and BCD is α. The point X is taken on AB such that the plane XCD is perpendicular to AB. Find the volume of the tetrahedron XBCD. 5. Solve the equation sin6x + cos6x = 1/4.
Solutions
1. A straightforward, if inelegant, approach is to multiply out and expand everything. All terms cancel except four and we are left with 2abcd ≤ a2d2 + b2c2, which is obviously true since (ad - bc)2 ≥ 0.
3. Let the ray AB' meet CD at X and the ray BA' meet CD at Y. If AB' and A'B intersect, then X = Y. Let L be the line through A' parallel to CD. Then L is perpendicular to AA'. Hence CD is perpendicular to AA'. Similarly, let L' be the line through B' parallel to CD. Then L' is perpendicular to BB', and hence CD is perpendicular to BB'. So CD is perpendicular to two non-parallel lines in the plane ABX. Hence it is perpendicular to all lines in the plane ABX and, in particular, to AB.
Question: How many women are there in the first scenario? Answer: 16 | 677.169 | 1 |
Assignment Method. Assigns values to this plane in such a way, that the plane will have a unit vector normal in the same direction as the first argument, and in such a way that the second argument lies in the plane represented by this plane equation.
Retrieves the shortest distance with sign from the specified point to this plane.
If the point lies on the side of the plane where the normal is pointing the sign would be positive. This side of the plane is known as the frontside. If the point was on the backside the sign would have been negative. The final possibility is that the point lies in the plane. In this case the reutrn value is zero.
Question: What is the primary function of the "Assignment Method" described? Answer: To assign values to a plane such that it has a unit vector normal in the same direction as the first argument and the second argument lies in the plane represented by the equation. | 677.169 | 1 |
Saturday, December 24, 2011
"I think I discovered a theorem, Mr. Karafiol!"
This past week, one of my students stopped me on my first pass around the class with the exciting words "I think I discovered a theorem! In a right triangle, the angles the median makes with the hypotenuse are twice the other angles!" We had discussed the median to the hypotenuse in the previous class, but I'd never thought about the angles before. So I started doing algebra in my head, confirming the relationship he was describing:
2 comments:
I tried to confirm the relationship without looking at the second diagram, and I came up with another way to see it - if you draw the bisector to either of the angles created by the median line, you get a triangle that is similar to the original triangle, which proves it as well. Neat!
Question: What type of triangle was the student referring to in the theorem? Answer: A right triangle | 677.169 | 1 |
The Golden Rectangle and the Golden Ratio
This
diagram shows a golden rectangle (roughly). I have divided the rectangle into a
square and a smaller rectangle. In a golden rectangle, the smaller rectangle is
the same shape as the larger rectangle, in other words, their sides are
proportional. In further words, the two rectangles are similar. This can be
used as the definition of a golden rectangle. The proportions give us:
a/b = (a+b)/a
This fraction, (a+b)/a, is called the golden ratio (or golden section or
golden mean).
Above I have defined the golden rectangle, and then said what the
golden ratio is, in terms of the rectangle. Alternatively, I could have defined
the golden ratio, using the above equation. And then a golden rectangle becomes
any rectangle that exhibits this ratio.
From our equation, we see that the ratio a/b=1/2+sqr(5)/2 (where
sqr() is the square root function), which in turn is about 1.61803398875 . . .
The symbol often used for the golden ratio is ø (phi). Sometimes,
1/ø (which is -1/2+sqr(5)/2 or 0.61803398875 . . .) is called the
golden ratio. Also, other mathematical quantities are called phi. The golden
ratio is also called tau. Some people call the bigger one (1.61803398875 . . .)
Phi (an uppercase phi) and the smaller one (0.61803398875 . . .) phi. By the
way, a more accurage value is 1.6180339887498948482045868343656 . . .
Supposedly, Pythagoras discovered this ratio. And the ancient Greeks
incorporated it into their art and architecture. Apparently, many ancient
buildings (including the Parthenon) use golden rectangles. It was thought to be
the most pleasing of all rectangles. It was not too thick, not too thin, but
just right (Baby Bear rectangles).
Because of this, sheets of paper and blank canvases are often somewhat
close to being golden rectangles. 8.5x11 is not particularly close to a golden
rectangle, by the way.
The golden ratio is seen in some surprising areas of mathematics. The
ratio of consecutive Fibonacci numbers (1, 1, 2, 3, 5, 8, 13 . . ., each number
being the sum of the previous two numbers) approaches the golden ratio, as the
sequence gets infinitely long. The sequence is sometimes defined as starting at
0, 1, 1, 2, 3 . . . Zero is the zeroth element of the sequence. See
Fibonacci Numbers.
Addendum #1:
Question: Who is credited with discovering the golden ratio? Answer: Pythagoras | 677.169 | 1 |
So, delta rho means that you have two concentric spheres, and you are looking at a very thin shell in between them. And then, you would be looking at a piece of that spherical shell corresponding to small values of phi and theta. So, because I am stretching the limits of my ability to draw on the board, here's a picture. I'm going to try to reproduce on the board, but so let's start by looking just at what happens on the sphere of radius a, and let's try to figure out the surface area elements on the sphere in terms of phi and theta. And then, we'll add the rho direction. And then we'll add the rho direction. OK, so --
So, let me say, let's start by understanding surface area on a sphere of radius a. So, that means we'll be looking at a little piece of the sphere corresponding to angles delta phi and in that direction here delta theta. OK, so when you draw a map of the world on a globe, that's exactly what the grid lines form for you. So, what's the area of this guy? Well, of course, all the sides are curvy. They are all on the sphere. None of them are straight. But still, if it's small enough and it looks like a rectangle, so let's just try to figure out, what are the sides of your rectangle?
OK, so, let's see, well, I think I need to draw a bigger picture of this guy. OK, so this guy, so that's a piece of what's called a parallel in geography. That's a circle that goes east-west. So now, this parallel as a circle of radius, well, the radius is less than a because if your vertical is to the North Pole, it will be actually much smaller. So, that's why when you say you're going around the world it depends on whether you do it at the equator or the North Pole. It's much easier at the North Pole. So, anyway, this is a piece of a circle of radius, well, the radius is what I would call r because that's the distance from the z axis. OK, that's actually pretty hard to see now.
So if you can see it better on this one, then so this guy here, this length is r. And, r is just rho, well, what was a times sine phi. Remember, we have this angle phi in here. I should use some color. It's getting very cluttered. So, we have this phi, and so r is going to be rho sine phi. That rho is a. So, let me just put a sine phi. OK, and the corresponding angle is going to be measured by theta. So, the length of this is going to be a sine phi delta theta.
Question: What is the radius of the circle (parallel) in terms of 'rho' and 'phi'? Answer: The radius is 'rho * sin(phi)'. | 677.169 | 1 |
G2.2 Relationships Between Two-dimensional and Three-dimensional
Representations
G2.2.1 Identify or sketch a possible three- When the teacher is demonstrating on the
dimensional figure, given two-dimensional mannequin the students will then copy the
views (e.g., nets, multiple views). Create a demonstration on their mannequin. The learner will
two-dimensional representation of a three- learn this concept during a demonstration by the
dimensional figure. teacher where the teacher will demonstrate on a
mannequin and the student will copy the concept on
C:\Docstoc\Working\pdf\321c15b9-8ed6-49ad-b9a9-dc5eb29b22c3.doc 12
08/08/2011
a paper head form. Wiggery with the measuring tape.
On a head form sheet. When the teacher is
demonstrating on the mannequin, the students will
copy the demonstrations on their mannequins.
G3 STANDARDS CTE APPLICATION and PRACTICES
TRANSFORMATIONS OF FIGURES IN THE PLANE
G3.1 Distance-preserving Transformations: Isometrics
G3.1.1 Define reflection, rotation, translation, and While practicing hairstyling and haircutting. The
glide reflection and find the image of a learner will learn this concept while practicing
figure under a given isometric. hairstyling and haircutting. Making a floor plan all the
rooms are 10 by 13--but it is the exact same shape
but it is turned around facing the opposite direction.
While practicing hairstyling and haircutting.
G3.1.2 Given two figures that are images of each Put zoom times one.
other under an isometric find the isometric
and describe it completely.
G3.2 Shape-preserving Transformations: Isometrics
G3.2.1 Know the definition of dilation and find the The students will take hair samples and look at them
image of a figure under a given dilation. through a microscope and see the hair cuticle. The
shaft and the bulb of the hair. The learner will use
this concept when we are analyzing the hair strand
under the microscope. The students will take hair
samples and look at them through a microscope and
see the cuticle of the hair strand and hair bulb.
Understand the Concept of Congruence and Basic Transformations
G.GS.06.02 Understand that for polygons, congruence When looking at facial shapes to match to a specific
means corresponding sides and angles hair cut or style. In haircutting, all sides of the clients
have equal measures. head should match evenly. (4 inches in length at top,
sides and back). The learner would use this concept,
face shape analysis when preparing for a client hair
cut. When looking at facial shapes to make a specific
haircut or style.
Question: What are some examples of two-dimensional views that can represent a three-dimensional figure? Answer: Nets and multiple views.
Question: What is the primary use of understanding congruence in haircutting? Answer: To ensure all sides of the client's head match evenly. | 677.169 | 1 |
30° 49′ 18.7″
40207.6
P3 P
174° 43′ 37.3″
17874.2
P3 P2
74° 37′ 29.0″
22499.4
P2 P
220° 09′ 24.7″
31093.3
Comparing the bearing and length of the side P2 P as. obtained from the two sets of triangles, we have—
220° 09′ 19-7″
31089-0 links; and
220° 09′ 24-7″
31093-3 "
giving differences of 5" and 4–3 links.
The application of the ordinary adjustment, resulting as it does in these differences, is therefore very unsatisfactory, and the question arises as to whether it is desirable in this and in similar cases to adopt some further adjustment to the observed angles so as to eliminate the discrepancies shown above.
Before discussing the further adjustment it may be as well to remark that the ordinary procedure would be to adopt the mean values of the bearing and distance of P P2. None of the other sides, however, would receive any correction; consequently if the calculation is repeated, using the mean value of P P2 as base, an entirely different set of values will be obtained for all the other sides of the triangles.
As the need for further adjustment is obvious, the method of applying it will now be indicated.
II. The Least-square Adjustment.
The problem to be solved is: Given the observed angles of the four triangles, corrected as shown in I., by applying one-third of the error of each triangle to each angle, what further corrections must be made to these angles so as to eliminate the discrepancies found above?
It is evidently desirable that the corrections should be as small as possible so that no undue alterations are made to the angles: this condition is satisfied when the sum of the squares of the corrections is a minimum.
The application of this condition is shown on the schedule, and is briefly as follows:—
In column No. 5 the natural sines of the angles in column No. 4 are given.
If the sines in No. 5 were correct we should have—
[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]
Sin A1 sin A2 sin A3 sin A4 / Sin B1 sin B2 sin B3 sin B4 = 1.
This equation shows that the length of P P2 calculated from P P4 by the first pair of triangles should be the same as the length calculated by the second pair of triangles.
This is not usually the case, so put
Sin A1 sin A2 sin A3 sin A4 / Sin B1 sin B2 sin B3 sin B4 = 1 + ε
Question: What is the length of side P2 P as obtained from the first set of triangles? Answer: 31089.0 links | 677.169 | 1 |
No, two pairs of corresponding sides are congruent, and one pair of corresponding angles is congruent, but the angle is not included in the sides, so the situation doesn't fit into SSS, SAS, or ASA. It is more like "SSA", which is not sufficient to prove
the congruence of triangles.
No, each triangle is equiangular, and therefore equilateral, but the sides of one triangle could be longer or shorter than the sides of the other. Just because both triangles are equilateral doesn't mean that they must be congruent.
Question: Are the triangles in the text equiangular? Answer: Yes. | 677.169 | 1 |
The applets below illustrate several purely geometrical properties of the parabola. For entirely idiosyncratic reasons, the parabola has been rotated 90o such that wherever a parabola had to be drawn, I used the equation y = x2/2p instead of (2). In the following, the feet of perpendiculars dropped from points A, B, etc. on the parabola to the directrix will be denoted A', B', etc. F will always denote the focus of the parabola at hand.
Theorem 1
Let A lie on a parabola. Then the tangent to the parabola at A makes equal angles with AF and AA'.
