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You can put this solution on YOUR website! let ABCD be your parallelogram
A is bottom left
B is top left
C is top right
D is bottom right.
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let ABCD lean to the right so that point B is slightly to the right of point A.
all you need is a little tilt to show that it's not a rectangle.
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draw diagonals AC and BD.
AC is the long diagonal and BD is the short diagonal.
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BC congruent to AD (opposite sides of a parallelogram are congruent)
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note:
AC and BD are diagonals of the parallelogram.
they are also transversals that intersect two parallel lines (BC and AD).
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angle ACB congruent to angle CAD (opposite internal angles caused by the intersection of a transversal with two parallel lines are congruent).
likewise, angle DBC congruent to angle BDA.
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you have triangles AED congruent to triangle CEB (ASA)
the ASA is formed by:
side BC congruent to side AD
angle ACB congruent to angle CAD
angle DBC congruent to angle BDA.
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CE is congruent to AE (corresponding parts of congruent triangles are congruent)
BE is congruent to DE (same reason).
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a sketch of the parallelogram can be found at this website:
look for 169616 parallelogram
shouldn't be too hard to find.
it's the only one there.
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Question: Which triangles are congruent in the given parallelogram? Answer: Triangles AED and CEB are congruent by the ASA (Angle-Side-Angle) postulate.
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What does an octahedron have to do with BATSE?
The octahedron represents the Large Area Detector (LAD) geometry when
BATSE is installed on the Compton Gamma-Ray Observatory.
The LADs on each of the eight BATSE modules are oriented so that when the
module is in a horizontal position, the face of each LAD is at an angle of
approximately 54.7 degrees with the vertical direction. This angle is required
so that the planes of the LADs intersect to form a regular right octahedron.
This configuration was chosen so that each gamma-ray burst would be observed
by four detectors simultaneously. There is a special case in which a
burst would be observed by only two detectors - can you think of when this
would occur?
A burst that is directly in front of the midpoint of the edge shared by two
detectors will be observed by only those two detectors. This is a relatively
rare occurrence for BATSE.
Question: In which case would a burst be observed by only two detectors? Answer: When the burst is directly in front of the midpoint of the edge shared by two detectors.
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constructing a line perpendicular to a line from a given point on the line
Answers
I don't really understand what you're asking, but if you're asking how to do it, then you take your compass, set it to some random length, and take that random length on both sides of the line. Then, at the intersections of the circle that you just drew and the line, take another larger random length, and draw circles where your previous circle intersected the line. Where the last two circles intersected, connect that point with the given point on the line, and there you have it. A line perpendicular to a line from a given poitn on the line.
Question: How many circles are drawn in total? Answer: Two.
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Inside this large rectangular border, draw a blob — yes, blob — with an area that's approximately 1/5 of the rectangle's area. No one will die if it's not quite 1/5.
Next, draw a dot anywhere inside the rectangle but outside the blob. Label this dot H.
Now, draw another dot — but listen carefully! — so that there's no direct path from this dot to the first dot H. Label this second dot B.
I asked the class if they knew what they just drew. After a few silly guesses, I told them it was a miniature golf course: blob = water, point B = golf ball, point H = hole location.
The challenge then was to get the ball into the hole. Since you can't putt the ball directly into the hole due to the water hazard, you need to make a bank shot.
(Some students may have drawn the blob and points in such a way that this was not really possible, at least not in one-bank shot. I let them just randomly pull from the stack of copies to pick a different one. I made a copy of their sketches first before they started their work.)
The discussions began as they started drawing in the paths. One student drew hers in quickly and asked, "Is this right?" I replied, "I'm not sure, but that's my challenge to you. You need to convince me and your classmates that the ball hitting the edge right there will bounce out and travel straight into the hole. Does it? What can you draw? What calculations are involved?"
What I heard:
The angle that the ball hits the border and bounces back out must be the same.
Because we're talking about angles, something about triangles.
This is like shooting pool.
Right triangles.
Similar right triangles.
Do we need to consider the velocity of the ball?
This is hard.
I can't figure out how to use the right triangles.
Similar right triangles because that'll make things easier.
Even though it's more than one bounce off the edges, I'm still just hitting the ball one time.
I think I got this.
I have an idea.
Wish my golfer is Happy Gilmore.
BIG struggles, so I was happy and tried not to be too helpful. (I struggled big time too on some of their papers! And I think this made them happy.)
Lauren explained in this 55-second video how she found the paths for the ball to travel. I also had her explain to the whole class later at the document camera.
Jack took a different approach. Instead of measuring the sides and finding proportions to find more sides to create similar triangles like Lauren did, he started with an angle that he thought might work [via eyeballing] and kept having the ball bounce off the borders at paired angle until it went into the hole. (His calculation was off — or his protractor use was inaccurate — as he had angles of 90, 33, and 63. Or maybe if he had a better teacher, he'd know the sum of the interior angles of a triangle was 180.)
Question: What geometric shapes and concepts are mentioned in the text that students might use to solve the challenge? Answer: Right triangles, similar right triangles, and angles are mentioned.
Question: Which student used a different approach to find the path for the ball? Answer: Jack
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It mentions subtracting the pivot point, but should I subtract the distance to the pivot point?
Since my pivot point is $(0,0)$ it sounds too easy, to just subtract 0 (and also doesn't give me the results I expect).
As an example, I have a point in $ (2328.30755138590, 1653.74059364716) $ (very accurate, I know).
I need to rotate it $ 5.50540590872993 $ degrees around $(0,0)$.
I would expect it to end in 2339.68319170805, 1878.18099075262(based on a rotation in my CAD software) but I don't really see how I can get it to do that.
Actually, I need to rotate it around $(0, 1884.25802838571)$. Sorry. I have gotten some coordinate systems mixed.
Please give a specific example of a situation, where you think the formula gives you a wrong result. Then it will be easier to correct misunderstandings. And, "NO!", you are supposed to subtract the pivot as a vector. How would you subtract a distance from a point anyway? They are incompatible data types. – Jyrki LahtonenSep 13 '12 at 10:33
I am sorry, if I am not being clear. I am asking here because my math skills are horrible.
When you say subtract the pivot point as vector, exactly what do you mean? The vector between 0,0 and my point? – NicolaiSep 13 '12 at 10:36
Question: What does Jyrki mean by "subtract the pivot as a vector"? Answer: Jyrki means to subtract the vector from the pivot point to the original point from the original point itself.
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Going Down - Angles Of Depression
Sorry, you need to install Flash and/or enable javascript in your browser to see this content. The latest version of Flash can be found at Adobe's website.
Angles of depression are angles that are measured from the horizontal downwards. Learners should be able to recognise situations where they would use angles of depression, for example, looking down from a height.
Question: What is the definition of angles of depression? Answer: Angles that are measured from the horizontal downwards.
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tan = sine/cosine
cotan = cosine/sine
arcfunctions are easy, they're 1/function (or is that inverse, or are arc and inverse the same?) or it might be arcfunction = function ^ -1? hell i barely remember trig, but thats what reference tables are for.
all that really matters is that you can form all of the trigonometric functions using the two provided and simple algebra.
Question: Can you explain the relationship between 'cotan' and 'tan'? Answer: cotan = cosine/sine, which is the reciprocal of tan.
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Question 68027: CATS is a square. CT = 6 times the square root of 2.
if CT is the diagnal of the square.
what is the distance of CA (any one side of the square).
My son has this problem and we've checked his Geometry book and see no samples that can assist us. Thank you. Click here to see answer by Nate(3500)
Question 69697: What is length of the line segment connecting (2, -2) and (-3, -1)?
Question 70327: to land, an airplane will aproach an airport at a 3 degree angle of depression. If the plane is flying at 30,000ft. fint the ground distance from the airport to the point directly below the plane when the pilot begins descending. give your answer to the nearest 10,000 ft.
Question 72637: Suppose
Question 73882: can someone please help me i have no idea how to solve this...thank very much...if you could help me...sect. 10.3 # 48
Construction. A garden area is 30 ft long and 20 ft wide. A path of uniform width is set around the edge. If the remaining garden area is 400 ft^2, what is the width of the path?
Question 73881: Please help me...step by step ....
sect. 9.6#48
Use the Pythagorean theorem to determine the length of each line segment. Where appropriate,round to the nearest hundredth.
the points in the line are: (1,-1) and (3,2)
Problem # 20 section 9.6
Directions: Find the distance between each pair of points.
(-3,0) and (4,0) Click here to see answer by stanbon(57307)
Question 75111: Find the distance between the points A(4, -3) and B(-4, 3). I came up with:
d=(4-(-4))^2 + (-3-3)^2=10
If points A, B, and C lie on a coordinate line and points A and B have coordinates 15 and 7 respectively, then which of the possible coordinates for point C satisfy(ies) d(A,C)< d(B, C)?
Question 76800: Can someone please help me with this question or a similar one Please?
Question: What is the distance between the points (2, -2) and (-3, -1)? Answer: The distance is 5 units.
Question: What is the distance between the points (1, -1) and (3, 2)? Answer: The distance is approximately 2.24 units.
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(edit added more explanation) I'm trying for the best possible fit. at each angle a different part of the coasts fit, pretty well, but then the other corresponding shorelines, further away, went out of position. All the shoreline shapes, of both continents, would fit together perfectly on a smaller globe.
Remember, the same size continents, on a smaller globe, would have more of a curve in them. For instance, a triangle curved over the surface of a globe, can have 270 degrees in it, instead of just 180 on a plain.
As a balloon is inflated, the curve of any given area decreases. Sort of strange to think, that as a balloon inflates, the curve in its' surface, flattens.
Question: What happens to the curve of any given area on a balloon as it is inflated? Answer: As a balloon is inflated, the curve of any given area decreases.
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where $A$, $B$, $C$ are the areas of the "leg-faces" and $D$ is the area of the "hypotenuse-face".
For right-corner simplices in higher Euclidean dimensions, we have that the sum of the squares of the content of leg-simplices equals the square of the content of the hypotenuse-simplex. (I don't happen to know the non-Euclidean counterparts of this generalization. Perhaps this makes for a good MO question!)
As generalizations of the Pythagorean Theorem for Triangles, I always found these (Euclidean) results to be more satisfying than the diagonal-of-a-box/distance formulas: instead of dealing only with segments, we have that, as the dimension of the ambient space goes up, so does the dimension of the objects involved in the relations.
Question: Which mathematical theorem is the described relationship a generalization of? Answer: The described relationship is a generalization of the Pythagorean Theorem for Triangles.
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It is not certain precisely what statements Euclid assumed for his postulates and
axioms, nor, for that matter, exactly how many he had, for changes
and additions were made by subsequent editors. There is fair evidence, however,
that he adhered to the second distinction and that he probably assumed
the equivalents of the following ten statements, five "axioms," or common
notions, and five geometric "postulates":
A1 Things that are equal to the same thing are also equal to one another. A2 If equals be added to equals, the wholes are equal. A3 If equals be subtracted from equals, the remainders are equal A4 Things that coincide with one another are equl to one another. A5 The whole is greater than the part. P1 It is possible to draw a straight line from any point to any other point. P2 It is possible to produce a finite straight line indefinitely in that straight line. P3 It is possible to describe a circle with any point as center and with a radius equal to any
to finite straight line drawn from the center. P4 All right angles are equal to one another. P5 If a straight line intersects two straight lines so as to make the interior
angles on one side of it together less than two right angles, these
straight lines will intersect, if indefinitely produced, on the side on
which are the angles which are together less than two right angles.
Question: What is the first axiom according to the text?
Answer: A1 - Things that are equal to the same thing are also equal to one another.
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Proposition 78
If from a straight line there is subtracted a straight line which is incommensurable in square with the whole and which with the whole makes the sum of the squares on them medial, twice the rectangle contained by them medial, and further the sum of the squares on them incommensurable with twice the rectangle contained by them, then the remainder is irrational; let it be called that which produces with a medial area a medial whole.
From the straight line AB let there be subtracted the straight line BC incommensurable in square with AB and fulfilling the given conditions.
I say that the remainder AC is the irrational straight line called that which produces with a medial area a medial whole.
Set out a rational straight line DI. Apply DE, equal to the sum of the squares on AB and BC, to DI producing DG as breadth. Subtract DH equal twice the rectangle AB by BC. Then the remainder FE equals the square on AC, so that AC is the side of FE.
Again, since twice the rectangle AB by BC is medial and equals DH, therefore DH is medial. And it is applied to the rational straight line DI producing DF as breadth, therefore DF is also rational and incommensurable in length with DI.
Question: What is the length of DE in the diagram? Answer: DE is equal to the sum of the squares on AB and BC.
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Angular mil
This article describes "mil" as a unit of angle. For alternative meanings, see Mil (disambiguation).
The mil (in full, angular mil) is a unit of angular measure common in the military of many countries. There are three different specifications for the unit, each roughly 1000th of a radian.
The name of the mil comes from the its definition as (approximately) a thousandth of a radian, or milliradian. Its relationship to the radian gives rise to the handy property that object of size S that subtends an angle A mils is at a distance D = 1000·S/A. Alternatively, if the distance is known, we can determine the size of an object by S = A·D/1000. The practical form of this that is easy to remember is: 1 mil at 1 km is about 1 metre. Another example: 100 mils at 2 km is about 200 metres. (No conversion to Imperial units is worthwhile here, for all armies use metric maps, even the U.S. Army.)
In the general case, where neither the distance nor the object size is known, the formulae may be of little use. In practice, sizes of observed objects are known with reasonable accuracy since they are often men, buildings and vehicles. Using the formulae, distances of the objects can be readily calculated without a calculator. In military terms, distances are of course essential for artillery bombardments and estimations of journey times.
Mathematically, recall that the arc that subtends one radian is one radius in length. Thus, one thousandth of this distance subtends a milliradian. The multiplication factor is introduced to avoid fractional measurements for common sizes of observed objects such as men, buildings and vehicles.
The three definitions of the mil
In NATO countries, including Canada, a mil is defined as 1/6,400 of a full circle. There are 1600 mils in 90 degrees, 17.8 mils in one degree. This mil is usually used in artillery discussion. It is also used in long-range precision rifle shooting, where the crosshairs on riflescopes are often calibrated in mils. This type of riflescope is usually referred to as a mil-dot scope.
The army of the Soviet Union used a mil that was 1/6000 of a full circle, which means that there were 1,500 of its mils in a right angle, which would be less accurate though easier to remember.
The military of Sweden during the Cold War desired to demonstrate its independence from both NATO and the Warsaw Pact, so they chose a size of greater accuracy. Because a right angle is more nearly 1.5708 radians than 1.600, their mil (which is called streck, literally "line") was 1/6,300 of a circle, so that there were 1,575 streck in a right
Question: What fraction of a full circle is one streck? Answer: 1/6,300
Question: What is the mil unit of measure used in NATO countries? Answer: 1/6,400 of a full circle
Question: What is the mil unit of measure used in the Soviet Union's army? Answer: 1/6000 of a full circle
Question: How many streck are there in a right angle? Answer: 1,575
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Triangles - Concurrent Lines - Altitudes and Medians
This geometry worksheet contains problems on concurrent lines in triangles. Concepts and vocabulary include points of concurrency, perpendicular bisectors, angle bisectors, altitudes, medians, and centroids. There are also problems on finding the center of a circle that you can circumscribe about a triangle.
# of Problems: 9
# of Pages: 3
Answer Key?: Yes
Sample Problems from Worksheet
(Worksheet pictures and diagrams not shown)
Geometry
Medians & Altitudes
1) The ___________________ of a triangle is a segment from a vertex to the midpoint of the opposite side.
2) The ___________________ of a triangle is a segment from a vertex perpendicular to the line that contains the opposite side.
For each triangle below, draw the median from A and altitude from A.
Complete.
6) If CN = NP, then _________ is a median of CAP
7) If AP is a median of ANB, then ______ = ______.
8) If ∠CAB is a right angle, then ______ and ______
are altitudes of CAB.
Question: What is the total number of problems in the worksheet? Answer: 9
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$ABCD$ is a square with a side of length 4. P is on AB, S is on CD and Q is on PS such that:
$AP = CS$
The triangles $PBR$ and $SDQ$ are both equilateral triangles. See the image below.
Calculate the combined area of the 2 triangles.
What would be the easiest way the solve this? I first tried to name SC x, and then make an equality involving x (something equalling 16, the area of the square, and then solving for that x) but that didn't work out.
Well, it satisfies the conditions of the problem to take $Q=R$ to be the center of the square, in which case the blue triangles have half the area of the square ($8$). So if the problem is well-defined, that must be the answer. – mjqxxxxFeb 15 at 22:07
1 Answer
Let's do it in the style that you did not complete. While it is not the easiest way, it is completely mechanical.
We write down faithfully the first calculation that we did. Then we describe a much more streamlined way of doing the same thing.
Let $a=AP=CS$. We find the equation of the "left" side of the bottom triangle. This passes through $(a,0)$ and has slope $\tan(60^\circ)=\sqrt{3}$.
So the equation is
$$\frac{y-0}{x-a}=\sqrt{3},$$
or equivalently $y=\sqrt{3}x-a\sqrt{3}$.
This line is the same as the "right" side of the upper triangle.
We find the equation of that right side. It has slope $\sqrt{3}$, and passes through the point $(4,4-a)$. So the equation of the right side of the upper triangle is
$$\frac{y-4}{x-4+a}=\sqrt{3},$$
or equivalently $y=\sqrt{3}x +4-4\sqrt{3}+a\sqrt{3}$.
Compare the two equations we have obtained. Their constant terms must be the same. This gives us a linear equation for $a$.
Another way: We want the line through $(a,0)$ and $(4-a,4)$ to have slope $\sqrt{3}$. That yields the equation
$$\frac{4}{4-2a}=\sqrt{3}.$$
It is slightly more convenient to let the side of the triangles be $b$. then we are talking about the line from $(4-b,0)$ to $(b,4)$. We get the equation
$$\frac{4}{2b-4}=\sqrt{3}.$$
Now that we have $b$, we can find the areas of the triangles by using a standard formula.
You are amazing. How do you know the answers to these types of questions so fast? I understand that a lot of math is just very hard work, but geometry is more intuition, and you just do this like it's nothing. Any short tips? – ZafarSFeb 15 at 22:28
Question: What is the side length of the square $ABCD$? Answer: 4
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About the CSEM Club
The purpose of the CSEM Club at CCBC-Catonsville is to develop and nurture interest in mathematics and Computer Science among the students at the Catonsville campus of the Community College of Baltimore County (CCBC) through meetings, informal mutual discussions, field trips, etc. The CSEM Club also pays for the prizes in the annual Catonsville Mathematics Competition. In the past the CSEM Club has hosted several graphing calculator workshops in order to develop skills in using TI-81, 82, 85, and 92 calculators.
The CSEM Club currently sponsors "First Friday", a Mathematics seminar which takes place on the first Friday of every month. The speakers are the faculty members of the Mathematics Department of CCBC-Catonsville and sometimes invited guest speakers from outside the college.
One of the many ways this home page intends to keep interest in mathematics alive in the community is by publishing every month interesting mathematical problems or puzzles that we challenge anyone on the internet to attempt to solve. We welcome you to e-mail your solution to us so that we can publish the names of the people with correct answers together with the solutions the next month.
Problems for the month of July
Let ABC be a triangle on the plane. If isosceles triangles ABD, BCE, and CAF
are drawn on the sides of the triangle ABC such that the angles ÐADB,
ÐBEC, and ÐCFA are all equal to 120°, then
show that the triangle DEF is an equilateral triangle.
Let ABC be a triangle inscribed in a circle. If D is any point on the arc BC, show that the distances of D from the vertices saisfy the relation
AD = BD + CD.