Proof
By the definition, FAA' is isosceles. Let T be the midpoint of FA'. Then the perpendicular bisector AT divides the plane into two parts: one consists of points that are nearer to F than they are to A'; the other consists of points that are nearer to A'. Except for A, all points of the parabola lie in the former half. Indeed, let B be a point on the parabola. Then, since BB' is the shortest segment from B to the directrix, FB = BB' < BA'. In particular, B does not belong to AT. We conclude that A is the only point of intersection of that line with the parabola. Therefore, AT is tangent to the parabola at A.
Corollary (Parabolic mirror)
If a light source is placed at the focus of parabola and the light is reflected from its inner surface, the reflected rays are all parallel to the axis. Radio telescopes are built on a reversed principle. Incoming signals parallel to the axes all pass through the focus.
If, at the outset, the parabola is not given, but only a point and a line, we may produce any number of creases by folding the paper so that the given point falls onto the given line. In time, a parabola will emerge as the envelope of paper lines. Note that it does not matter whether the goal of a particular folding is to place a point on a line, or make the line pass through the point. As a practical matter, if the given line coincides with a paper edge then it is much easier to pursue the latter goal.
The parabola divides the area of an Archimedes triangle in the ratio 2:1. In other words, the area of the parabolic segment AB equals 2/3 of the area of the Archimedes triangle ABS.
Proof
Let area(ABS) = 1. Two thousand years before the invention of Calculus, Archimedes filled the parabolic segment with triangles, whose areas are easily arranged into a geometric series whose sum he already knew.
The first triangle in the series is the "inner triangle" ABO. From Lemma, area(ABO) = 1/2 — the first term of the series. Area(A1B1S) = 1/4. Therefore, area(AA1O) + area(BB1O) = 1/4.
Question: What is the area of the parabolic segment AB in relation to the area of the Archimedes triangle ABS? Answer: The area of the parabolic segment AB equals 2/3 of the area of the Archimedes triangle ABS.
Question: How can a parabola be produced without being given initially? Answer: By folding paper so that a given point falls onto a given line, a parabola will emerge as the envelope of paper lines.
Question: What does the letter 'F' represent in the context of the text? Answer: The focus of the parabola
Question: What is the principle behind the construction of radio telescopes, according to the text? Answer: Incoming signals parallel to the axes all pass through the focus. | 677.169 | 1 |
(4)-(5) prove the theorem.
Parabola as Envelope II
Assume a parabola with two points A and B and their tangents AS and BS are given. Pick a number n and divide AS and BS into n equal intervals. Label division points on AS with numbers 1, 2, 3, ... counting from S, and mark those on BS counting from B. Connect the points with the same labels. From Apollonius theorem, the lines will envelope the parabola [Dörrie, pp. 220-222, Wells, p. 171].
If one starts with just two segments AS and BS, the emerging parabola will touch them at points A and B November 2003, his site has welcomed its 8,000,000th visitor. The site is a recipient of the 2003 Sci/Tech Award from the editors of Scientific American.
Question: What are points A and B in the context of the text? Answer: Points A and B are two given points on a parabola. | 677.169 | 1 |
If a circle touches a line at exactly one point, that means it's tangent to it. Since this circle does so for both the x- and y-axes, its center must be equidistant from both. So, for example, it could be at (5, 5). If it was at (5, 5), though, that'd only eliminate choice (A). To eliminate the other incorrect choices, you need to consider where else the center could be. It could also be at (5, -5), (-5, 5), or (-5, -5). If you evaluate each of those points, you'll eliminate every choice but the correct one. :)
See the image below for a chart of Sharon and Michelle's water balloon game.
19/20 !! For, QN 20, i thought it was A, Can you please explain? And for QN18, It says the shaded part...So, (3/4)(4)(2.pi.) (1/2) is for those four squares..What about the perimeter of square? square is shaded as well..So, don't we need to add 4(the perimeter of square) as well?
Nice work! You can eliminate (A) on #20 with a quick plug in. Pick a score that should NOT work (like 10) and see if the inequality in (A) is still true: |10 - 39| ≤ 75. That's true, which means the inequality in choice (A) doesn't represent the same range as the scores in the class. See my post on absolute values for a quick way to solve a question like this.
The square in #18 is important, as you figured out, but it doesn't count towards the perimeter because it's surrounded by the larger shape. When I adapted this drill for my book, I made the square a dotted line, to make that more clear. :)
Hi, can you please explain number 5? When I first did it, I somehow didn't have issues with it. A month later I'm reviewing it and I don't understand it. I made the table just like how you make it. My issue is where did you get -10 from in the sum column. I understand where the -36 is coming from but if the sum of the four numbers is 10, then why is it put as negative? If it was positive the answer would be -26. Sorry, but this really confused me >.<
The negative 10 comes from the fact that we are TAKING OUT the 4 numbers from the 6 that add up to -36. Think about it: if 6 numbers add up to -36, but 4 of them add up to positive 10, the other 2 must add up to -46.
Question: What does it mean for a circle to be tangent to a line? Answer: A circle is tangent to a line if it touches the line at exactly one point.
Question: If the sum of the four numbers in question 5 was positive 10, what would the final answer be? Answer: If the sum of the four numbers was positive 10, the final answer would be -26.
Question: Which of the following is NOT a possible center of a circle tangent to both axes: (A) (5, 5), (B) (5, -5), (C) (-5, 5), (D) (0, 0)? Answer: (D) (0, 0) is not a possible center as it would not be equidistant from both axes. | 677.169 | 1 |
Label the angles in one triangle , , and . Label the angles in the other triangle , , and . Arrange the labels so that and are adjacent and and are adjacent. Then label the angle that is the combination of and with a and the angle that is the combination of and with a . The four interior angles of the quadrilateral should then be , , , and and we want to prove that
The interior angles of a triangle sum to 180 degrees (you find the theorem that says so; it is in your Geometry book somewhere.)
You have got to be kidding. This is Algebra.com. It is NOT the Psychic Hot Line. We can't see your diagram or whatever other information is provided in your textbook. Use your head for something besides a hat rack, please.
No idea how your diagram is numbered, though I assume that the angles you are talking about are formed by a transversal. Can't help you without a diagram.
John
My calculator said it, I believe it, that settles it
Probability-and-statistics/526332: A manager of a grocery store took a random sample of 100 customers. Their average checkout time is 3 minutes. The standard deviation of the population is 2 minutes. What is the probability that the average time is greater than 3.6 minutes? Less than 3.4 minutes?
I would like to identify a z score for each of the questions, but I don't have a population mean? Can I solve this without that number?
Graphs/526311: In the coordinate plane, the graphs of the equations x^2 + y^2 – 4x + 6y –12 =0 and y = ax^2 + bx + c have exactly 3 points in common . Two of these points are (3, -3) and (7, -3). What are all possible coordinates of the third point?
On the other hand (-3,-3) IS on , so on the off chance you made the obvious typo:
First complete the square on each of the variables to obtain the standard form of the equation of a circle.
(Verification left as an exercise for the student)
Note that (-3,-3) and (7,-3) are the endpoints of a horizontal diameter. Hence, if a parabola, which has a vertical line axis of symmetry passes through those two diameter endpoints, the third point of commonality, given exactly three points in common, must be the vertex of the parabola which could intersect the circle at either endpoint of a vertical diameter segment.
Find both points on the circle where by evaluating:
John
My calculator said it, I believe it, that settles it
Linear-equations/526306: Write the equation of the line in slope-intercept form with a slope of 2 and a y-intercept of -3
Question: What is the standard deviation of the checkout times? Answer: 2 minutes
Question: What is the angle formed by the combination of and? Answer: a | 677.169 | 1 |
Similarity In A Right Triangle When An Altitude Is Drawn From The Hypotenuse To The Right Angle.
So far, we know how to solve similarity in triangles, but how about having a right triangle with the height from the hypotenuse? Can you see the embedded triangles in the figure? May you determine which sides are corresponding for them to setup proportions? Do you know why they are similar?
In the problem solving of these examples, you will explore the examples by separately drawing the embedded triangles and then relating the corresponding sides to build the proportions that will enable you to find the unknown segment lengths. It is a very visual approach, You will like it!
Angle-angle-angle (AAA) similarity
The angle-angle-angle (AAA) similarity test states that given two triangles that have corresponding angles that are congruent, then the triangles are similar. As we know the sum of the interior angles in a triangle is 180°, so if two corresponding are congruent, then the other ones should be as well.
Proportion
A statement that two ratios are equal.
Proportional
One of four numbers that form a true proportion.
Pythagorean theorem a2 + b2 = c2.
The Pythagorean theorem states that if you have a right triangle, then the square built on the hypotenuse is equal to the sum of the squares built on the other two sides.
Ratio
A quotient of 2 numbers.
Right triangle
A triangle that has a 90 degree angle.
Side-angle-side (SAS) similarity
The side-angle-side (SAS) similarity test states that given two triangles that have two pairs of sides that are proportional and the included angles are congruent, then the triangles should be similar.
Side-side-side (SSS) similarity
The side-side-side (SSS) similarity test states that for two triangles to be similar; all corresponding sides should be proportional.
Similar
Two polygons are similar polygons if corresponding angles have the same measure and corresponding sides are in proportion.
Similar triangles
Similar triangles are triangles which have the same shape but probably different size. Corresponding angles need to be congruent, and corresponding sides are in proportion.
Triangle
A polygon with three sides.
Interactive Geometric Applets: Relevant Theorems.
This great applet shows you, how you may draw three similar triangles, after you
draw the altitude from the hypotenuse to the right angle in a right triangle.
The triangles have been drawn with the same orientation. You may identify them
by the colors, the labels in the vertices, and the colors and measures of the side segments.
Drag point "B" to see how they update angles and side lengths to keep their
proportionality and similarity as shown in the provided extended proportions.
Question: What is the name of the interactive tool mentioned at the end of the text? Answer: Interactive Geometric Applets | 677.169 | 1 |
Analytical Geometry Definition
posted on: 20 Jun, 2012 | updated on: 31 Aug, 2012
Most of us would not be familiar with the term analytical Geometry in the context of the Math. So in this article we will try to define Analytical Geometry in a very precise way to have a good understanding. The analytical geometry is also sometimes called as analytic geometry.
Before we define analytical geometry we should mention that the analytical geometry possesses two distinct types of the meanings in the context of the math. One of those two meanings of the analytical geometry is the modern meaning and the other meaning refers to the classical meaning of the analytical geometry. The definition of analytical geometry in two of its forms can be given as follows.
The definition of analytical geometry in its modern form or we can say that the advanced understanding of the term analytical geometry generally emphasizes on the geometry of such type of the varieties which are analytic.
Now we will define analytical geometry in its classical form but before we do that we should first mention that the meaning of analytical geometry in its classical form is same as the definition of the coordinate geometry or the definition of the Cartesian geometry. In context to classical math, the analytical geometry is defined as the study of the geometry with the help of the system of a coordinate and also by the use of various principles of the analysis and the principles of the Algebra. The analytical geometry has got many applications in the fields of engineering and Physics.
Question: What is the classical definition of analytical geometry? Answer: The classical definition of analytical geometry is the study of geometry using a coordinate system and principles of analysis and algebra. | 677.169 | 1 |
Find the point of intersection of three planes
Find the point of intersection of three planes
1. The problem statement, all variables and given/known data
The plane P1 contains the points A,B,C, which have position vectors a=(0,0,0), b=(1,1,8) and c=(0,1,5) respectively. Plane P2 passes through A and is orthogonal to the line BC, whilst plane P3 passes through B an is orthogonal to the line AC. Find the coordinates of r, the point of intersection of the three planes
3. The attempt at a solution
I have found the equation of plane P1 using the vector product of vectors AB and AC, but I can't find the equation of the line perpendicular to BC which passes through A to find plane P2.
Can anyone help?
Thanks
Question: What is the relationship between plane P2 and line BC? Answer: Plane P2 is orthogonal to the line BC | 677.169 | 1 |
Given triangle ABC with side AB identified as the base, there is only one line through C parallel to AB, thus determining the height of the triangle. Proposition I.38 asserts that two triangles with equal bases and equal heights will in fact have equal area. Another result equivalent to the Euclidean parallel postulate is the theorem stating that any triangle has angle sum 180° [2], which is in turn equivalent to a rectangle having angle sum 360°. The web resource offers further history of the parallel postulate.