Problems for the month of August
Show that the three medians of a triangle are concurrent.
Show that the bisectors of the angles of a triangle are concurrent.
Show that the perpendicular bisectors of the three sides of a triangle are concurrent.
Show that the three altitudes (perpendiculars from vertices to opposite sides) of a triangle are concurrent.
Problems for the month of September
Show that the circle that passes through the midpoints of the three sides of a triangle also passes through the feet of the perpendiculars from the vertices to the opposite sides.
The orthocenter of a triangle is the point where the altitudes intersect.
Show that the circle in problem 1 also passes through the midpoints of the line segments joining a vertex and the orthocenter. This circle is called the Nine-Point circle of the triangle.
Show that the radius of the nine-point circle of a triangle is half that of the circum-circle of the triangle. The circum-circle of a triangle is the circle that passes through the vertices of the triangle.
Problems for the month of October
Show that the center of the Nine-Point circle, the orthocenter,
and the circum-center of a triangle lie on the same line.
The circum-center is the center of the circum-circle.
The centroid of a triangle is the point where the three medians intersect.
Question: Which months do the club publish mathematical problems or puzzles? Answer: Every month, with the solutions and solvers' names published the next month.
Question: What is the relation between the distances of point D from the vertices in the July problem? Answer: AD = BD + CD
Question: What is the name of the monthly mathematics seminar hosted by the CSEM Club? Answer: "First Friday"
Question: What is the name of the circle that passes through the midpoints of the three sides of a triangle and the feet of the perpendiculars from the vertices to the opposite sides? Answer: The Nine-Point circle
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some ideas By cptjway
You could have the students work in pairs to do a "scavenger hunt" in your room, to find examples of each type of angles. Example: the squares on the floor are right angles, the letter A on the bulletin board shows an acute angle. They could write each example on a sticky, then stick them on the board under labels: right, acute, obtuse, straight.
Or, I have students stand up and show the angles with their arms as I call out different angles.
They could do a sorting activity, either on a smart board or on individual worksheets.
I also teach angles using a "new, handy-dandy highly technical" measuring tool: the index card! It's a right angle if the corner of the card fits exactly; it's acute if you line up one edge on one ray of the angle, and the other edge is hidden under the card; it's obtuse if you line up one edge on a ray, and the other line is visible and open.
Question: What are some methods suggested by the author to teach students about different types of angles? Answer: The methods suggested are a scavenger hunt, using students' arms to demonstrate angles, and a sorting activity.
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Let t and s (t > s) be the sides of the two inscribed squares in a right triangle with hypotenuse c. Then s2 equals half the harmonic mean of c2 and t2.
Let a trapezoid have vertices A, B, C, and D in sequence and have parallel sides AB and CD. Let E be the intersection of the diagonals, and let F be on side DA and G be on side BC such that FEG is parallel to AB and CD. Then FG is the harmonic mean of AB and DC. (This is provable using similar triangles.)
Crossed ladders. h is half the harmonic mean of A and B
In the crossed ladders problem, two ladders lie oppositely across an alley, each with feet at the base of one sidewall, with one leaning against a wall at height A and the other leaning against the opposite wall at height B, as shown. The ladders cross at a height of h above the alley floor. Then h is half the harmonic mean of A and B. This result still holds if the walls are slanted but still parallel and the "heights" A, B, and h are measured as distances from the floor along lines parallel to the walls.
In an ellipse, the semi-latus rectum (the distance from a focus to the ellipse along a line parallel to the minor axis) is the harmonic mean of the maximum and minimum distances of the ellipse from a focus.
The harmonic mean also features in elementary algebra when considering problems of working in parallel.
For example, if a gas powered pump can drain a pool in 4 hours and a battery powered pump can drain the same pool in 6 hours, then it will take both pumps
hours to drain the pool working together.
Another example involves calculating the average speed for a number of fixed-distance trips. For example, if the speed for going from point A to B was 60 km/h, and the speed for returning from B to A was 40 km/h, then the average speed is given by
Question: If a gas-powered pump can drain a pool in 4 hours and a battery-powered pump in 6 hours, how long will it take for both pumps to drain the pool together? Answer: It will take both pumps 4/3 hours to drain the pool working together.
Question: What is the semi-latus rectum in an ellipse in relation to the harmonic mean? Answer: The semi-latus rectum is the harmonic mean of the maximum and minimum distances of the ellipse from a focus.
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repeated with the fixed point outside the circle. The result is an
envelope of tangent lines of the hyperbola with foci at the fixed point
and the center of the circle.
The lemniscate
An interesting extension for the hyperbola is to consider an envelope of
circles generated by taking a variable point on the hyperbola as the center
of a circle that passes through the midpoint of the segment connecting the
foci of the hyperbola
Now, as the variable point (center of the circle) moves on the hyperbola,
an envelope of circles is generated. Click here
to see a GSP animation of this.
The envelope identifies a lemniscate. The lemniscate is the inversion
of the hyperbola in circle. Thus the construction of an envelope of circles
accomplishes an inversion of the curve that is the path of the centers of
the circles.
Moving the fixed point of the envelope of circles to some point other than
the midpoint of the segment connecting the loci of the hyperbola produces
a modified "lemniscate."
To experiment with a GSP sketch where other "centers" can be
explored, click here or click here
for a movie illustrating the same idea. For example, locating the fixed
point of the envelope at either foci generates an envelope that defines
a circle. A kidney shape results when the fixed point is on the hyperbola
but not on the line segment joining the foci.
Envelopes of Circles generated by an Ellipse
Take a variable point on an ellipse and a fixed point not on the ellipse.
Using the variable point on the ellipse as the center of a circle that passes
through the fixed point, an envelope of circles is traced as the variable
point moves around the ellipse.
When the fixed point is on the orthogonal to the center of the segment connecting
the two loci of the ellipse, the following figure results.
Click here for a GSP sketch to experiment with various locations of the
fixed point. The above figure is symmetric. When the fixed point is moved
to a point above the line of foci of the ellipse and to the left of the
perpendicular to the center of the line of foci, then the envelope generates
a non-symmetric figure:
Another symmetric figure results by placing the fixed point on the line
of the loci:
Special cases of this figure result when the ellipse is replaced by a
circle. If the fixed point is outside the circle, the resulting figure is
a limaçon.
If the fixed point is on the circle, the resulting envelope is a cardioid:.
The Nephroid
The nephroid is formed by an envelope of circles with centers on a given
circle and each tangent to a diameter of the circle.
Click here to obtain a GSP script for
generating the nephroid. Interesting variations occur when the line is a
chord other than a diameter:
or a tangent line:
or a line that does not intersect or touch the circle:
Click here for an animations of these
transformations of the envelope of circles.
Question: What is the shape formed by the envelope of circles generated when the variable point is on a hyperbola and the fixed point is the midpoint of the segment connecting the foci? Answer: A lemniscate.
Question: What shape is formed when the fixed point is on the hyperbola but not on the line segment joining the foci? Answer: A kidney shape.
Question: What happens to the lemniscate if the fixed point is moved to a point other than the midpoint of the segment connecting the loci of the hyperbola? Answer: A modified "lemniscate" is formed.
Question: What is the resulting envelope when the fixed point is on the circle and the given shape is an ellipse? Answer: A cardioid.
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I don't even know where to begin with this. Someone please help me step by step?
Thanks! 1 solutions Answer 209906 by richwmiller(9135) on 2010-04-08 12:25:22 (Show Source):
Triangles/289968: C. If the triangle has one angle that measures 35 degrees and another angle that measures 105 degrees, what is the measure of its third angle? 1 solutions Answer 209903 by richwmiller(9135) on 2010-04-08 12:10:36 (Show Source):
Question: What is the measure of the first angle mentioned in the solution with the answer 209903? Answer: 35 degrees
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These names follow from the fact that they are customarily written in terms of the haversine function, given by haversin(θ) = sin2(θ/2). The formulas could equally be written in terms of any multiple of the haversine, such as the older versine function (twice the haversine). Historically, the haversine had, perhaps, a slight advantage in that its maximum is one, so that logarithmic tables of its values could end at zero. These days, the haversine form is also convenient in that it has no coefficient in front of the sin2 function.
On the left side of the equals sign d/r is the central angle, assuming angles are measured in radians (note that φ and λ can be converted from degrees to radians by multiplying by π/180 as usual).
Solve for d by applying the inverse haversine (if available) or by using the arcsine (inverse sine) function:
where h is haversin(d/r), or more explicitly:
In the era before the digital calculator, printed tables for the haversine/inverse-haversine and its logarithm (to aid multiplications) saved navigators from squaring sines, computing square roots, etc., an arduous process and likely to exacerbate small errors (see also versine).
When using these formulae, ensure that h does not exceed 1 due to a floating point error (d is only real for h from 0 to 1). h only approaches 1 for antipodal points (on opposite sides of the sphere) — in this region, relatively large numerical errors tend to arise in the formula when finite precision is used. Because d is then large (approaching πR, half the circumference) a small error is often not a major concern in this unusual case (although there are other great-circle distance formulas that avoid this problem). (The formula above is sometimes written in terms of the arctangent function, but this suffers from similar numerical problems near h = 1.)
As described below, a similar formula can be written using cosines (sometimes called the spherical law of cosines, not to be confused with the law of cosines for plane geometry) instead of haversines, but if the two points are close together (e.g. a kilometer apart, on the Earth) you might end up with cos (d/R) = 0.99999999, leading to an inaccurate answer. Since the haversine formula uses sines it avoids that problem.
Question: Which function's use might lead to inaccurate answers for close points on Earth? Answer: The spherical law of cosines, as it might result in cos(d/R) = 0.99999999.
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There is something wrong with the third problem, if 2 of the three angels of the triangles formed are the same it means the third angles are the same meaning so why are the arcs different lengths?
Answers
Nothing is wrong with this problem because no central angles are present. If the vertex of the third angle in each triangle rested on the center of the circle, then you would be correct: congruent central angles must intersect congruent arcs. However, nothing in the problem states that the intersection of those two chords is at the center of the circle.
Question: What would happen if the vertex of the third angle in each triangle rested on the center of the circle? Answer: If the vertex of the third angle in each triangle rested on the center of the circle, then congruent central angles would intersect congruent arcs.
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The same construction can also be applied to the hyperbola. If P0 is taken to be the point (1,1), P1 the point (x1,1/x1), and P2 the point (x2,1/x2), then the parallel condition requires that Q be the point (x1x2,1/x11/x2). It thus makes sense to define the hyperbolic angle from P0 to an arbitrary point on the curve as a logarithmic function of the point's value of x.[1][2]
Whereas in Euclidean geometry moving steadily in an orthogonal direction to a ray from the origin traces out a circle, in a pseudo-Euclidean plane steadily moving orthogonal to a ray from the origin traces out a hyperbola. In Euclidean space, the multiple of a given angle traces equal distances around a circle while it traces exponential distances upon the hyperbolic line.[3]
The quadrature of the hyperbola is the evaluation of the area swept out by a radial segment from the origin as the terminus moves along the hyperbola, just the topic of hyperbolic angle. The quadrature of the hyperbola was first accomplished by Gregoire de Saint-Vincent in 1647 in his momentous Opus geometricum quadrature circuli et sectionum coni. As expressed by a historian,
[He made the] quadrature of a hyperbola to its asymptotes, and showed that as the area increased in arithmetic series the abscissas increased in geometric series.[4]
The upshot was the logarithm function, as now understood as the area under y = 1/x to the right of x = 1. As an example of a transcendental function, the logarithm is more familiar than its motivator, the hyperbolic angle. Nevertheless, the hyperbolic angle plays a role when the theorem of Saint-Vincent is advanced with squeeze mapping.
Circular trigonometry was extended to the hyperbola by Augustus De Morgan's in his textbook Trigonometry and Double Algebra.[5] In 1878 W.K. Clifford used hyperbolic angle to parametrize a unit hyperbola, describing it as "quasi-harmonic motion". In 1894 Alexander Macfarlane circulated his essay "The Imaginary of Algebra", which used hyperbolic angles to generate hyperbolic versors, in his book Papers on Space Analysis.[6]
When Ludwik Silberstein penned his popular 1914 textbook on the new theory of relativity, he used the rapidity concept based on hyperbolic angle a where tanh a = v/c, the ratio of velocity v to the speed of light. He wrote:
It seems worth mentioning that to unit rapidity corresponds a huge velocity, amounting to 3/4 of the velocity of light; more accurately we have v = (.7616) c for a = 1.
... the rapidity a = 1, ... consequently will represent the velocity .76 c which is a little above the velocity of light in water.
The hyperbolic angle is often presented as if it were an imaginary number. In fact, if x is a real number and i2 = −1, then
Question: In what year did Gregoire de Saint-Vincent accomplish the quadrature of the hyperbola? Answer: 1647
Question: Who extended circular trigonometry to the hyperbola? Answer: Augustus De Morgan
Question: Who used the rapidity concept based on hyperbolic angle in his 1914 textbook on the new theory of relativity? Answer: Ludwik Silberstein
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In mathematics, a curve (also called a curved line in older texts) is, generally speaking, an object similar to a line but which is not required to be straight. This entails that a line is a special case of curve, namely a curve with null curvature. Often curves in two-dimensional (plane curves) or three-dimensional (space curves) Euclidean space are of interest.
Various disciplines within mathematics have given the term different meanings depending on the area of study, so the precise meaning depends on context. However many of these meanings are special instances of the definition which follows. A curve is a topological space which is locally homeomorphic to a line. In everyday language, this means that a curve is a set of points which, near each of its points, looks like a line, up to a deformation. A simple example of a curve is the parabola, shown to the right. A large number of other curves have been studied in multiple mathematical fields.
The term curve has several meanings in non-mathematical language as well. For example, it can be almost synonymous with mathematical function (as in learning curve), or graph of a function (as in Phillips curve).
Question: Is a line considered a special case of a curve? Answer: Yes
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In mathematics, domain is an important concept in a function. Domain is defined as a
set of input value or arguments. The set of all the output values is called as a range.
The range is an interval ...
In geometry, a Cuboid is a solid figure bounded by six faces, forming a convex polyhedron. There are
two competing (but incompatible) definitions of a cuboid in mathematical literature. In the more ...
In our daily to daily life, we use many objects that are circular in shape. Such as wheels,
flower beds etc. because of this problem of finding areas related to circular shapes have a big
practical ...
Parabola is a mathematical conic shape which is formed by intersection of a right circular plane and line on a surface. In the simple we can say that parabola is a kind of arc which is specified by aContent Preview
Share Modeling and Simulation of a Four Bar Mechanism and a Crane using ProM - PDF to:
Download Modeling and Simulation of a Four Bar Mechanism and a Crane using ProM - PDF
Question: What is the domain of a function in mathematics? Answer: The domain of a function is the set of input values or arguments.
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measure, side length, perimeter, and area learn this concept to design a 45*, 90* or 180*
of all types of triangles. haircut.
G1.2.5 Solve multi-step problems and construct Salon design project.
proofs about the properties of medians,
altitudes, and perpendicular bisectors to the
sides of a triangle, and the angle bisectors
of a triangle. Using a straightedge and
compass, construct these lines.
G1.4 Quadrilaterals and Their Properties
G1.4.1 Solve multi-step problems and construct Design a floor plan in a beauty salon. The students
proofs involving angle measure, side length, are required to design it to scale. Salon Business
diagonal length, perimeter, and area of project requires students to draw and build a scale
squares, rectangles, parallelograms, kites, model of a salon they have designed. Construct a
and trapezoids. salon floor plan to scale. The learner would learn this
concept while creating a floor plan in salon design.
Salon project to scale and angles to construct a floor
plan.
G1.6 Circles and Their Properties
G1.6.1 Solve multi-step problems involving We use circle placement in pins curls and rollers and
circumference and area of circles. perm rods. Using the circumference of the pin curls
or roller, determine the size needed. The different
shape bases of the pin curls arc, half moon and
triangle and rectangle base. Using the circumference
of the pin curl, or roller to determine.
G1.6.4 Know and use properties of arcs and The learner will learn this concept in roller placement,
sectors and find lengths of arcs and areas arc and other shaped bases in the designing of pin
of sectors. curls.
G1.8 Three-dimensional Figures
G1.8.1 Solve multi-step problems involving surface When we are going into wiggery we measure the
area and volume of pyramids, prisms, head forms to customize the wigs for the perfect fit.
cones, cylinders, hemispheres, and Using a 3D head, form measure the circumference of
spheres. the head for wiggery. The learner will learn this
concept in wiggery by measuring the mannequin for
a proper fitting wig. Wigs with a 3-D head form with a
tape measure with a head form. Using a 3D head
form to measure the circumference of the head.
G1.8.2 Identify symmetries of pyramids, prisms, When the student is involved in hair cutting each side
cones, cylinders, hemispheres, and needs to be the same length. When performing a
spheres. comb out the set need to be symmetrical. Hair is
distributed evenly/symmetrically between the
quadrants of the client"s head.
G2 STANDARDS CTE APPLICATION and PRACTICE
RELATIONSHIPS BETWEEN FIGURES
Question: What is the main concept that students learn while creating a floor plan in salon design? Answer: The concept of scale.
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A sketch of the proof can be found below. Hence, we know that if a weeble of uniform density exists then it cannot be effectively two-dimensional. What we can observe is that the 2D oval has two stable equilibria and for any number greater or equal to three the regular polygon with "n" sides will have "n" stable equilibria.
But what about three dimensions? Sadly, things are not so simple. However, when logical proofs are not forthcoming mathematicians are just as open to experimentation as any other scientist.
Below are a number 3D shapes, have a go at counting the number of stable equilibria they have. The answers can be found if you scroll down past the theorem proof.
Figure 2. A cube, a square based pyramid and a cylinder with edges sliced at an angle.
Now that you have got a feel for finding stable equilibria, have a think about what kind of shape would have one stable equilibrium point and one unstable only. Are you sure it even exists? Before we answer this question we take a detour next week and show how we can make our eggs not so lazy!
Consider a 2D convex object that has only one stable and one unstable equilibrium point. Define a function, R(ɵ), that is the distance of the perimeter away from the centre of gravity as shown in the figure on the left. The function depends on the angle ɵ (measured in radians) around the centre of gravity and so as ɵ increases from 0 to 2π the function, R, will do one revolution of the shape.
Now consider the graph of R. By assumption there are only two equilibria, one stable and one unstable, thus, the corresponding graph has just one maximum and one minimum, as shown in the figure on the right. This is how the equilibria where defined last week.
Suppose we now drop a horizontal line across the graph. In particular let the points at which the horizontal line touches the graph be separated by an angular distance of π radians. This will correspond to a straight cut through the shape. Call this corresponding value R0.
By definition everything above this horizontal would be the part of the shape that is further than R0 away from the centre of gravity and everything below would the part of the shape closer than R0 to the centre of gravity. This means that one half of the shape would be bigger than the other half, and, thus, heavier. But by definition the cut goes through the centre of gravity and so each side weighs the same. As we come to a contradiction, our original hypothesis (that such a shape exists) must be wrong.
________________________________________________________________
________________________________________
Answers
The cube has 6 stable equilibria; one on each of the flat faces, as shown.
The cube has 5 stable equilibria; one on each of the flat faces, as shown.
The sliced cylinder only has 1 stable equilibrium. However, it is not a shape that satisfies our need because it has 3 unstable equilibria; one on the top of the cylinder and one at each of the tips.
Monday, 7 January 2013
Question: What is the angular distance in radians between the points where a horizontal line touches the graph of R(θ), assuming the hypothesis of a 2D convex object with only one stable and one unstable equilibrium point? Answer: π radians.
Question: What does the value R0 represent in the context of the graph of R(θ)? Answer: The distance of the perimeter away from the centre of gravity where the shape is cut, which corresponds to a straight cut through the shape.