While Euclid follows a step-by-step model of deductive reasoning with every statement justified by a previous proposition, definition, or postulate, the Chinese method is a bit more intuitive, particularly when identifying what today would be called congruent triangles. The ease by which similarity results are then proven (as a modern exercise) is appealing. Moreover, the argument founded on Chinese principles does not require a comparison of possibly two incommensurable lengths for the bases in similar triangles, as Euclid must consider. Two lengths L1, L2 are commensurable if a whole-number multiple of L1 can be constructed on a whole-number multiple of L2, or in modern language [3, p. 30], if there are positive integers n1, n2 with n1L1 = n2L2, or equivalently, [(L1)/(L2)] is a rational number. See for detailed information about the history of Greek mathematics.
The curricular presented materials in Section 4 are ideal for a course in geometry, taught either in college or high-school, or for a course that draws prospective teachers of secondary mathematics. For use in the classroom, the instructor should present the results of Section 3, although Section 2 may be omitted from class discussion, depending on course direction and time constraints. When assigning the material in class, the instructor may delete or rearrange certain parts of the teaching module to fit the course.
The Ancient Greek Contribution
Similarity has its roots in antiquity, and ideas of proportion were likely known to the Pythagoreans in the fifth century [4, p. 82-83], while Eudoxus of Cnidus (ca. 408-355 ) established a notion of what today would be called the equality of ratios [3, p. 28], crucial in the case of incommensurable magnitudes. Euclid (third century ) devotes Book V of The Elements to a study of proportion, incorporating many of the ideas developed earlier in ancient Greece, particularly those practiced by Pythagoras and Eudoxus. Proposition Four in Book VI of The Elements [1] states:
VI.4. In equiangular triangles the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles.
Question: Which of the following is NOT equivalent to the Euclidean parallel postulate? A) Two triangles with equal bases and equal heights have equal area. B) The angle sum of a triangle is 180°. C) A rectangle has an angle sum of 360°. Answer: C) A rectangle has an angle sum of 360°.
Question: What is the angle sum of a triangle according to the text? Answer: 180°
Question: What is the area of two triangles with equal bases and equal heights according to Proposition I.38? Answer: They have equal area. | 677.169 | 1 |
What name is given to numbers with no factors apart from itself and 1?
What type of triangle has 3 equal sides?
What type of triangle has 2 equal sides?
What type of triangle has no equal angles?
There are 360 of which unit of angle measure in a circle?
What unit of angle measure is broken into 400 parts?
What unit of angle measure must be used in calculus?
What is the 'posh' name for the inverse function to #9
What is the 'posh' name for the inverse function to #10
What is the 'posh' name for the inverse function to #11
What is the expression obtained after differentiation called?
What is the correct term for a 'top heavy' fraction?
What is the name of equation expressed in terms of an additional parameter (usually 't')
Name for the x and y co-ordinate quiz should really have to be answered in order, otherwise e.g. you could guess all the trig functions without knowing which is which. You need to capitalize 'cosine'. If you write 'Natural numbers' and 'Prime numbers' you should also write 'Rational numbers', 'Complex numbers' and 'Real numbers' rather than 'Rational', 'Complex' and 'Real'. The inverse function of the exponential is the 'logarithmic function' not the 'logarithm'. Just answering 'isosceles' doesn't seem to work for the two-equal-sides question, and I have no idea why you put 'right-angled' as an alternative answer - that has nothing at all to do with 2 equal sides. Degrees are not broken into 360 parts, it's just that 360 degrees is the number of degrees in a full rotation.
I agree, a right angled triangle can be isosceles. And imaginary numbers aren't the same as complex numbers. Imaginary numbers aconsist of only imaginary components, while complex numbers can have real components too. That's why it doesn't accept imaginary I think
Very tricky with the gradians there. Haven't seen those since I broke my TI-30! One beef: Careful with the '!' sign. I know it's most common use is for the factorial, but I found myself typing "unique," as in the proof shorthand that involves an upside-down 'A', a backwards 'E', etc. Maybe you could be more specific.
for equals, I typed in equal, equal to like 20 times and didnt think about "equals"
"4 is equal to 6", "the two expressions are equal"
I didn't know it had to be "equals"
I like how the quiz gets into harder topics as it progresses.
and answering in order would make it harder
for logarithmic, could you not except "log" or "logx" or lnx or something?
also i dont like the inconsistency with plurals - you accept "degree" for degrees but for radians and gradians only the plurals were accepted.
Question: What is the expression obtained after differentiation called? Answer: Derivative
Question: What unit of angle measure must be used in calculus? Answer: Radians
Question: There are 360 of which unit of angle measure in a circle? Answer: Degrees
Question: What unit of angle measure is broken into 400 parts? Answer: Gradians | 677.169 | 1 |
I realize that this is highly unfair to compare the edifice erected by Morley with the ad hoc cabins of proofs of what is now known as his Trisector theorem. Taken out of context, the theorem indeed becomes a miracle that inspires admiration but loses the sparkle of intrinsic consistency. To have a fighting chance to a claim for motivation, an elementary proof should probably handle the whole set of 27 triangles simultaneously, which means proving existence of three sets of parallel lines cutting at the 60o angles. But even then an elementary framework will not attain the broad outlook of Morley's theory that includes the angle bisectors and trisectors (and more) under a single roof.
So this is the end of an endeavor to understand Frank Morley's work and particularly his Trisector theorem. I began writing this column with a quiet satisfaction that, after diligently plowing through Morley's papers, I finally reached a stage where it became possible to assert my inderstanding of Morley's theorem. At this point, I completed the first half of the column, inserted the applet and ran my browser to see how things looked so far. This presented me with the first opportunity to play with the applet. For, while writing it, I did not really experiment with it. Just made sure there were no bugs.
Toying with the applet gave me an additional insight. The axes of a 3-line serve as a sort of index into how a cardioid may touch the three lines. (I remove my hat to Frank Morley who saw everything so clearly without the benefit of modern day gadgets.) Well, you can say, is it not what you wrote above, when outlining Morley's reasoning in the first place? Yes and no. Is it internalization that I lacked before? Probably. Whatever it is called, this something had clearly enhanced my understanding of the theorem. The meaning of it may be different from what Donald Newman meant when he asked "Were we to give up, forever, understanding the Morley Miracle?" But I certainly did not get it upon reading his and other proofs.
Alex Bogomolny has started and still maintains a popular Web site Interactive Mathematics Miscellany and Puzzles
to which he brought more than 10 years of college instruction and, at least as much, programming experience. He holds M.S. degree in Mathematics from the Moscow State University and Ph.D. in Applied Mathematics from the Hebrew University of Jerusalem. He can be reached at [email protected]
Question: What is Alex Bogomolny's email address? Answer: [email protected] | 677.169 | 1 |
Constructions
This page is dedicated to teaching you how to make constructions. This page would have been easier to understand if I was able to include our animation Java applet. Unfortunately, Dave never had a chance to complete this (even with the later deadline and working for 24 hours straight at a time). Kind of makes me glad I don't know Java. So, that'll have to wait until next year. Of course, this page would still be the same for those of you who can't read Java yet (i.e., most of us).
A construction is a specialized type of drawing. It is used to draw mathematical figures that require exactness. In a drawing you can use any tool. However, in a construction, you can only use two tools. They are the unmarked straightedge and the compass.
The straightedge is simply a flat tool with a straight edge or two. It is unmarked because specific lengths aren't used in a construction. That is, figures are of unspecified size in constructions - just bigger or smaller.
A compass is a tool used to draw circles. They are NOT the kind used to determine north or south. Compasses used for construction are tools that have one end with a sharp point and one end with a pencil or pen attached. It's sort of like two pens or pencils attached to each other at the erasers. A good compass is marked to determine the radius of the circle drawn, but that isn't necessary for a construction (as I've noted several times above). Anyways, you place the sharp point on the spot that you want to be the center of the circle, open the compass to the correct radius, and then move the pencil end in a circle. If you don't have one, it is easy to make one from scratch. Just tie a piece of string to the end of a pencil.
Question: Why was the Java applet not included? Answer: Dave never completed it, even after working for 24 hours straight
Question: What is a compass used for in a construction? Answer: To draw circles | 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: What is the relationship between the triangles ABD and FBC? Answer: They are congruent
Question: What is the area of the square GB compared to the triangle FBC? Answer: Double | 677.169 | 1 |
Curvature
circle, curve and radius
CURVATURE. The curvature of a plane curve at any point is its tendency to depart from a tangent to the curve at that point. In the circle this deviation is constant, as the curve is per fectly symmetrical round its eentre. The curva ture of a circle varies, however, inversely as the radius—that is, it diminishes at the same rate as the radius increases. The reciprocal of the radius is therefore taken as the measure of the curvature of a circle. A straight line may be considered a circle of infinite radius and as hav ing no curvature, since = 0. The constancy of curvature in the circle suggests an absolute measure of curvature at any point in any other curve, for whatever be the curvature at that point a circle can be found of the same curva ture. The radius of this circle is called the radius of curvature for that point; and the circle itself the osculating circle. By means of this radius we may compare the curvatures at different points of the same curve or of different curves. In simple cases, as in the conic sections,
the measure or radius of curvature may be de termined geometrically, but it is usually neces sary to employ the calculus. The expression for the radius of curvature at any point (,v, y) of a curve is 3 [ 1 ±
d.c 2 If the curve, instead of lying in a plane, twists in space, it is sometimes called a gauche curve or a curve of double curvature, and its curvature at any point may be measured by the radius of its osculating sphere at that point. The centre of the osculating circle or sphere is called the centre of curvature. The curvature of surfaces is de termined similarly to that of curves. Thus the measure of the curvature of the earth, commonly taken as the deviation of the line of apparent level from the line of true level—that is, from a line everywhere parallel to the surface of still water—is approximaIely eight inches per mile.
Question: What is the centre of curvature? Answer: The centre of the osculating circle or sphere is called the centre of curvature. | 677.169 | 1 |
It is numerically less efficient than Newton's method but it much less prone to odd behavior.
In geometry, bisection refers to dividing an object exactly in half, usually by a line, which is then called a bisector.
The most often considered types of bisectors are segment bisectors and angle bisectors.
A segment bisector passes through the midpoint[?] of the segment.
Particularly important is the perpendicular bisector of a segment, which, according to its name, meets the segment at right angles.
The perpendicular bisector of a segment also has the property that each of its points is equidistant[?] from the segment's endpoints[?].
An angle bisector divides the angle into two equal angles.
An angle only has one bisector.
Each point of an angle bisector is equidistant from the sides of the angle.
(Please add figures to this entry. Should ruler-and-compass constructions be included?)
Question: What is the property of each point on an angle bisector? Answer: Each point on an angle bisector is equidistant from the sides of the angle. | 677.169 | 1 |
Florida - Mathematics: Geometry
This correlation lists the recommended Gizmos for this state's curriculum standards. Click any Gizmo title below to go to the Gizmo Details page.
MA.912.G: Geometry
MA.912.G.1: Understand geometric concepts, applications, and their representations with coordinate systems. Find lengths and midpoints of line segments, slopes, parallel and perpendicular lines, and equations of lines. Using a compass and straightedge, patty paper, a drawing program or other techniques, students also construct lines and angles, explaining and justifying the processes used.
MA.912.G.1.1: Find the lengths and midpoints of line segments in two-dimensional coordinate systems.
MA.912.G.1.2: Construct congruent segments and angles, angle bisectors, and parallel and perpendicular lines using a straight edge and compass or a drawing program, explaining and justifying the process used.
MA.912.G.2: Identify and describe polygons (triangles, quadrilaterals, pentagons, hexagons, etc.), using terms such as regular, convex, and concave. Find measures of angles, sides, perimeters, and areas of polygons, justifying the methods used. Apply transformations to polygons. Relate geometry to algebra by using coordinate geometry to determine transformations. Use algebraic reasoning to determine congruence, similarity, and symmetry. Create and verify tessellations of the plane using polygons.
MA.912.G.2.2: Determine the measures of interior and exterior angles of polygons, justifying the method used.
MA.912.G.2.4: Apply transformations (translations, reflections, rotations, dilations, and scale factors) to polygons to determine congruence, similarity, and symmetry. Know that images formed by translations, reflections, and rotations are congruent to the original shape. Create and verify tessellations of the plane using polygons.