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Given a line and a point not on it, at most one parallel to the given line can be drawn through the point.
It is equivalent to Euclid's parallel postulate and was named after the Scottish mathematicianJohn Playfair. It is only required to state "at most" because the rest of the postulates will imply that there is exactly one. It could perfectly be assumed to write it saying "there is one and only one parallel". It is important to remark that in the Euclid book, two lines are said to be parallel if they never meet. It does not matter if their distance is always the same or not.[1][2]
This axiom is used not only in Euclidean geometry, but also in a broader study called affine geometry where the concept of parallelism is central. In the context of affine geometry the axiom has been called Euclid's parallel axiom,[4] but for Euclidean geometry the parallel postulate which refers to angles is the traditional expression of parallelism.
Contents
In 1795 John Playfair published an alternative, more stringent formulation of Euclid's parallel postulate, which is now called Playfair's axiom; though the axiom bears Playfair's name, he did not create it, but credited others, in particular William Ludlam, with the prior use of it.[5] However, Proclus (410–485 A.D.) clearly makes the statement in his commentary on Euclid I.31 (Book I, Proposition 31)[6]
If the sum of the interior angles α and β is less than 180°, the two straight lines, produced indefinitely, meet on that side.
Playfair's axiom is not exactly equivalent to Euclid's Fifth Postulate[7] because on an elliptical geometry, such as the surface of a sphere, Euclid's postulate in its original version:
If a line segment intersects two straight lines forming two interior angles on the same side that sum to less than two right angles, then the two lines, if extended indefinitely, meet on that side on which the angles sum to less than two right angles.
holds because two lines on a sphere always meet. Playfair's postulate is therefore stronger and prevents elliptic geometries. Therefore, it is not possible to derive Playfair's postulate from Euclid Fifth alone.
Euclid's fifth postulate with the other four implies Playfair's postulate[edit]
The easiest way to show this is using the Euclid theorem (based in the fifth postulate) that states that the angles of a triangle are two right angles. Given a line and a point, construct a line perpendicular to the given one by the point, and a perpendicular to the perpendicular. This line is parallel because it cannot form a triangle. Now it can be seen that no more parallels exist because any line that forms an angle with the second one will cut the first one.[8]
Question: In which context is Playfair's axiom used besides Euclidean geometry? Answer: Affine geometry
Question: Which geometry does Playfair's axiom prevent? Answer: Elliptic geometries
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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: How does an angle bisector divide an angle? Answer: An angle bisector divides the angle into two equal angles.
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In this context (a,b) changes meaning; it means a is party to the same information as b which implies that b is party to the same
information as a. Again, in this context (a,b) is the same as
(b,a). An arc (a,b) or a → b in digraph G has changed because of the perspective or interpretation into an edge (a,b) or a ↔ b. To achieve this effect, the set of arcs in digraph G(V,A) is symmetrized into the set of edges E to result in just a graph G(V,E).
Question: Is the relationship between 'a' and 'b' in (a,b) the same as in (b,a)? Answer: Yes, the relationship is symmetric.
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A new era in determining the size of Earth began through the introduction of triangulation. The idea of triangulation was apparently conceived by the Danish astronomer Tycho Brahe before the end of the 16th century, but it was developed as a science by a contemporary Dutch mathematician, Willebrord van Roijen Snell. Snell used a chain of 33 triangles to determine the length of an arc essentially in the way customarily done today. The resulting size of Earth, however, was 3.4 percent too small. The idea of triangulation is to establish a network of stations that form connecting triangles. One side of the first triangle in the chain, called the baseline, and all angles of the triangles are accurately measured. Using the law of sines from spherical trigonometry, the lengths of all sides thus can be computed starting from a known baseline. When the lengths and angles are known, coordinates can be computed for each point, provided the coordinates of one point and one azimuth are known. Triangulation points are usually placed on the tops of hills because the neighbouring points must be clearly visible. Commonly, more complicated figures such as quadrilaterals with diagonals are used in triangulation.
In 1669 Jean Picard, a French astronomer, first used a telescope in determining latitude and in measuring angles in triangulation that consisted of 13 triangles and extended from Paris 1.2° northward. His observations and results were extremely important because his length of arc on a great circle corresponding to 1° was used by the English physicist and mathematician Sir Isaac Newton in his theoretical calculations to prove that the attraction of Earth is the principal force governing the motion of the Moon in its
Question: What is the law used to compute the lengths of sides in triangulation? Answer: The law of sines from spherical trigonometry
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where $A$, $B$, $C$ are the areas of the "leg-faces" and $D$ is the area of the "hypotenuse-face".
For right-corner simplices in higher Euclidean dimensions, we have that the sum of the squares of the content of leg-simplices equals the square of the content of the hypotenuse-simplex. (I don't happen to know the non-Euclidean counterparts of this generalization. Perhaps this makes for a good MO question!)
As generalizations of the Pythagorean Theorem for Triangles, I always found these (Euclidean) results to be more satisfying than the diagonal-of-a-box/distance formulas: instead of dealing only with segments, we have that, as the dimension of the ambient space goes up, so does the dimension of the objects involved in the relations.
Question: What are the "leg-faces" and "hypotenuse-face" in the context of the text? Answer: "Leg-faces" refer to the faces of a right-corner simplex that are not the hypotenuse, while the "hypotenuse-face" is the face that is opposite the right angle.
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In this lesson, students develop the area formula for a triangle. Students find the area of rectangles and squares, and compare them to the areas of triangles derived from the original shape. Student handouts are included here.
Congruence of Triangles (Grades 6-8)
With this virtual manipulative, students arrange sides and angles to construct congruent triangles. They drag line segments and angles to form triangles and flip the triangles as needed to show congruence. Options include constructing triangles given three sides (SSS), two sides and the included angle (SAS), and two angles and an included side (ASA). But the option that will motivate most discussion is constructing two triangles given two sides and a nonincluded angle (SSA). The question in this case is: Can you find two triangles that are not congruent?
Transformations—Reflections
Here students can manipulate one of six geometric figures on one side of a line of symmetry and observe the effect on its image on the other side. A triangle may be selected and then translated and rotated. The line of symmetry can be moved as well, even rotated, giving more hands-on experience with reflection as students observe the effect on the image of the triangle.
The Pythagorean Theorem
This site invites learners to discover for themselves "an important relationship between the three sides of a right triangle." The site's author, Jacobo Bulaevsky, speaks directly to students, encouraging them throughout five interactive exercises to delve deeper into the mystery. Within each exercise he gives hints that will motivate and entice your students 12/09/2011.
How could we organize a math fair? And what kinds of projects would our students present? I'm not thinking here of projects that would be judged, as in a science fair, but rather investigations and activities that would engage middle school students and be presented for the whole school as well as parents. One idea comes from a 7th grade class at Frisbie Middle School in Rialto, California.
Multicultural Math Fair
Ten activities for the fair, each based on a different cultural heritage, are well described in both Spanish and English. Included here are tips on how to set up a math fair as well as student handouts and free software for specific activities, such as the Tower of Hanoi. You will also find links to resources for related activities, such as studying symmetry and patterns in Navajo rugs. A unique teacher-created site!
If you are looking for more project ideas, here are some I think would make great fair presentations and involve students in learning sound math:
Pascal's Triangle
Here are three ways to explore the famous triangle: by finding patterns and relations within the triangle, solving a pizza toppings problem in Antonio's Pizza Palace, or working with an interactive web unit. The set of three investigations could work well as one fair project.
The Noon Day Project: Measuring the Circumference of the Earth
Question: Who is the author of the interactive exercises on the Pythagorean Theorem? Answer: Jacobo Bulaevsky.
Question: What can students do with the "Transformations—Reflections" tool? Answer: Manipulate one of six geometric figures on one side of a line of symmetry and observe the effect on its image on the other side.
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Question 146950: THIS IS THE MOST IMPORTANT QUESTION ON MY HOMEWORK AND I JUST CAN'T FIGURE IT OUT!!! please help !
A 25 foot ladder is placed against a building. The bottom of the ladder is 7 feet from the building. If the top of the ladder slips down by 4 feet, by how many feet will the bottom slide out? Click here to see answer by scott8148(6628)
Question 147048: Pat and Chris were out in their row boat one day, and Chris spied a water lily. Knowing that Pat liked a mathmatical challenge, Chris announced that, with the help of the plant, it was possible to calculate the depth of the water under the boat. When pulled taut, the top of the plant was origionally 10 inches above the water surface. While Pat held the top of the plant, which remained rooted to the lake bottom, Chris gently rode the boat 5 feet. This forced Pat's hand to the water surface. Use this information to calculate the depth of the water. jim_thompson5910(28595) solver91311(16897)
Question 146953: Consider the triangle defined by P=(1,3), Q=(2,5, and R=(6,5). The transformation defined by T(x,y)=(x+2,y-1) is mathematical notation for translating a point by the vector [2,-1]. The point P(1,3) becomes T(1,3)=(1=2,3-1)=P'(3,2). Find Q' and R'. Graph both the original triangle and its image. Click here to see answer by psbhowmick(529)
Question 147275: given an arbitrary triangle,what can you say about the sum of 3 exterior angles, 1 for each vertex of the triangle??
Question 147277: Ive tried this problem 6 different times, and cannot seem to find the answer.
Given triangle ABC with AB = AC, extend segment AB to a point P so that B is between A and P and BP = BC. In the resulting triangle APC, show that angle ACP is exactly three times the size of angle APC. (By the way, notice that extending segment AB does NOT mean the same thing as extending segment BA.)
thnx!!! Click here to see answer by vleith(2825)
Question 147268: PLEASE HELP! I don't know how to find the length of any altitude. There are several problems on my homework like this, so if i understand the steps for this problem, i can figure out the others. THANKS!
A triangle that has a 13-inch side, a 14-inch side, and a 15-inch side has an area of 84 square inches. Accepting this fact, find the lengths of all three altitudes of this triangle.