MA.912.G.4: Identify and describe various kinds of triangles (right, acute, scalene, isosceles, etc.). Define and construct altitudes, medians, and bisectors, and triangles congruent to given triangles. Prove that triangles are congruent or similar and use properties of these triangles to solve problems involving lengths and areas. Relate geometry to algebra by using coordinate geometry to determine regularity, congruence, and similarity. Understand and apply the inequality theorems of triangles.
MA.912.G.5: Apply the Pythagorean Theorem to solving problems, including those involving the altitudes of right triangles and triangles with special angle relationships. Use special right triangles to solve problems using the properties of triangles.
MA.912.G.5.1: Prove and apply the Pythagorean Theorem and its converse.
MA.912.G.6: Define and understand ideas related to circles (radius, tangent, chord, etc.). Perform constructions, and prove theorems related to circles. Find measures of arcs and angles related to them, as well as measures of circumference and area. Relate geometry to algebra by finding the equation of a circle in the coordinate plane.
Question: What is the name of the theorem that relates the sides of a right triangle? Answer: Pythagorean Theorem
Question: What is the difference between the sum of the measures of the interior angles of a triangle and the sum of the measures of the exterior angles of the same triangle? Answer: 180 degrees | 677.169 | 1 |
January 1, 2013
Triangle Centers
I am in the process of working through a series of lessons on triangle centers. So far, I've made it through the perpendicular and angle bisector theorems, as well as perpendicular and angle bisectors of triangles. I am posting what I have accomplished here in an effort to get input on whether these seem to be too difficult. Are my expectations too high? My students are average level sophomores. I am wavering on including more basic skill level types of questions along with the problems currently on the worksheet.
I have considered updating parts of it by incorporating some sort of technology investigation using either TI-84s or 92s, but would have to incorporate some sort of training on how to use the Cabri environment before looking at triangle centers. I think this would be especially useful in portions of the investigation into angle bisectors of a triangle.
Question: What is the average level of the author's students? Answer: Sophomores | 677.169 | 1 |
Gudmundsson et al. (2004) consider the problem of finding a pseudotriangulation of a point set or polygon with minimum total edge length, and provide approximation algorithms for this problem.
Pointed pseudotriangulations
A pointed pseudotriangulation can be defined as a finite non-crossing collection of line segments, such that at each vertex the incident line segments span an angle of at most π, and such that no line segments can be added between any two existing vertices while preserving this property. It is not hard to see that a pointed pseudotriangulation is a pseudotriangulation of its convex hull: all convex hull edges may be added while preserving the angle-spanning property, and all interior faces must be pseudotriangles else a bitangent
Bitangent
In mathematics, a bitangent to a curve C is a line L that touches C in two distinct points P and Q and that has the same direction to C at these points...
line segment could be added between two vertices of the face.
A pointed pseudotriangulation with v vertices must have exactly 2v − 3 edges. This follows by a simple double counting
Double counting (proof technique)
In combinatorics, double counting, also called counting in two ways, is a combinatorial proof technique for showing that two expressions are equal by demonstrating that they are two ways of counting the size of one set...
In mathematics, and more specifically in algebraic topology and polyhedral combinatorics, the Euler characteristic is a topological invariant, a number that describes a topological space's shape or structure regardless of the way it is bent...
: as each face but the outer one is a pseudotriangle, with three convex angles, the pseudotriangulation must have 3f − 3 convex angles between adjacent edges. Each edge is the clockwise edge for two angles, so there are a total of 2e angles, of which all but v are convex. Thus, 3f − 3 = 2e − v. Combining this with the Euler equation f − e + v = 2 and solving the resulting simultaneous linear equations gives e = 2v − 3.
Similarly, since any k-vertex subgraph of a pointed pseudotriangulation can be completed to form a pointed pseudotriangulation of its vertices, the subgraph must have at most 2k − 3 edges. Thus, pointed pseudotriangulations satisfy the conditions defining Laman graph
Laman graph
In graph theory, the Laman graphs are a family of sparse graphs describing the minimally rigid systems of rods and joints in the plane. Formally, a Laman graph is a graph on n vertices such that, for all k, every k-vertex subgraph has at most 2k −3 edges, and such that the whole graph...
Question: What is a bitangent to a curve? Answer: A line that touches the curve in two distinct points and has the same direction at these points.
Question: Which of the following is NOT a property of a pointed pseudotriangulation? A) It has exactly 2v - 3 edges. B) It can be completed to form a pointed pseudotriangulation of its vertices. C) It has crossing line segments. Answer: C) It has crossing line segments.
Question: What is a Laman graph? Answer: A family of sparse graphs describing the minimally rigid systems of rods and joints in the plane, where every k-vertex subgraph has at most 2k - 3 edges. | 677.169 | 1 |
Before we get down to our initial question, what about some small functions in order to warm up? Let's say we wanna completely exclude the use of Scala's object oriented features and hence don't wanna call a Circles getter methods in order to retrieve its center or radius. Hugh? But how we're supposed to get those fields then? Well, we could leverage Pattern Matching in order to retrieve a circles components. Observe:
What we have here is a so called selector function. In fact, you could see the constructor as a function which takes some arguments and creates a new value of the given type. Then, a selector is nothing else than another function which takes such a constructed value and picks a certain component from within that value. In our case, we did that with a certain form of pattern matching: we just introduced a new value by revealing all its components with a name (center and radius). Those are then bound to the actual values of the given circle. In fact, we could've make use of our famous underscore for the first component, since we're only interested in the radius here.
With that knowledge on board, it's rather easy to come up with a selector function for the center of a circle:
Given those two key players and our two selectors, we only need to come up with an idea on how to detect if a certain point is within the area of a certain circle. There's a simple solution, thanks to the genius of Pythagoras …
Two points and a distance to go
It turns out, that the Pythagorean theorem is a perfect model for calculating the distance of two points (with the center of our circle as the one and the point in question as the other point) in a cartesian coordinate system (what a fortunate coincidence, since our game's based on such). If the distance is shorter than the radius of our circle, then the point in question is clearly within the circle (otherwise not).
As you can see from the picture, we simply need to calculate the difference of our two points both x-coordinates and the same with both y-coordinates to come up with the length for the adjacent and the opposite leg (with the distance between our two points as the hypotenuse then). Then the only thing left to do is to solve the famous equation a² + b² = c². If the point in question is within the circle (or on the circles edge), then a² + b² need to be smaller or equals than radius².
Aaahh, a neat function which takes two arguments – a circle and a point, resulting into a Boolean value which indicates if the given point lies within the circles area. Again we're leveraging pattern matching to reveal the x- and y-coordinates of the given center and point. We then simply utilize our knowledge about Phytagoras for the final answer.
Spicing up a tasteless Function …
Question: Which feature of Scala is not used in this approach to retrieve a circle's components? Answer: Object-oriented features, specifically getter methods.
Question: Which theorem is used to calculate the distance between two points in a Cartesian coordinate system? Answer: Pythagorean theorem
Question: What is the final function created in this context? Answer: A function that takes a circle and a point as arguments and returns a Boolean value indicating whether the point lies within the circle's area. | 677.169 | 1 |
The triangle on the right has height 'h' a side of (80-x) and a hypotenuse of 70. Again using the Pythagorean Theorem, 702 =h2 plus (80-x)2. Since h2 appears in both equations, we obtain an equation of: 2,500 - x2 = 702-(80-x)2 This reduces to x equaling 25. By substitution, we obtain h=43.30127019. From trigonometry we look up the arctangent of (h/x) = arctan(1.732050808) which equals 60 degrees exactly. The angle at the other side of the base equals arctan(43.30127019/55) or 38.2132107 degrees. Since all triangles have 180 degrees, the 3rd angle is calculated to be 81.7867893. Wow - another longwinded explanation!!!
***************************************
Puzzle #6       10/17/99
One pipe can fill a water tank in 50 minutes.
Another pipe can fill the same
water tank in 40 minutes.
How much time would it take
if both pipes were to fill the same water tank at the
same time ?
ANSWER
There is a formula for this type of problem but just by using logic, the answer is easily obtained. Assume the tank is 50 gallons. So the 'slow' pipe can fill it up at a rate of 1 gallon per minute. The 'fast' pipe can fill it up at a rate of1.25 gallons per minute. Therefore, with both pipes working together, the tank is filling at a rate of 2.25 gallons per minute. Since it is a 50 gallon tank, then it will fill up in (50/2.25) minutes or 22 minutes and 13 and a third seconds.
If you want to use the formula, it is (Rate1 x Rate2)/(Rate1 + Rate2) = Time.
***************************************
Puzzle #5       10/10/99
An employee is to be rewarded with a $250 bonus AFTER TAXES have been deducted.
If the federal tax is 15%, the state tax is 10% and the Social Security is 7.5 % what
must the GROSS PAY be in order for the net pay to be $250 ?
Twelve is the smallest positive integer
that can be divided by the integers 1, 2, 3 and 4 with no remainder. What is the smallest
positive integer that can be divided by 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 without any remainder?
Question: If both pipes in Puzzle #6 were to fill the same water tank at the same time, how long would it take? Answer: 22 minutes and 13 and a third seconds | 677.169 | 1 |
Geometry Performance Task: Area, Perimeter, and Angle Measures task, students are asked to interpret a mapping scenario and distinguish appropriate types of measurement strategies. In order to do this effectively, they must recognize that both shapes and angles need to be broken down into familiar pieces.
This assessment works well with students who have just learned about the Pythagorean Theorem and are familiar with area and circumference of circles. I have also used it in the introductory week of a high school geometry course. Students should already be familiar with:
This task can be completed individually or cooperatively. It also works well as either an in-class assessment or a take-home assignment. I typically give students all three pages and review the tasks and rubric with them before letting them dive in.
Included in the file download are the student task, full answer key, scoring rubric, and a cover page with essential questions, teacher notes and Common Core standards alignment.
PDF (Acrobat) Document File
Be sure that you have an application to open this file type before downloading and/or purchasing.
3928.1 should have known it was one of yours! You are my number one rockstar and life saver this year... how you do it all astonishes me - I'm always two steps behind (you bring me almost up to pace - and I really can't thank you enough!)
Just moved to a new state and their state standards are radically different from what I'm used to. This is a perfect product to help students master the concept and to help me regain confidence in teaching the material! :D
Question: What is included in the file download besides the student task?
Answer: Full answer key, scoring rubric, and a cover page with essential questions, teacher notes and Common Core standards alignment. | 677.169 | 1 |
sin µ / cos µ = tan µ / 1. Which is correct. Tangent comes from the latin word tangere by the way, which means 'to touch'. That is exactly what the tangent does: it touches the side of the circle.
Math.atan() is used with y/x: Math.atan(ydistance/xdistance). This is because you need to provide the tangent to calculate the angle from, and the tangent is sin/cos, where sin is on the y (vertical) axis and cos is on the x (horizontal) axis.
But, there's a but. Two angles can have the same tangent:
When calculating the arc tangent ( == atan == tan^-1), Flash will return the angle to the right side of the triangle. But that means that if our point is at the left part of the circle, Flash will return the angle at the right side that has the same tangent.
Therefore, we need to check if our point is at the left or at the right of our circle. If it's at the left, we will have to add 180°, so Math.PI, because Flash uses radians.
Another thing to be aware of is that Flash doesn't go counterclockwise around the circle, but clockwise. And, that the _rotation property doesn't go from 0 to 360, but from -180 to 180. So we can't have a clip rotate to 225° using _rotation = 225. We have to substract 360 from that value to make it negative, and then rotate it to that negative number.
Question: When calculating the arc tangent, what does Flash return? Answer: The angle to the right side of the triangle. | 677.169 | 1 |
Measure of Arcs and Angles Formed by Intersecting Secants and Tangents. Inside and Outside the Circle.
Should you be given a problem with a secant and a tangent intersecting in the exterior of the circle; would you be able to find the measure of one of the intersected arcs, if you are given the other intersected arc and the angle at the point of intersection? What if they ask you to find the angle when the two referred arcs are given? If they give you two intersecting tangents ¿would you be able to do the same?
In this lesson we strife to give you a detailed presentation of the involved theorems when you have a tangent and a secant intersecting at the point of tangency, a tangent and a secant intersecting at an exterior point, and two tangents, or two secants intersecting at an exterior point; followed by a step by step solution of several examples. We are sure your knowledge will not escape through the tangent!