Question: How far did Chris ride the boat, forcing Pat's hand to the water surface?
Answer: 5 feet
Question: What is the transformation T(x, y) that translates a point by the vector [2, -1]?
Answer: T(x, y) = (x + 2, y - 1)
Question: What is the depth of the water under the boat?
Answer: 15 inches
Question: What was the original height of the water lily above the water surface?
Answer: 10 inches
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Question 147271: If it is know that one pair of alternate interior angles is equal, what can be said about...
-the other pair of alternate interior angles
-either pair of alternate EXTERIOR angles
-any pair of corresponding angles
-either pair of non-alternate interior angles.
Question 147488: Let A = (1,1), B = (3,5), and C = (7,2). Explain how to cover the whole plane with non overlapping triangles, each of which is congruent to triangle ABC.
In the pattern of lines produced by your tessellation, you should see triangles of many different sizes. What can you say about their sizes and shapes?7838: I got an answer for this, but I'm not sure if i'm right. thanks!!!!
Standing on a cliff 380 meters above the sea, Sue sees an approaching ship and measures its angle of depression, obtaining 9 degrees. How far from the shore is the ship?
Now Sue sights a second ship beyond the first. The angle of depression of the second ship is 5 degrees. How far apart are the 2 ships?
Question 148385: Taylor lets out 20 meters of kite string then wonders how high the kite has risen. Taylor is able to calculate the answer after using a protractor to measure the 63-degree angle of elevation that the string makes with the ground. How high is the kite, to the nearest meter? What unrealistic assumptions did you make in answering this question?
Question 148446: 10. If the sides of a triangle are 13, 14, and 15 cm long, then the altitude drawn to the 14-cm side is 12 cm long. How long are the other two altitudes? Which side has the longest altitude?
Question: What is the distance between the two ships? Answer: 380 tan(5°) - 380 tan(9°) ≈ 17.6 meters.
Question: If one pair of alternate interior angles is equal, what can be said about the other pair of alternate interior angles? Answer: They are also equal.
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From the Delaunay triangulation one can compute a list of adjacent triangles and the three angles of each triangle. Table 1 displays a list of the angles and the adjacent tringles for each triangle in FIG. 3.
TABLE 1
Adjacent Triangles
angles
Triangle 1
0 0 2
30 31 119
Triangle 2
3 1 4
89 61 30
Triangle 3
0 2 5
45 88 46
Triangle 4
2 0 6
34 85 61
Triangle 5
3 6 0
38 45 97
Triangle 6
5 4 0
56 53 71
The sum of the angles for each triangle in Table 1 is not always 180 degrees because the angles have been rounded to the nearest integer. A '0' in the 'Adjacent Triangles' columns of Table 1 indicates that side of the triangle has no neighbor. For the Delaunay triangulation shown in FIG. 3, only Triangle 2 has three adjacent triangles. FIG. 4 displays a close up view of Triangle 2 from FIG. 3. In FIG. 4 the nodes are labeled, as are their associated angular values and the side opposite each node. If a side of a triangle is shared by another triangle, then those triangles are adjacent. FIG. 5 displays Triangle 2 and its adjacent triangles. FIG. 5 demonstrates that adjacency relationships are defined geometrically and that each node has an associated angle and adjacent triangle. The phrase neighboring triangles is used to refer to those triangles that are near to a triangle, but not adjacent to it. For example, in FIG. 3, Triangle 5 and Triangle 6 are neighbors of Triangle 2.
To match an input object with a template in memory, the first step is to begin with a triangle in the list of triangles of the input, and begin looking for a similar triangle in the list of triangles from the template. Triangles are similar if there is a sufficient similarity in their angles and if there is sufficient similarity in their node labels. Once a single triangle match has been found, the next step is to examine the appropriate adjacent triangles. For example, if Triangle 2 of FIG. 3 has a match with another triangle in a template stored in memory, the next step is to examine the appropriate adjacent triangles in both the input and the template. Thus the triangles opposite the node associated with 89 degrees are compared, then the triangles opposite the nodes associated with 61 degrees are compared, and then the triangles opposite the nodes associated with 30 degrees are compared. If all three of the appropriate adjacent triangles match their appropriate counterparts triangles in the template, then six nodes, their labels, and their spatial relationships (twelve angles) match. Depending on the conditions used for matching individual triangles, there is the possibility that matching one triangle and its three appropriate adjacent triangles is exceptionally good evidence for a positive match between the input and the template.
Question: What are the angular values associated with the nodes of Triangle 2 in FIG. 4? Answer: 89, 61, and 30 degrees
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Arc BE measures 118 because it is twice the measure of the inscribed angle D which is 59 degrees. Therefore, angle y is 1/2 the measure of arc BE because it is also an inscribed angle. Angle y is the same measure as angle D because they both intercept the same arc BE.
To find the measure of arc DF we have to subtract the other arcs from 360. So 360 - 134 - 118 - 30 = 78 degrees for arc DF. And angle z is 1/2 of that.
Question: Is angle y the same measure as angle D? Answer: Yes
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OK, we all know average people tend to use either degrees or turns to measure angles, whereas mathematicians use radians, some other people use gradians, mils, or both, and there are other standards besides (points for boxing the compass, quadrants for studying your geometry, etc.).
My question is, are any of these more or less popular based on region, the way SI units give way to Imperial and/or US units based on where you go in the worldTwo realms where angles are expressed differently are in grade (usually in % slope but other systems are used) and in roof slopes using a ratio. But they seem to be fairly universal in Western society. Anyone know if other cultures use their own versions for theseQuote:
Originally Posted by Derleth
I suppose the 'turn' system is so informal nobody recognizes it as a system of angle measure.
That's what happened to me, yes. Not only is it informal, but extremely imprecise. Quarter turns and half turns are on the same metrological league as pinches of salt and small amounts of sugar. I might use it to give instructions on how much to rotate something, but not to measure how much it was actually rotated.
So are you actually going to tell us what this "turn" system is, or is it a secret?I hadn't really thought about it, but we do something similar around the lab and my house too. Except we do it in terms of "hours". as in "I turned it about 4 hours clockwise/forward". We find it works better for things more than a 6 hour turn to decrease ambiguity and works well in breaking up the circle into mentally recognized fractions.
My car's navigation system gives turn directions as "Turn left" for a turn to 270 degrees, "Turn right" for a turn to 90 degrees, and the face of the clock for other turns (e.g., "Turn right in the direction of One O'clock" for a turn to 30 degrees).
Question: Are turns a more precise way to measure angles compared to degrees or radians?
Answer: No, turns are extremely imprecise.
Question: In the turn system, what is a half turn equivalent to in degrees?
Answer: 180 degrees
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Points, Lines and Curves
The most practical branch of mathematics is geometry. The term 'geometry' is derived from the Greek word 'geometron'.
Lesson Demo
The most practical branch of mathematics is geometry. The term 'geometry' is derived from the Greek word 'geometron'. It means Earth's measurement. The fundamental elements of geometry are given below:
Point:
In geometry, dots are used to represent points. A point is used to represent any specific location or position. It neither has any size, nor dimensions such as length or breadth. A point can be denoted by a capital letter of the English alphabet. Points can be joined in different ways.
Line segment:
A line segment is defined as the shortest distance between two points. For example, if we mark any two points, M and N, on a sheet of paper, then the shortest way to join M to N is a line segment. It is denoted by Points M and N are called the end points of the line segment.
Line:
A line is made up of an infinite number of points that extend indefinitely in either direction. For example, if a line segment from M to N is extended beyond M in one direction and beyond N in the other, then we get a line, MN. It is denoted by A line can also be represented by small letters of the English alphabet.
Ray:
A ray is a portion of a line. It starts at one point and goes on endlessly in one direction. For example, if a line from M to N is extended endlessly in the direction of N, then we get a ray, MN. It is denoted by and can be read as ray MN.
Plane:
A plane is said to be a very thin flat surface that does not have any thickness, and is limitless. For
example, this sheet is said to plane PQR. An infinite
number of points can be contained within a plane.
Intersecting lines:
If two lines pass through a point, then we say that the two lines intersect at that point. Thus, if two lines have one point in common, then they are called intersecting lines. For example, two lines
pass though point P. These two lines are called intersecting lines.
Parallel lines or non- intersecting lines :
In a plane, if two lines have no point in common, then they are said to be parallel or non- intersecting lines.
Parallel lines never meet, cut or cross each other. In the figure, it can be observed that two lines are parallel. We write .
Curves:
Curves can be defined as figures that flow smoothlywithout a break. A line is also a curve, and is called a straight curve. Curves that do not intersect themselves are called simple curves. The end points join to enclose an area. Such curves are called closed curves. For example, (i), (ii) and (iii) are simple curves, whereas (iv) and (v) are closed curves.
For a closed curve, we can identify three regions:
The interior of the curve. Here, point P is in the interior of the circle.
Question: What are the three regions identified for a closed curve? Answer: The interior of the curve, the boundary of the curve, and the exterior of the curve
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Question 150882: Hello, can you please help me solve this problems:
1.) The sum of the measures of two angles is 180 degrees. Three times the measure of one angle is 24 less than the measure of the other angle. What is the measure of each angle?
2.) The length of a rectangle is 17cm larger than its width. When its width is decreased by 5 cm and its length is increased by 7cm the area of the new rectangle is 22cm² biggerthan the original rectangle. Find the dimension of the original rectangle. Click here to see answer by stanbon(57246)
Question 151018: A RECTANGLE IS 4 TIMES AS LONG AS IT IS WIDE A SECOND RECTANGLE IS 5 CENTIMETERS LONGER AND 2 CENTIMETERS WIDER THAN THE FIRST. THE AREA OF THE SECOND RECTANGLE IS 270 SQUARE CENTIMETERS GREATER THAN THE FIRST. WHAT ARE THE DIMENSIONS OF THE ORIGINAL RECTANGLE Click here to see answer by jojo14344(1512)
Question 151167: write true or false after the statement.