Question: What is the main goal of this lesson? Answer: To provide a detailed presentation of the involved theorems and step-by-step solutions for various examples of intersecting secants and tangents. | 677.169 | 1 |
In another math region we can find the hyperbolic geometry (or geometry of
Lobachevsky), one of the non-Euclidean geometries, based on the axiom that there exist at
least two lines through a point P that are parallel to a line not through P. The hyperbolic functions have
not much in common with the hyperbola. Their name is based on the fact that two hyperbolic
functions can be drawn geometrically in a hyperbola.
The wave equation, a partial differential equation of second order, is also named the hyperbolic
differential equation. Reason is the resemblance between the equations
6).
The hyperbolism of a curve f (x,y)
= 0 is a curve with equation f (x, xy) = 0.
Question: Which of the following is a non-Euclidean geometry mentioned in the text? A) Euclidean geometry B) Spherical geometry C) Hyperbolic geometry Answer: C) Hyperbolic geometry | 677.169 | 1 |
H = AD * (AB sin angle DAB) /2
so
sin angle DAB = 2H / (AD * AB)
so the data determine sin angle DAB, but different values for angle DAB give the same area if those angles have the same sine. In such a case, one angle is acute and its cosine is positive; the other angle is obtuse and its cosine is negative. The length of the third side BD can be calculated using the cosine rule:
BD² = AB² + AD² – 2 AB AD cos angle DAB
Thus, where angle DAB is acute, BD² will be less than AB² + AD², and, where angle DAB is obtuse, BD² will be greater.
To answer the quesion "when do the area and lengths of two sides of a triangle determine the length of the other side?": it's when the sine of the angle determines the angle, namely when the sine is 1 so the angle is a right angle, so we have the lengths of the legs of a right-angled triangle, and the formula for the length of the other side, the hypotenuse, simplifies to the formula that comes from Pythagoras's theorem:
BD² = AB² + AD²
Your neighbour used Heron's formula for the area of a triangle in terms of the lengths of the sides, so let's solve the problem using that formula. Suppose we have a triangle of area H and sides of lengths a, b, c, and H, a and b are known, and c is unknown. The question is: under what circumstances is there only one possible value of c?
a quadratic in c². Now, the product of the roots of the quadratic equation Ax²+Bx+C = 0 is C/A. Here, that is (a²-b²)² + 16H². The first of those terms is the square of a real, and thus cannot be negative; the second is positive because H is positive. Thus the product of the roots of the quadratic equation is positive, so those roots have the same sign. If they are positive, both apply to the geometrical problem; if negative, neither applies. Thus, if the geometrical problem has a unique solution, the roots must be equal. (It cannot be that the roots are different but the geometrical problem has a unique solution because one root is applicable, and the other root must be rejected. The only way that could happen is if one root were positive and the other weren't.)
The roots of the quadratic equation Ax²+Bx+C = 0 are equal iff the discriminant B²-4AC is zero. Here, that is true iff (substituting coefficients from equation [1]):
The area of a triangle is half the base times the height. Thus the condition for there to be one and only one solution is equivalent to the condition that, where one of the sides whose length we know is the base, the height of the triangle is the other side length we know. This is true iff the triangle is right-angled and the sides whose lengths we know meet at the right angle.
Question: What is the product of the roots of the quadratic equation in terms of the sides of the triangle? Answer: (a² - b²)² + 16H²
Question: True or False: If the product of the roots of the quadratic equation is positive, both roots apply to the geometrical problem. Answer: False. If the product is positive, both roots are positive or both are negative. If one is positive and the other is negative, only the positive root applies.
Question: What is the condition for the roots of the quadratic equation to be equal? Answer: The discriminant (B² - 4AC) is zero.
Question: When does the area and lengths of two sides of a triangle determine the length of the other side? Answer: When the sine of the angle determines the angle, i.e., when the sine is 1, making the angle a right angle. | 677.169 | 1 |
In this
exploration, we want to look at all the conditions in which the
three vertices of the pedal triangle are collinear (that is, it is a degenerate triangle).
So first, let's look at a the
Pedal triangle. We know that if point P is any point in the plane,
then the triangle formed by constructing perpendiculars to the
sides of ABC form the points Z, Y, Z, which are the intersections.
below is a basic construction of the pedal triangle.
Now, as the point P approaches
any of the vertices on triangle ABC, the pedal triangles points
Z, Y, and X form a straight line which means that they are collinear.
SEE
GRAPH HERE !(play with
the pedal point P and see if a line is formed when it is on top
of the vertices of the triangle ABC)
Now what happens if the pedal
point is outside the graph? Below, I have constructed a circumcircle
with the pedal triangle.
Now if we let p be a point
on the circumcircle, we have the following:
So wlog, let PABC be a cyclic
quadrilateral. From the right angles, PYXC, PZAY and PXBZ are
cyclic like PABC.
Question: What happens to the points Z, Y, and X of the pedal triangle when point P approaches any of the vertices of triangle ABC? Answer: They become collinear, forming a straight line. | 677.169 | 1 |
no. the definition of a rectangle is a quadrilateral with 2 sets of opposite sides congruent and parallel and 4 right angles. a trapezoid has only one set of sides that are congruent (those on the side) and only one set that are parallel (those on the top and bottom). there are also no right angles in a trapezoid. therefore, a rectangle cannot be a trapezoid.
I'm sorry. i failed to mention that i was talking about isosceles trapezoids in my response. trapezoids can be like rectangles in that the top and bottom are always parallel, the 2 sides do not have to be congruent (they can be and if they are it turns into an isosceles trapezoid) but they cannot be parallel.
Question: Can a trapezoid have two sets of opposite sides that are congruent and parallel? Answer: Yes, but only if it is an isosceles trapezoid. | 677.169 | 1 |
prove that that the function has a reasonable shape when a(x) is decreasing and b(x) is increasing. Is there some easy proof I have overlooked?
By convex I mean that any shortest line connecting the points on the ellipse is 'inside' the elipseellipse (i.e. the distance |AX| + |XB| is smaller or equal then the distance defining the elipseellipse for any point X of the line).
Is 'small enough' ellipse projected on asurfaceof
Is ellipse on a sphere convex? (proof)
Is 'small enough' ellipse on
Question: What is the distance condition for a point X on a line connecting points A and B on the ellipse? Answer: The distance |AX| + |XB| must be smaller or equal to the distance defining the ellipse for any point X on the line. | 677.169 | 1 |
Question 177732: okay, i am doing proofs in class and i have had three tests and failed them all. Not to mention im a transfer student and have never taken Geometry before. I had taken Algebra and Algebra 2 but no math this year, so they put me in this Geometry class half way through a semester. I need to know what to expect and how to learn it FAST.
p.s. i had already asked for help for peers and teachers, im un teachable.! Click here to see answer by Mathtut(3670)
Question 178000: I'm trying to help my niece with an isoceles triangle question. It's been years and I have forgotten the formulas! The vertex is 20 degress larger than a base angle. What is the measurement of each angle? Thank you!! Click here to see answer by stanbon(57982)
Question 178457: you are building a ramp that will be in the shape of a perfect right-angle triangle. The vertical height of the ramp will be 10 feet. the horizontal base of the ramp will be 15 feet. What will be the length of the downward sloping side of the ramp? Click here to see answer by cartert(27)
Question 179556: Hi in math class we are doing stuff with triangles. Here is my word problem: Scott wants to swim across a river that is 400 meters wide. He begins swimming perpendicular to the shore he started from but ends up 100 meters down river from where he started because of the current. How far did he actually swim from his starting point?
Thank you so much! Click here to see answer by stanbon(57982)
Question 179555: Hi. In math class we are doing word problems with right triangles. Here is the problem: Suppose a roller coaster's height is 105 feet off the ground. The distance on the ground from the start of the hill to the top of the hill is 200 feet. How fat do you actually travel on the roller coaster's track? Click here to see answer by Earlsdon(6294)
Question 179756: How do you measure a base triangle? I'm helping my son and I've forgotten some steps. Triangle MNP (acute Isosceles) is shown on a protractor, where angle M (base angle) measures 70 degrees. How do I find out the measure of angle N (base angle)? Click here to see answer by solver91311(17077)
Question 179729: Here is my question:
Triangle ABC has coordinates A(-3,0) B(0,3)and C(3,0). The number of sqaure unites in the area of ABC is :
1)6
2)9
3)12
4)18
(which of the above is the answer to the question?)
Question: What is the student's previous math background? Answer: The student has taken Algebra and Algebra 2 but no math this year, and they are a transfer student who has never taken Geometry before. | 677.169 | 1 |
Let's learn a little bit about the Dot product. The Dot product frankly out of the two ways of multiplying vectors, I think it's the easier one. So, what is the Dot product do? One I'll give you definition and then I'll give you the intuition. So, if I have two vectors. Let say vector A and vector B. That's how I draw my arrows and I can draw my arrows like that. That is equal to the magnitude of vector A times the magnitude of vector B times the cosine of the angle between them. Now where this is cosine come from? This might seem a little arbitrarily but I think with the visual explanation, I'll make a little bit more sense.
So, let me draw arbitrarily these two vectors. So, that is my vector A, nice big and fat vector. It's good for showing the point. And let me draw of vector b like that and then let me draw the cosine or let me start angle between them and this is θ. So, there is two way of viewing this. You can view it and let me label them. This is vector A. I'm trying to be color consistent and this is vector B. So, there is two ways of viewing this product and you can view it as vector A. Because multiplication associative, you can twist the order. So, this could also be written as the magnitude to the vector A times cosine θ times and now I'll do it in color appropriate times vector B.
And these times, this is the Dot product. I mean you almost have to write it. This is regular multiplication because these are all scalar quantities. When you see the dot between the vectors you are talking about the vector of that product.
So, if we are just rearrange this expression this way and what is it mean? What is a cosine θ? Let me ask you a question, if I were to draw A right angle right here perpendicular to B. So, I just drop right angle there cosine θ, SOCATOA. SOCA is equal to adjacent of a hypotenuse, right. Well let see adjacent is equal to this and the hypotenuse is equal to the magnitude of A, right. So, let me rewrite that. So, cosine θ and all of this applies to the A vector and cosine θ of this angle is equal to adjacent which is, I don't what you can call this. Let's call this the projection of A on to B, right. It's like if you where to shine a light perpendicular to B. If there was a light show what where straight down, it would be the shadow of A on to B.
And you could almost think of it is the part of the A that goes in the same direction of B. So, this projection they call it and at least the way I get intuition of what a projection is I can if this shadow, if you had a light source that came up perpendicular and what would be the shadow of that factor on to this one. So, if you think about it this shadow right here. We can call that the projection of A on to B or I don't know, let just call it A sub B and it's magnitude of it, right.
Question: How is the dot product defined in the given text? Answer: The dot product of two vectors A and B is defined as the magnitude of A times the magnitude of B times the cosine of the angle between them.
Question: Where does the cosine come from in the dot product formula? Answer: The cosine comes from the definition of the dot product and provides a measure of how similar the directions of the two vectors are.
Question: Which of the following is NOT a way to represent the dot product of vectors A and B in the given text? A) A · B = |A||B|cosθ B) A · B = |A|cosθ B C) A · B = |A||B|cosθ D) A · B = |A|cosθ B Answer: B) A · B = |A|cosθ * B | 677.169 | 1 |
The Cross product is actually almost the opposite. You're taking the orthogonal components, right. The difference was this was a sine θ. And I don't want to mess you this picture too much but you should review the Cross product videos. And I'll do another video where you compared and contrast them. But the Cross product as your saying, let's multiply the magnitude of the vectors that are perpendicular to each other. That isn't going in the same direction. That they are actually orthogonal to each other. And then you have to pick a direction since you are not saying the same direction that they are going in. So, you are picking the direction that is orthogonal to both vectors. And that's why the orientation matters and you have to take the right hand rule because there are actually two vectors that go, that are perpendicular to any other two vectors in three dimensions.
Anyway I'm all out of time. I will continue this hopefully you are not too confusing discussion. In the next video I'll compare and contrast the Cross product and that the Dot product. See you in the next video.
Question: What does the Cross product do with the magnitudes of the original vectors? Answer: The Cross product multiplies the magnitudes of the two original vectors that are perpendicular to each other. | 677.169 | 1 |
11 to both sides.