1. All radii of two or more circles are congruent.
2. Radii of the same circle are congruent.
3. The endpoint of the radius lies in the interior of the circle.
4. Diameters of the same circle are congruent.
5. The endpoints of any chord of any circle lies on the circle.
6. A diameter is a chord.
7. A chord is a diameter.
8. Twice the length of the diameter is equal to the length of the radius.
9. There exists a point x whose distance from the center of a circle is greater
than the radius, so point x lies on the interior of the circle.
10. There exists a point y whose distance from the center of a circle is less
than the radius, so point y lies on the interior of the circle. Click here to see answer by Edwin McCravy(8882)
Question 151410: can someone help me with this word problem? i have tried everything, but i just cant seem to figure it out...
three circles have to be cut into a piece of metal. the specifications state that each of the diameters must be within 0.001 centimeter of the given measurements. let D represent the given measurement and let x respresent the actual diameter of the circle. write an absolute value inequality that descrides the acceptible diameters of the circle. if the circles are to be 13 centimeters, 9 centimeters, nad 6 centimeters, describe the acceptable diameters of each circle.
Question: What is the acceptable diameter for a circle with a given measurement of 13 cm in the fourth problem? Answer: 12.999 ≤ x ≤ 13.001
Question: What is the measure of the smaller angle in the first problem? Answer: 60 degrees
Question: What is the measure of the larger angle in the first problem? Answer: 120 degrees
Question: What is the width of the original rectangle in the second problem? Answer: 9 cm
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The coordinate geometry is not
difficult. The best way to do well on the coordinate geometry is to know the formulas very
well. Here we list the most commonly used formulas on coordinate geometry. By knowing
these formulas, we guarantee you will do well on the coordinate geometry.
The Distance Formula:
The distance d between P1(X1, Y1) and P2(X2, Y2) is given by the formula d=Ö (X2-X1)2 + (Y2-Y1)2
.
The Midpoint Formula:
Given P(X1, Y1) and Q(X2, Y2), the coordinates (Xm, Ym) of M, the midpoint of PQ, are
[(X1+X2)/2, (Y1+Y2)/2].
When comes to determine a line on a
coordinate system, it is important to know the slope of the line. The slope of a line is
the difference between the Y-Coordinates divided by the difference between the
X-Coordinates of the two points that are used to determine the line. So the equation for
the slope of a line is (Y2-Y1)/(X2-X1). Here are some additional concepts about the slope
of a line:
All segments of a
non-vertical line have equal slopes.
The slope of a
non-vertical line is the slope of any segment of the line.
You can also jump to the
chapter of your choice by using the drop-down list at below.
Question: What does the slope of a line represent? Answer: The slope of a line represents the ratio of the change in y (rise) to the change in x (run).
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Facebook Daily Deal - Parts of a Circle Charts
$9.00
Add to Cart:
Help your students remember the different parts of a circle and brighten up your classroom with these 9 colourful posters. Each card contains a diagram and description of the following: Semi-circle, arc, radius, diameter, circumference, centre, chord, quadrant and concentric circles.
Question: Which of the following is NOT a part of a circle mentioned in the text? A) Sector B) Tangent C) Chord Answer: A) Sector
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Polar Coordinates
Remember
that when r is negative, you go to the opposite side of the
graph.
Remember
that each point has many different possible sets of polar
coordinates. (This is different from rectangular coordinates, where
each point has a unique (x, y) pair.)
To
find the limits, you often have to draw the graph and find out
what angle θ makes r = 0, or when the graph comes
back to meet itself.
In
finding areas, you often have to integrate sin²θ
or cos²θ.
To integrate these, use the half-angle formulas from the section on
trigonometric integrals.
To
find the area between the two graphs, subtract the two area
formulas just as you would with rectangular coordinates. To find
the limits, set the two functions equal to each other and solve for
the angles θ.
To
remember the arclength formula, it helps to recall that it comes
from the distance formula between two points, which in turn comes
from the Pythagorean Theorem.
Don't
make the common algebraic mistake of thinking that
reduces to a + b! This is extremely wrong, and your
teacher will likely be merciless if you do it
When
it's feasible, check that your answers make sense. Unlike area
integrals in rectangular coordinates, which can be negative if a
curve goes below the x-axis, areas and arclengths in polar
coordinates should always be positive. You might also be able to
check geometrically that the area or length of your curve looks
approximately right.
Polar Coordinates
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: How are polar coordinates different from rectangular coordinates in terms of representing points? Answer: Each point has many different possible sets of polar coordinates, unlike rectangular coordinates where each point has a unique (x, y) pair.
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Math: Trigonometric Identities
These sites are about trigonometric identities. The topics cover trigonometric identities, functions, and properties. Learn how to apply trigonometric identities to solve trigonometric problems. There are also interactive illustrations on Proof of the Pythagorean Theorem and Triangles and Law of Sines. Includes examples, formulas, graphs, worksheets, exercises, and interactive activities and games.
This site provides information about trigonometric functions and identities. Learn about their important identities, properties, and examples of their application. Includes practice exercises at the end of each module. NOTE: This site includes ads and links to external websites.
Learn about trigonometric functions and identities, sums and difference formulas, double angle and half angle, inverse functions, and Sin and Cos laws. Includes how to plot a graph and how to find triangle sides by applying trigonometric rules.
Question: Can you plot a graph and find triangle sides by applying trigonometric rules using this site? Answer: Yes, the site includes information on how to plot a graph and how to find triangle sides by applying trigonometric rules.
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Chapter Fifteen
With the Canon of Sines for Hundredths or Thousandths of Degrees, the Canons of Tangents, Secants, and of Logarithms are being provided with the same parts.
Prop.1. Tangents and Secants are most conveniently being found by the Rule of Proportion. For any Sine is to the Sine of the Complement : as The Radius to the Tangent of the same Complement.
From this Proposition alone any of the whole Quadrant of Tangents will be able to be found.
Prop. 2. The Radius is the mean proportional between the Sine of the Arc and the Secant of the Complement of these as you please.
From this Proposition any Secants you wish will be able to be found.
Prop. 3. The Radius is to the Sine of any Arc you please: as the Secant of the same to the Tangent.
Prop.4. The Radius is the mean proportional between the Tangents and of the Complement of these Arcs as you please.
If by dividing the Quadrant into 144 parts, by the first proposition of these, the tangents of half the Quadrant or with the first 72 equal parts being found appropriately; the Tangents of the remaining parts will be able to be found, and the Secants of all the others, by addition alone. As by the following Propositions being demonstrated.
Prop. 5 The Secant of any Arc you please, being equal to the [sum of the] tangents of the same Arc and half of the Complement1.
Let the Angle EAD be 23:0', and with the line GEF touching the periphery in the point D. GE, EF being taken equal to the line EA. GAF will be right, and DAF, EGA, EAG equal among themselves, and EAG half the Compliment [of EAD] EAB. But EF is equal (equal from the construction of the line AE to the secant of the angle EAD) to [the sum of] the Tangents ED of the angle EAD given, and DF of the angle FAD of half the Complement EAD 23:0', EAB 67:0' the Complement , DAF 33:30'.
Prop. 6 The Secant of any Arc you wish, being added to the Tangent of the same, being equal to the Tangent of the Arc being composed from the given arc and half of the Complement. [As AE + ED = DG: Figure 15-3].
For let the Angle EAD be 23:0', the Complement 67:0'; half of the Complement 33:30'. The Arc being composed 56:30'
Prop. 7. The Tangent of any Arc you wish by being taken from the Secant of the same, there is left the Tangent of half the Complement. Prop. 7.
Question: What is the relationship between the Radius, the Sine of an Arc, and the Secant of the Complement of that Arc, according to Proposition 2? Answer: The Radius is the mean proportional between the Sine of the Arc and the Secant of the Complement of that Arc.
Question: What is the main topic of this chapter? Answer: This chapter discusses the Canon of Sines and related trigonometric functions, specifically focusing on Tangents, Secants, and Logarithms.
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Circle Graphs
A circle graph is a way to organize data using the sectors of a circle.
Example:
Suppose you take a poll of the students in your class to find out their favorite foods, and get the following results:
Pizza – 41%
Ice Cream – 24%
Raw Mushrooms – 9%
Dog Food – 11%
Chicken Livers – 15%
Organize this data in a circle graph.
The data are given as percentages. To make the sectors the right size, you need to use the fact that there are 360 degrees in a circle, and solve some proportions. For example, to find the number of degrees in the sector for chicken livers, you need to solve:
It turns out that x = 54, so you need to measure a 54 degree sector to represent the portion of the class that loves chicken liver.
Question: Which food has the highest percentage of students liking it? Answer: Pizza (41%)
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Graph Angles in a Standard Position
In trigonometry and most other math topics, you draw angles in a standard, universal position, so that mathematicians around the world are drawing and talking about the same thing.
An angle in standard position.
An angle in standard position has its vertex at the origin of the coordinate plane, as shown in the preceding figure. Its initial ray (starting side) lies along the positive x-axis. Its terminal ray (ending side) moves counterclockwise from the initial side.
If the terminal ray moves clockwise instead of counterclockwise, then the measure is a negative value. You often name this type of angle with a Greek letter.
The lengths of the rays that create the angle have nothing to do with the angle size. You can extend rays as long as you need them to be, and the angle measure won't change. Only the direction of the terminal ray determines the angle.
Question: Can you change the size of an angle in standard position by extending the rays? Answer: No
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Below you have been given figure A. Draw figure B by reflecting figure A in the given line. Draw figure C by translating figure B 8 units right and 2 units down. Then rotate figure C 180 ° around the point marked X in figure A to give figure D. We can say that figure D is a complex transformation of figure A, as we needed several steps to draw it.