Combine like terms.
So the equation that goes through the points and is
Pythagorean-theorem/551114: In a 3-D setting:
Given 1 leg and the Hypotenuse, calculate the missing leg. HOW TO DO!?!
I was out sick from school right before winter break and I missed an exam! I completely forgot! Then on the Thursday that I returned on, I did not have Math class that day and Friday was a short day with no academics. Now, I had all week to study for this test but I'm buckling down right now! If i don't get a good grade on this exam I will look foolish and be disappointed! After all, A week to study for a test?!? Along with extra time before this week!?! I have to get a good grade other wise that is just not acceptable! 1 solutions Answer 359402 by jim_thompson5910(28595) on 2011-12-30 14:49:19 (Show Source):
Since the center is the midpoint of the line segment with endpoints (3,5) and (1,2), we need to find the midpoint.
X-Coordinate of Midpoint =
Since the x coordinate of midpoint is , this means that
Y-Coordinate of Midpoint =
Since the y coordinate of midpoint is , this means that
So the center is the point (2, 7/2)
---------------------------------------------------
Now let's find the radius squared
Use the formula , where (h,k) is the center and (x,y) is an arbitrary point on the circle.
In this case, and . Also, and . Plug these values into the equation above and simplify to get:
So because , , and , this means that the equation of the circle that passes through the points (3,5) and (1,2) (which are the endpoints of the diameter) is
.
Linear-equations/551090: 1. Write the equation of the line passing through the points (0, -4) and (-2, 2. Write the equation of the line with a slope of -5 and a y-intercept of (0,3).
3. Write the equation of the line passing through the points (0, -4) and (-2, 4. Write the equation of the line passing through the points (-6,1) and (-4,2).
Question: What is the formula to find the midpoint of a line segment? Answer: (X-Coordinate of Midpoint = (x1+x2)/2, Y-Coordinate of Midpoint = (y1+y2)/2)
Question: What is the radius squared of the circle? Answer: 25/4 | 677.169 | 1 |
So when mathematicians talk about a Möbius strip, or any surface in general there is no "side" of the surface rather it is a set of points that looks locally like the plane. So if you make a paper model of a Möbius band, mathematically we consider a point on one side of the paper to be the same point as the one on the other side of the paper. If you consider them different, we might say you are actually working on the orientable double cover of the Möbius strip. The Möbius strip is an example of a non-orientable surface, which formalizes the notion of being "one sided".
Also, since the Möbius strip is usually defined in the context of topology (as opposed to having a specific geometry) there is no real notion of lines, and thus no good notion of "parallel".
That's one point of view. Another perfectly legitimate point of view is that the midline is of length 1. The difference is that one point of view (midline is of length 2) thinks of the midline as being painted on the surface of the strip, and the other point of view (midline is of length 1) thinks of the midline as being baked into the strip itself.
Well, a definition of parallel in Euclidean geometry is two lines that do not intersect at any points. Since you'd be taking the same line and checking for intersections to itself then it's definitely not parallel since it intersects itself infinitely many times.
However, I'm not sure if ideas of Euclidean geometry are valid for a Mobius strip.
Almost surely no because the point on the flipside isn't on the same parallel line (If you do it by starting with a rectangle and define parallel in this case for loops by being the ones that are perpendicular to the edge). The only way it'd be only that same line is if it was in the center of the strip.
Well i can't really ask if the line would be parallel to itself, even though that's what i really mean. so pretty much a segment is what i mean.
imagine this: They're building a Halo for us to live on, but to increase the naturalism they turn it in to a large mobius strip that has no space under the land.
you obviously can't see what's on the "other side" of where you're standing, but you have a cabin set up on the midpoint of a segment of the Halo and you want one exactly underneath it. Would you ask if the new cabin underneath the one you already had was parallel to your old cabin or would you treat it as a point on a segment being a certain distance from the other cabin?
That's why I'm so interested with Mobius strips. They are a three dimensional representation of a 2d object, i can't call the cabins anything but collinear, even though they behave like they are coplanar, and even if you rationalized that you were at the exact opposite point of a small segment of the strip, it is only true to that local area.
Question: What is the relationship between two cabins placed at the midpoint and directly below each other on a Möbius strip? Answer: They are collinear, meaning they lie on the same line.
Question: What is the key difference between the two points of view regarding the length of the midline of a Möbius strip? Answer: One point of view considers the midline as being painted on the surface of the strip (length 2), while the other considers it as being baked into the strip itself (length 1). | 677.169 | 1 |
To make them same, b^2 must turn negative, which is not possible... unless b is complex! Is that what you mean?Take an ellipse. Fix one focus, and drag the other focus away (keeping the same directrix).
When the prodigal focus reaches infinity, you have a parabola …
now let the prodigal focus come back from infinity, on the other side of the directrix …
you have a hyperbola.
Now rejoice, and kill the fatted calf!
You get the same thing if you produce the figures with a plane cutting a conic … swivel the plane round, keeping the "apex" of the conic in the same place, and you see the ellipse turning into a parabola and then a hyperbola.
And you can also do it using caustic curves (of light), but I can't remember how off-hand.Yes … if you use the inversion z goes to 1/z, then the origin goes to the whole of the "circle at infinity", so if you define that circle to be one point, then the inversion is one-to-one.
I think this is a an exampe of a general technique (which, once again , I can't remember the details of ).
might be something to do with conformal maping or projective geometry …
Question: What is the name of the curve that results from the intersection of a plane and a conic section? Answer: The curve is called a conic section. | 677.169 | 1 |
Circles
In the Circles lesson, you will first learn what are circles as well as the equation of a circle and how to graph a circle. The special case of when the center is not at the origin is also investigated before four video examples round out this lecture.
This content requires Javascript to be available and enabled in your browser.
Circles
Use symmetry to help you graph a circle.
Review completing the square and do so to put the equation of a circle in standard form.
Circles
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
Question: What is suggested to help in graphing a circle? Answer: Use symmetry | 677.169 | 1 |
Question:
Is there a name for the shape that is created when two
circles intersect? It looks to be a pointed ellipse - but does an
official geometric name for the shape exist?
Replies:
Betty,
I believe you are referring to the "VESICA". Vesicas are the "eye-ball"
shaped patterns formed when two or more circles intersect each other. They
are also known to be present when looking at the "UFO - style crop circles."
I believe the shape of the vesica also appears in 2-dimensional space when
two waves of the same phase and amplitude collide. These "vesica" appear as
the crests and the valleys (maxs and mins) of the CONstructive and
DEstructive interference of two similar waves.
Question: In which other scenario can this shape appear? Answer: When two waves of the same phase and amplitude collide | 677.169 | 1 |
I knew dot and cross product, but because I write code so I need the simplest way to boost performance. As in 2D case I don't need to calculate anything, just use the trick. Also that it works for normalized vectors which doesn't need square root, a slow operation. I hope there are some tricks like that in 3D
Find orthogonal vector to current vector in 3D
There isn't a unique vector orthogonal to a given vector in 3D. If the vector doesn't need to have any other properties, the same "trick" works. A vector orthogonal to (a, b, c) is (-b, a, 0), or (-c, 0, a) or (0, -c, b).
But if you want a unit orthogonal vector, you will have to use something like a square root.
Question: Is it necessary to use a square root operation to find a unit orthogonal vector? Answer: Yes. | 677.169 | 1 |
Example of Riks Method with a Geometric Nonlinearity Finite Element Problem This is a 2d ***ysis of a curved beam fixed at its ends. The dimensions of the cross section are 1 mm x 1mm. The Radius of the inner curve is 20mm. The total angle of the beam is 60 degrees. The Force in the graph is the load in the center of top 3 nodes of the beam. The Displacement is the displacement of the center of the beam. The reason this problem has significance is because it is geometrically nonlinear. If you try to solve this in your FEA program, it may fail because Force vs Displacement is not monotonically increasing. Force in most mechanically nonlinear-static finite element problems is similar to time in numerically solving Differential Equation. Time is always monotonically increasing so this type of numerical problem is not an issue for a dynamic solver. Most FEA programs incrementally add force which is fine until the the stiffness of the beam goes to zero (ie slope of force vs displacement is zero); here the FEA solver will normally fail unless you use displacement as your monotonically increasing variable or you use Riks Method. Riks Method uses the arc length of Force vs Displacement as the monotonically increasing variable. Note that in this ***ysis, time has no real meaning. It is a quasi-static ***ysis.
Evil explained geometrically - 6 Simple explanation of evil with the use of geometric principles found in the Metatron's Cube.
Precalculus - The Geometric Representation of Vectors Free Math Help at Brightstorm! How to understand what vectors are, and how they can be represented geometrically.Janosh: Meditation | Rebirth Janosh is a Dutch artist who creates contemporary, geometric art and gets inspired by ancient civilizations, philosophy and geometric proportions as these appear all over the world. The entire universe is shaped according to set, geometric proportions; from the alignment of the stars and the leafs on trees to architectural structures and the human body. Because our cells are geometrically structured too, focussing on geometric shapes has a profound effect. And, as Greek philosopher Plato stated: "Geometry will draw the soul towards the truth." With his art and accompanying multimedia presentations, Janosh wants to inspire people to discover their power and follow their passion in order to turn dreams into actions and actively determine the course of their life. More info:
Question: What is the variable used in Riks Method that is monotonically increasing? Answer: The arc length of Force vs Displacement | 677.169 | 1 |
It is a data sufficiency problem, so you don't have to solve. You only have to know what you need in order to solve. So far it sounds like you have realized that you need both statements in order to deduce that x is the 90 degree corner. You know you have the length of the two legs, so you know you can solve the hypotenuse; so you can solve perimeter.
If you were trying to solve for the actual value of the hypotenuse, use the pythagorean theorem. a^2 +b^2 = c^2
Question: What is already known about the legs of the triangle? Answer: The length of the two legs is known. | 677.169 | 1 |
Projective Geometry/Classic/Projective Transformations/Transformations of the projective line
Projective transformation
Let X be a point on the x-axis. A projective transformation can be defined geometrically for this line by picking a pair of points P, Q, and a line m, all within the same x-y plane which contains the x-axis upon which the transformation will be performed.
Points P and Q represent two different observers, or points of view. Point R is the position of some object they are observing. Line m is the objective world which they are observing, and the x-axis is the subjective perception of m.
Draw line l through points P and X. Line l crosses line m at point R. Then draw line t through points Q and R: line t will cross the x-axis at point T. Point T is the transform of point X [Paiva].
Analysis
The above is a synthetic description of a one-dimensional projective transformation. It is now desired to convert it to an analytical (Cartesian) description.
Let point X have coordinates (x0,0). Let point P have coordinates . Let point Q have coordinates . Let line m have slope m (m is being overloaded in meaning).
The slope of line l is
so an arbitrary point (x,y) on line l is given by the equation
,
On the other hand, any point (x,y) on line m is described by
The intersection of lines l and m is point R, and it is obtained by combining equations (1) and (2):
Joining the x terms yields
and solving for x we obtain
x1 is the abscissa of R. The ordinate of R is
Now, knowing both Q and R, the slope of line n is
We want to find the intersection of line n and the x-axis, so let
The value of λ must be adjusted so that both sides of vector equation (3) are equal. Equation (3) is actually two equations, one for abscissas and one for ordinates. The one for ordinates is
Solve for lambda,
The equation for abscissas is
which together with equation (4) yields
which is the abscissa of T.
Substitute the values of x1 and y1 into equation (5),
Dissolve the fractions in both numerator and denominator:
Simplify and relabel x as t(x):
t(x) is the projective transformation.
Transformation t(x) can be simplified further. First, add its two terms to form a fraction:
Inverse transformation
It is clear from the synthetic definition that the inverse transformation is obtained by exchanging points P and Q. This can also be shown analytically. If P ↔ Q, then α → α′, β → β′, γ → γ′, and δ → δ′, where
Therefore if the forwards transformation is
then the transformation t′ obtained by exchanging P and Q (P ↔ Q) is:
Then
.
Dissolve the fractions in both numerator and denominator of the right side of this last equation:
.
Question: What is the inverse transformation of t(x)? Answer: t'(x) = (α' x - β' x1) / (γ' x - δ' x1)
Question: What are the coordinates of point X in the given problem? Answer: (x0,0) | 677.169 | 1 |
Self-Check Quizzes
randomly generates a self-grading quiz correlated to each lesson in your
textbook. Hints are available if you need extra help. Immediate feedback
that includes specific page references allows you to review lesson skills.
Choose a lesson from the list below.
Evaluate numerical expressions and simplify algebraic expressions by applying the correct order of operations and the properties of rational numbers (e.g., identity, inverse, commutative, associative, distributive). Justify each step in the process.
Know and understand the Pythagorean Theorem and use it to find the length of the missing side of a right triangle and the lengths of other line segments. Use direct measurement to test conjectures about triangles.
Students compare units of measure and use similarity to solve problems. They compute the perimeter, area, and volume of common geometric objects and use the results to find measures of less regular objects.
Use formulas for finding the perimeter and area of basic two-dimensional shapes and the surface area and volume of basic three-dimensional shapes, including rectangles, parallelograms, trapezoids, triangles, circles, right prisms, and cylinders.
Question: Which property of rational numbers is NOT mentioned? A) Commutative B) Associative C) Distributive D) Transitive Answer: D) Transitive | 677.169 | 1 |
An ellipse, informally, is an oval or a "squished" circle.
In "primitive" geometrical terms, an ellipse is the figure you can draw in the sand by
the following process: Push two sticks into the sand. Take a piece of string and form a loop that
is big enough to go around the two sticks and still have some slack. Take a third stick, hook it
inside the string loop, pull the loop taut by pulling the stick away from the first two sticks,
and drag that third stick through the sand at the furthest distance the loop will allow. The resulting
shape drawn in the sand is an ellipse.
Each of the two sticks you first pushed into the sand is a "focus" of the
ellipse; the two together are called "foci" (FOH-siy). If you draw a line in the sand
"through" these two sticks, from one end of the ellipse to the other, this will mark
the "major" axis of the ellipse. The points where the major axis touches the ellipse
are the "vertices" of the ellipse. The point midway between the two sticks is the "center" of the
ellipse.
If you draw a line through this center, perpendicular to the major
axis and from one side of the ellipse to the other, this will mark the "minor" axis.
The points where the minor axis touches the ellipse are the "co-vertices". A half-axis, from
the center out to the ellipse, is called a "semi-major" or a "semi-minor" axis,
depending on which axis you're taking half of.
ADVERTISEMENT
The distance from the center to either focus is the fixed value
c. The distance
from the center to a vertex is the fixed value a. The values of a and c will vary from one ellipse to another, but they are fixed for any given ellipse. I keep
the meaning of these two letters straight by mispronouncing the phrase "foci for c" as "FOH-ciy
foh SEE", to remind me that c relates to the focus. Then the other letter (a) is for the other type of point (the vertex).
The length of the semi-major axis is a and the length of the whole major axis is
2a, and
the distance between the foci is 2c. Okay, so now we've got relationships for a and c, which leads one to wonder, "What happened to b?" The three letters are related by the
equation b2 = a2
– c2 or, alternatively (depending on your
book or instructor), by the equation b2
+ c2 = a2.
(Proving the relationship requires pages and pages of algebraic
computations, so just trust me that the equation is true. It can also be shown — painfully — that
b is also
the length of the semi-minor axis, so the distance across the ellipse in the "shorter"
direction is 2b.
Question: What is the line drawn from one end of the ellipse to the other through the foci called? Answer: Major axis | 677.169 | 1 |
Proposition 25
The rectangle contained by medial straight lines commensurable in square only is either rational or medial.
Let the rectangle AC be contained by the medial straight lines AB and BC which are commensurable in square only.
I say that AC is either rational or medial.
Describe the squares AD and BE on AB and BC. Then each of the squares AD and BE is medial.
Set out a rational straight line FG. Apply the rectangular parallelogram GH to FG equal to AD, producing FH as breadth, apply the rectangular parallelogram MK to HM equal to AC, producing HK as breadth, and further apply similarly NL to KN equal to BE, producing KL as breadth. Then FH, HK, and KL are in a straight line.
Since each of the squares AD and BE is medial, and AD equals GH while BE equals NL, therefore each of the rectangles GH and NL is also medial.
And they are applied to the rational straight line FG, therefore each of the straight lines FH and KL is rational and incommensurable in length with FG.
But the rectangle FH by KL is rational, therefore the square on HK is also rational.
Therefore HK is rational. And, if it is commensurable in length with FG, then HN is rational, but, if it is incommensurable in length with FG, then KH and HM are rational straight lines commensurable in square only, and therefore HN is medial.
Question: Are the straight lines FH and KL commensurable in length with FG? Answer: Incommensurable | 677.169 | 1 |
Using Matrices
posted on: 14 May, 2012 | updated on: 07 Sep, 2012
Matrices are a Set of Numbers or we can say it is a collection of number which is arranged in rows and column. The numbers which we are using in matrices are Real Numbers. Generally in the matrices the complex numbers are used. With the help of matrices we can solve many problems. Here we will focus on Area of Triangle by using matrices.
We can find the area of triangle by using matrices, which is one of the major applications of matrices.
Now we see how to find the area of a triangle using matrices?
We have to follow some steps for finding the area of triangle by use of matrices:
Step1: First we take a triangle.
Step2: Then we find all the coordinates of vertices of a triangle.
Step3: Then we put all the coordinates in the matrices form and then solve Using Matrices to find the coordinates of a triangle.
We know that the area of triangle is:
Area of triangle = ½ | p1 q1 1|
| p2 q2 1|
| p3 q3 1|
Now we see how to find the area of a triangle using matrices:
Suppose we have the values of vertices of triangle are: (2, 3), (4, 2), (3, 7), here we have to find the area of triangle using matrices method.
For finding the area of triangle using matrices we have to follow above steps:
We know that the vertices of a triangle are:
(p1, q1), (p2, q2) and (p3, q3).
Step1: First we have triangle coordinates are: (2, 3), (4, 2), (3, 7);
Step2: Now put these coordinates in the given formula:
And we know that the area of triangle is:
Area of triangle = ½ | p1 q1 1|
| p2 q2 1|
| p3 q3 1|
Now putting the value of all coordinates in the given formula:
Area of triangle = ½| 2 3 1|
| 4 2 1|
| 3 7 1|
Now we have to solve the determinant values:
Area of triangle = ½ [2(2 – 7) - 3(4 – 3) + 1(28 – 6)];
Area of triangle = ½ [2(-5) – 3(1) + 1(22)];
Area of triangle = ½ [-10 - 3 + 22];
Area of triangle = ½ [-13 + 22];
Area of triangle = ½[9].
Area of triangle = 4.5 Square inch.
So we get area of triangle equals to 4.5 by using matrices.
Question: Which vertices of a triangle are represented by (p1, q1), (p2, q2), and (p3, q3)? Answer: The coordinates of the vertices of the triangle.
Question: What is the formula given in the text to calculate the area of a triangle using matrices? Answer: Area of triangle = ½ | p1 q1 1| | p2 q2 1| | p3 q3 1|
Question: What is one of the major applications of matrices mentioned in the text? Answer: Finding the area of a triangle.
Question: What is the definition of a matrix according to the text? Answer: A matrix is a set or collection of numbers arranged in rows and columns, typically consisting of real numbers, but can also include complex numbers. | 677.169 | 1 |
I've been reading about stereographic projections. I did a problem about finding the stereographic projection of a cube inscribed inside the Riemann sphere with edges parallel to the coordinate axes. This was simple since the 8 vertices have coordinates $(\pm a,\pm a,\pm a)$, with $3a^2=1$.
Trying it with a regular tetrahedron is a little tougher for me. If a regular tetrahedron is inscribed in the Riemann sphere in general position, with two vertices $(x_1,x_2,x_3)$ and $(x'_1,x'_2,x'_3)$, is there some way to compute the coordinates of the other 2 vertices in terms of $x_1,x_2,x_3,x'_1,x'_2,x'_3$ in order to compute the stereographic projection of the vertices? How could this be done otherwise, if not?
One vertex won't quite determine the coordinates of the other vertices—the tetrahedron can be rotated about the altitude from the known vertex to the opposite face. The coordinates will all lie on the circle that is the intersection of the sphere with a plane orthogonal to the diameter of the sphere through the known vertex, some specific distance from that vertex (which can be found using right triangles, but I don't have scratch paper in front of me at the moment). – IsaacJan 18 '12 at 23:40
@Isaac, Thanks, I see that now, darn it. If you know the coordinates of two points are then, is it possible to find the other two in relation to them? – JPanJan 19 '12 at 0:05
Yes, and probably a good bit easier, though I'm guessing a little off the top of my head. Any two vertices are the endpoints of a single edge. The other two vertices are the endpoints of an edge that lies in the plane that perpendicularly bisects the first edge, so that puts the points on the circle that is the intersection of that plane with the sphere. The two unknown vertices are the same distance apart as the two known vertices and together with the midpoint of the known-vertices-edge form an isosceles triangle. I think that's sufficient to find them... – IsaacJan 19 '12 at 0:10
Note that the vertices of a regular tetrahedron form a subset of the vertices of a cube. So, if you just want some tetrahedron, then take the coordinates $(\pm a, \pm a, \pm a)$ of your cube such that an even number of "$\pm$"s are "$-$". Granted, the tetrahedron isn't general position; but, then, neither was the cube in the first problem. – BlueJan 22 '12 at 6:11
2 Answers
Question: How can the coordinates of the other two vertices of a regular tetrahedron be found, given the coordinates of two points? Answer: By using the fact that the unknown vertices lie on the circle that is the intersection of the plane that perpendicularly bisects the known edge with the sphere, and that they form an isosceles triangle with the midpoint of the known edge.
Question: If two vertices of a regular tetrahedron are known, can the other two be found? Answer: Yes, given the coordinates of two points, the other two vertices can be found. | 677.169 | 1 |
Given your two vertices of the tetrahedron, $A=(x_1,x_2,x_3)$ and $B=(x_1',x_2',x_3')$, the plane that perpendicularly bisects the edge determined by $A$ and $B$, which contains the other two vertices, has equation $$X\cdot(A-B)=\left(\frac{A+B}{2}\right)\cdot(A-B)$$ (where $\cdot$ is the vector dot product and $X$ is a general point in 3-space). The other two vertices must also lie on the unit sphere, $$|X|=1.$$ And, the other two vertices must be the same distance from both $A$ and $B$ as $A$ and $B$ are from one another: $$|A-X|=|A-B|$$ $$|B-X|=|A-B|.$$
Solving all four (or perhaps any three) of these equations simultaneously should give the coordinates of the other two vertices. I strongly doubt that this yields anything nice in the fully general case. Even putting one vertex at $(0,0,1)$ and letting a second vertex have $y$-coordinate $0$, the expressions for the coordinates of the vertices weren't pretty. (Numerical approximations: $(0.970984, 0, -0.239146)$, $(-0.485492, 0.840896, -0.239146)$, and $(-0.485492, -0.840896, -0.239146)$.)
Given tetrahedral vertices $P$ and $Q$ on a unit sphere centered at the origin $O$, let $M$ be the midpoint of edge $PQ$, and let $N$ be the reflection of $M$ in the origin. (That is, $N := -\frac{1}{2}(P+Q)$.) Then, the tetrahedron's other vertices, $R$ and $S$, determine vectors $NR$ and $NS$ that are (1) perpendicular to plane $OPQ$ (and, hence, are parallel to the vector $P\times Q$), and (2) of length $\frac{1}{2}|PQ|$.
Noting that $\cos\angle POQ =-\frac{1}{3}$ and $|PQ| = \frac{2\sqrt{2}}{\sqrt{3}}$ [*], we deduce that vertices $R$ and $S$ have the form
You can sanity-check this using your cube coordinates, as in my comment on your question. If $P := (a,a,a)$ and $Q := (a,-a,-a)$ with $3a^2=1$, then the formula gives vertices $(-a,a,-a)$ and $(-a,-a,a)$.
[*] You can derive the values of $\cos\angle POQ$ and $|PQ|$ using the cube coordinates.
Question: What is the length of the vectors NR and NS? Answer: The vectors NR and NS have a length of $$\frac{1}{2}|PQ|$$.
Question: What is the equation of the plane that perpendicularly bisects the edge determined by vertices A and B in the tetrahedron? Answer: The equation of the plane is $$X\cdot(A-B)=\left(\frac{A+B}{2}\right)\cdot(A-B)$$ where X is a general point in 3-space, and A and B are the given vertices. | 677.169 | 1 |
circle
A circle is a simple shape of Euclidean geometry consisting of those points in a plane which are equidistant from a given point called the centre (British English) or center (American English). The common distance of the points of a circle from its centre is called its radius.
Circles are simple closed curves which divide the plane into two regions, an interior and an exterior. In everyday use, the term "circle" may be used interchangeably to refer to either the boundary of the figure (also known as the perimeter) or to the whole figure including its interior. However, in strict technical usage, "circle" refers to the perimeter while the interior of the circle is called a disk. The perimeter of a circle is also known as the circumference, especially when referring to its length.
Further terminology
The diameter of a circle is the length of a line segment whose endpoints lie on the circle and which passes through the centre of the circle. This is the largest distance between any two points on the circle. The diameter of a circle is twice its radius.
As well as referring to lengths, the terms "radius" and "diameter" can also refer to actual line segments (respectively, a line segment from the centre of a circle to its perimeter, and a line segment between two points on the perimeter passing through the centre). In this sense, the midpoint of a diameter is the centre and so it is composed of two radii.
A chord of a circle is a line segment whose two endpoints lie on the circle. The diameter, passing through the circle's centre, is the longest chord in a circle. A tangent to a circle is a straight line that touches the circle at a single point, thus guaranteeing that all tangents are perpendicular to the radius and diameter that stem from the corresponding contact point on the circumference. A secant is an extended chord: a straight line cutting the circle at two points.
An arc of a circle is any connected part of the circle's circumference. A sector is a region bounded by two radii and an arc lying between the radii, and a segment is a region bounded by a chord and an arc lying between the chord's endpoints.
History
The etymology of the word circle is from the Greek, kirkos "a circle," from the base ker- which means to turn or bend. The origin of the word "circus" is closely related as well.
The circle has been known since before the beginning of recorded history. Natural circles would have been observed, such as the Moon, Sun, and a short plant stalk blowing in the wind on sand, which forms a circle shape in the sand. The circle is the basis for the wheel, which, with related inventions such as gears, makes much of modern civilization possible. In mathematics, the study of the circle has helped inspire the development of geometry, astronomy, and calculus.
Length of circumference
The ratio of a circle's circumference to its diameter is π (pi), an irrationalconstant that takes the same value (approximately 3.141592654) for all circles. Thus the length of the circumference (c) is related to the radius (r) by
Question: If the radius of a circle is 5 units, what is the approximate length of its circumference? Answer: Approximately 31.4159 units.
Question: Which of the following is NOT a part of a circle? A) Chord B) Tangent C) Square D) Arc Answer: C) Square | 677.169 | 1 |
draw a point along selected polyline you can use Divide command or Measure command or use the Auto Lips to combine the steps to a new Lisp routine. To give the length of polyline you can refer this following file: Hope this helps Cheer!!
I want to know the length of a line from the begin till the point where
I click or where the line cross a other line.
I can get the line and i can find where the intersection points are
with VBA but I can NOT find a function that calculate the the exact
length from begin to the intersection.
Can anyone help me to find the good way.
Question: What specific function is the user unable to find in VBA to calculate the length from the beginning to the intersection? Answer: A function to calculate the exact length from the beginning to the intersection. | 677.169 | 1 |
Conic Section Explorer
Explore the different conic sections and their graphs. Use the Cone View to manipulate the cone and the plane creating the cross section, and then observe how the Graph View changes.
Instructions
Use the tools along the bottom to manipulate the parameters of the double-napped cone and the plane creating the cross section:
Increase or decrease Height to make the cone taller or shorter.
Increase or decrease Slant to make the cone wider or narrower.
Increase or decrease m to change the angle of the plane.
Increase or decrease b to change the distance between the cone's center and the plane.
Click Reset Parameters to reset all values to 1.
In the Cone View:
Click and drag in the window to change the perspective.
Click the magnifying glasses to zoom in and out, or Reset to see the default view.
In the Graph View:
Click and drag in the window to move the center of the graph.
Click the magnifying glasses to zoom in and out, or Reset to see the default view.
Exploration
Change the parameters and manipulate the Cone View and Graph View to find the different conic sections. As you explore, try to answer these questions:
How many different conic sections are there?
How do the different conic sections look different in the Cone View? in the Graph View?
What range or values do the parameters have for each conic section?
What are the key features of each conic section?
Can you create two identical Cone Views with different sets of parameters? Is there a relationship in the parameters for these two views? Do the Graph Views also look the same?
When the cross section is an X, a line, or a point, it is called a "degenerate" case because the graphs are not unique to conic sections. Can you create these graphs? What are the parameters to create them
Question: What is the effect of increasing the 'Slant' parameter on the cone's width? Answer: Increasing the 'Slant' parameter makes the cone wider.
Question: Which parameter can be adjusted to create a 'degenerate' case where the cross section is a point? Answer: Adjusting the 'b' parameter to a very large value can create this case. | 677.169 | 1 |
Florida - Mathematics: Geometry
This correlation lists the recommended Gizmos for this state's curriculum standards. Click any Gizmo title below to go to the Gizmo Details page.
MA.912.G: Geometry
MA.912.G.1: Understand geometric concepts, applications, and their representations with coordinate systems. Find lengths and midpoints of line segments, slopes, parallel and perpendicular lines, and equations of lines. Using a compass and straightedge, patty paper, a drawing program or other techniques, students also construct lines and angles, explaining and justifying the processes used.
MA.912.G.1.1: Find the lengths and midpoints of line segments in two-dimensional coordinate systems.
MA.912.G.1.2: Construct congruent segments and angles, angle bisectors, and parallel and perpendicular lines using a straight edge and compass or a drawing program, explaining and justifying the process used.
MA.912.G.2: Identify and describe polygons (triangles, quadrilaterals, pentagons, hexagons, etc.), using terms such as regular, convex, and concave. Find measures of angles, sides, perimeters, and areas of polygons, justifying the methods used. Apply transformations to polygons. Relate geometry to algebra by using coordinate geometry to determine transformations. Use algebraic reasoning to determine congruence, similarity, and symmetry. Create and verify tessellations of the plane using polygons.
MA.912.G.2.2: Determine the measures of interior and exterior angles of polygons, justifying the method used.
MA.912.G.2.4: Apply transformations (translations, reflections, rotations, dilations, and scale factors) to polygons to determine congruence, similarity, and symmetry. Know that images formed by translations, reflections, and rotations are congruent to the original shape. Create and verify tessellations of the plane using polygons.
MA.912.G.4: Identify and describe various kinds of triangles (right, acute, scalene, isosceles, etc.). Define and construct altitudes, medians, and bisectors, and triangles congruent to given triangles. Prove that triangles are congruent or similar and use properties of these triangles to solve problems involving lengths and areas. Relate geometry to algebra by using coordinate geometry to determine regularity, congruence, and similarity. Understand and apply the inequality theorems of triangles.
MA.912.G.5: Apply the Pythagorean Theorem to solving problems, including those involving the altitudes of right triangles and triangles with special angle relationships. Use special right triangles to solve problems using the properties of triangles.
MA.912.G.5.1: Prove and apply the Pythagorean Theorem and its converse.
MA.912.G.6: Define and understand ideas related to circles (radius, tangent, chord, etc.). Perform constructions, and prove theorems related to circles. Find measures of arcs and angles related to them, as well as measures of circumference and area. Relate geometry to algebra by finding the equation of a circle in the coordinate plane.
Question: Which of the following is a special right triangle? A) Isosceles B) Scalene C) Equilateral D) All of the above Answer: C) Equilateral
Question: Which of the following is NOT a topic covered in the Massachusetts geometry standards for grade 9? A) Congruent segments B) Gravity C) Tessellations D) Symmetry Answer: B) Gravity
Question: What is the name of the theorem that relates the sides of a right triangle? Answer: Pythagorean Theorem
Question: Which of the following can be used to construct lines and angles? A) Ruler only B) Compass only C) Both a ruler and compass or a drawing program D) None of the above Answer: C) Both a ruler and compass or a drawing program | 677.169 | 1 |
Unit Circle: One of the most basic things you need to know regarding trigonometry if how to find the Sine and Cosine of angles which are multiples of π/6 (30º) or π/4 (45º). The easiest way to accomplish this is to learn the unit circle.
Graphing: You need to be able to recognize and find equations for graphs of all the trig functions. This requires understanding amplitude and phase shift as well as having the basic graphs memorized. Notice that if you know the unit circle, you can use this information to find the important points on the basic graphs
Question: What can the knowledge of the unit circle help with in finding the basic graphs? Answer: It can help in finding the important points on the basic graphs. | 677.169 | 1 |
A parallelogram is a quadrilateral (polygon with four straight sides) with the opposite sides parallel. The typical parallelogram is shown in the Figure to the left along with other types of quadrilaterals.
A rectangle, a rhombus, a square are all parallelograms, too. They are special kinds of parallelograms. The bottom figures show a trapezoid, a kite, and a general quadrilateral that are NOT parallelograms, just for example.
Lessons under this topic consider different properties of parallelograms
Question: What are the opposite sides of a parallelogram? Answer: The opposite sides of a parallelogram are parallel to each other. | 677.169 | 1 |
Geometry
Geometry helps people describe the world around them through the study of geometric shapes, structures, and their characteristics and relationships. Through the study of geometry, students will learn about shapes and dimensions around them and how to analyze spatial relationships in everyday life. Spatial visualization—building and manipulating mental representations of two-and three-dimensional objects and perceiving an object from different perspectives—is an important aspect of geometric thinking. An understanding of the attributes and relationships of geometric objects can be applied to diverse contexts.
In the third grade, students begin a study of the area for a particular region that can be covered by a certain quantity of same-sized units of area. They also study perimeter. Through the study of perimeter, students not only can practice measuring in various units, but in adding and subtracting those units. Just as area connects to multiplication and division, perimeter connects to addition and subtraction. Letting students measure items in the classroom or around the school, sketch the item and label the measurements they made, and then solve problems involving those measurements is one way students can experience and gain a richer understanding of these geometric concepts. The use of geoboards is a way students can compare and contrast areas and perimeters.
Students begin to develop an initial understanding of a family of shapes, such as quadrilaterals. They apply their prior knowledge as well as new knowledge of the attributes that certain shapes possess. This will help students develop precision in describing the properties of geometric objects. They can articulate reasons and informal arguments about how certain shapes are related.
Question: What is one of the key aspects of geometric thinking? Answer: Spatial visualization. | 677.169 | 1 |
Geometry 1.1Calculate volume and surface area to blast away rocks and save your space ship! Galactic Geometry is an engaging 3D environment for learning about geometric figures and their equations. Students practice math with vivid animation and sound. Get into the pilot seat and learn about volume and surface area while blasting away space debris to save your ship! Galactic Geometry is a 3D educational game that offers an engaging environment for serious learning about geometric figures. AsGeo-Calc is a geometry calculator which computes the area, perimeter, sides, angles and the height of common 2D figures such as Ellipses, Rectangles, Triangles, Trapeziums, Parallelograms and Polygons, and the volume, surface, and sides of 3D figures such as Prisms, Cylinders, Cones, Pyramids and Spheres. The application allows you to execute the calculations with every possible combination of entered parameters.
Calculate geometry problems with this tool. Calculate geometry problems with this tool. Geometry Solver 3D will solve analytic geometry problems easily. It will provide tools for calculations in 3D as well as graphic OpenGL demonstrations. Targeted platforms are Linux and Windows both x86...
Far away in the Universe near Andromeda Galaxy Ambient Light born. Far away in the Universe near Andromeda Galaxy Ambient Light born. Few Cosmic objects came across Ambient Light and began fluctuating in his loving rays. The huge harmonic energy released during this process were spreading in all directions and Earn a best score result and send it directly to TopChart section of our website.
Lines is a variation of the famous logic game, presented in a new entertaining way. It is exciting and simple with a diversity of three-dimensional figures and colors. The goal of the game is to score points by aligning figures of the same shape. Try it and you'll love it!
Board logical game. You have found the entertaining board game. This game for everybody. Your task - to move red figures on a place of green. You play by red figures the computer - green. The figures move on rules chess knight. The game develops logic thinking.
Question: What is the name of the geometry calculator mentioned in the text? Answer: AsGeo-Calc. | 677.169 | 1 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.