Figure 8
ACTIVITY 4
To enjoy transformations in the form of tilings and tessellations
[LO 3.2, 3.7]
The most remarkable and widely spread use of tessellations can be found in the decoration applied to buildings in the Islamic world. Islam forbids the making of images, so the builders concentrated on shapes. The Persians were competent mathematicians, and this helped to establish the rules of tessellation they used to such brilliant effect in the mosques and other important cultural centres. Even more interesting is the fact that the surfaces were often curved, not flat, which makes the principles of tessellation even trickier.
When you can make tiles of a certain shape with the property that you can place them next to each other on a surface so that they don't overlap, and don't leave any gaps, then we call this a tessellation.
You can experiment with this by cutting shapes carefully out of cardboard, and fitting them together.
You can also do this as a drawing on paper, by combining the principles of transformation (translation, reflection and rotation) to a starting shape until you have tessellated the surface completely.
The shapes can be simple, without any transformation except translation, or complicated with complex transformations. When you use more than one shape in a tessellation, you can produce some very beautiful designs.
Below you can see a few tessellations. Discuss (in your group) what you see and then try to write down exactly what was done to each shape (translation, reflection and rotation), to produce the final result. Complete any incomplete ones.
Figure 9
Figure 10
Figure 11
Figure 12
Figure 13
Figure 14
Assessment
Table 1
LO 3
Space and Shape (Geometry)The learner will be able to describe and represent characteristics and relationships between two-dimensional shapes and three–dimensional objects in a variety of orientations and positions.
We know this when the learner:
3.2 in contexts that include those that may be used to build awareness of social, cultural and environmental issues, describes the interrelationships of the properties of geometric figures and solids with justification, including:
3.2.2 transformations.
3.3 uses geometry of straight lines and triangles to solve problems and to justify relationships in geometric figures;
3.4 draws and/or constructs geometric figures and makes models of solids in order to investigate and compare their properties and model situations in the environment
Question: What are the two methods mentioned to experiment with tessellations? Answer: Cutting shapes out of cardboard and fitting them together, or drawing on paper using transformation principles.
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Imagine you had a robot, consisting of series of arms. They would be of same length, and each one would be attached to the end of the previous one, so it could rotate in the same 2d plane. If each joint were given some angular speed, what would the curve traced by the end-most point look like?
Well, here are several examples. The ellipse is expected, and boring, but some other curves are beatiful, and complicated. Some look like they were traced with a spirograph.
Some of these curves are epitrochoids, but I'm not sure what their general classification would be.
Question: What is the shape of the curve traced by the end-most point of the robot if each joint is given the same angular speed? Answer: An ellipse
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VBForums - Maths Forum
By popular request, a place for you to discuss Maths of all forms. Somewhere to think about algorithms and the applications of maths to programming too.enWed, 22 May 2013 00:28:12 GMTvBulletin60 - Maths Forum
Mon, 06 May 2013 03:35:55 GMT
I just saw the above on FaceBook.. Using the order of operations, the answer I came to was 1. However, a number of commenters insist passionately that the answer is 9. Some of them offer some semblance of an explanation why but I must admit, I was lost. Some people said they used something called PEDMAS which I honestly never heard of before (I know it as BOMDAS) and some people talked about some distribution principle behind their conclusions.
I'd be the first to admit that math really isn't my strong suit. I know that sounds strange coming from a programmer.
My question is, who is right ?
Attached Images
]]>Maths ForumNiya
Tue, 23 Apr 2013 17:24:12 GMThey is this true even when two balls meet?
is this the right way to get the answer?
To get the final angle:
AngleIncidence = AngleReflection
AngleIncidence = (180° + InitialAngle) - (NormalAngle)
AngleReflection = (NormalAngle) - (FinalAngle)
Solving for FinalAngle:
180 + Init - Norm = Norm - Final
180 + Init - 2 Norm = - Final
Final = 2 * Norm - 180 - Init
is the InitialAngle the angle between the two balls and if so does it matter if i get the angle from A to B first or B to A first?
i have the angle between B and A of 134 by 314 degrees when touching each other
ball B is moving at 23 degrees
ball A is moving at 229 degrees
how do i get the right reflextion angle for each ball?
thanks verry muchhey is this true even when two balls meet?
is this the right way to get the answer?
Question: What is the purpose of the Maths Forum on VBForums? Answer: The Maths Forum on VBForums is a place for discussing mathematics, algorithms, and the applications of math to programming.
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This page consists of Animations showing how the conic sections are generated and lecture
notes on the derivation of their algebraic equations.
The conic sections are the curves of intersection of a double cone and a plane.
In the following Animation We start with a horizontal plane intersecting the double cone in a circle. As the plane rotates
it the curve of intersection becomes an ellipse followed by a parabola and finally a hyperbola as the plane approaches the
verticaL.
Question: What are the subsequent curves of intersection as the plane rotates? Answer: An ellipse, followed by a parabola, and finally a hyperbola.
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Question 638291: In my question there is a diagram which shows a circle with a circumference of 1cm being rolled around an equilateral triangle with sides of length 1cm.
The circle is positioned haalfway between the top and bottom of one side of the triangle
How many COMPLETE TURNS does the circle make as it rolls around the triangle ( without slipping) to its original starting position. I know that the answer is 4, please could someone explain in simple terms why it is 4.
Thankyou Click here to see answer by MathLover1(6627)
Question 638641: I am not sure how to solve the following word problem. I have more difficulty answering word problems than number problems. Words seem to overwhelm me more than numbers do. Any guidance that you can offer on this subject and the provided math problem would be greatly appreciated.
Problem: In a triangle the two larger angles differ by 10°. The small angle is 50° less than the largest angle. Find the measure of each angle.
Question 639293: In a triangle, one side is three times as large as the smallest side and the third side is 40 feet more than the smallest side. The perimeter of the triangle is 184 feet. Find the measurement of all three sides. Click here to see answer by aquyu214(10)
Question 637783: the perimeter of an equilateral triangle is 79 ft. Find the length of each side. stanbon(57307) MathLover1(6627)
Question 642338: Using the Pythagorean Theorem, find the missing side for these right triangles.
If the answer is not a perfect square, leave it in radical form.
Here's the problem:
C = 5
B = 4
A = (blank space and has a radical sign over it)
Thank's so much for your help!!!
Question 647407: The sum of lengths of any two sides of triangle must be greater than third side. If a triangle has one side that is 14 inches and a second side that is one inch less than twice the third, what are the possible lengths for second and third? Click here to see answer by aaronwiz(69)
Question: How many complete turns does the circle make as it rolls around the equilateral triangle to its original starting position? Answer: 4
Question: Why does the circle make 4 complete turns? Answer: The circle's diameter (1cm) is equal to the side of the equilateral triangle (1cm). When the circle rolls around the triangle, it covers a distance equal to its circumference (3.14cm) in one complete turn. To return to its original position, it needs to cover a distance equal to the perimeter of the triangle (3 sides of 1cm each, total 3cm). Therefore, it takes 3/3.14 ≈ 0.96 turns to cover one side of the triangle. Since there are three sides, the circle makes 3 * 0.96 ≈ 2.88 turns to cover the entire triangle. However, since we are looking for complete turns, we round up to the nearest whole number, which is 3. But since the circle started halfway along one side, it needs to make an additional half turn to reach its starting position, making it 3 + 0.5 = 4 complete turns.
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Example 2
What is the height of the triangle in Figure 3.10?
Even though the height is labeled h, it is not the hypotenuse. The longest side has length 10 feet, and thus must be alone on one side of the equation.
h2 + 32 = 102
h2 = 100 – 9
h = √91 ≈ 9.54
With the help of a calculator, we can see that the height of this triangle is about 9.54 feet.
We can use the Pythagorean theorem on triangles without illustrations. All we need to know is that the triangle is right and which side is the hypotenuse.
Example 3
If a right triangle has a hypotenuse length of 9 feet and a leg length of 5 feet, what is the length of the third side? We use the Pythagorean theorem with the hypotenuse, 9, by itself on one side, and the other two lengths, 5 and x, on the other.
52 + x2 = 92
x = √81 – 25 = √56 = 2√14 ≈ 7.48
The third side is about 7.48 feet long.
Pythagorean Word Problems
Many word problems involve finding a length of a right triangle. Identify whether each given length is a leg of the triangle or the hypotenuse. Then solve for the third length with the Pythagorean theorem.
Example 1
A diagonal board is needed to brace a rectangular wall. The wall is 8 feet tall and 10 feet wide. How long is the diagonal?
Having a rectangle means that we have a right triangle, and that the Pythagorean theorem can be applied. Because we are looking for the diagonal, the 10-foot and 8-foot lengths must be the legs, as shown in Figure 3.19.
The diagonal D must satisfy the Pythagorean theorem:
D2 = 102 + 82
D2 = 100 + 64 = 164
D = √164 = 2√41 ≈ 12.81 feet
Example 2
A 100-foot rope is attached to the top of a 60-foot tall pole. How far from the base of the pole will the rope reach?
We assume that the pole makes a right angle with the ground, and thus, we have the right triangle depicted in Figure 3.20. Here, the hypotenuse is 100 feet.
The sides must satisfy the Pythagorean theorem:
x2 + 602 = 1002
x = √6,400 = 80 feet
Pythagorean Triples
Question: Is the longest side in Figure 3.10 the hypotenuse? Answer: No.
Question: In Example 1, what are the lengths of the legs of the right triangle? Answer: 8 feet and 10 feet.
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Loci: Convergence
Approximate Construction of Regular Polygons: Two Renaissance Artists
by Raul A. Simon
Albrecht Durer
Albrecht Dürer (1471-1528), considered the father of modern German painting, was also a great mathematical amateur. He wrote a book titled Unterweysung der Messung..., which deals with all sorts of geometrical problems. In this book he gives an exact construction of the regular pentagon, but he also gives an approximate construction which is quick to execute in practical drawing. (His book was aimed at artisans, stonemasons, etc., who cared more for simple procedures than for geometrical accuracy.)
This construction is shown below:
Take a fixed opening of the compass: AB = a. Draw circles of radius a, centered at A and B; let these circles intersect at C and D. Then AB = AD = BD, as we know. Draw the circle of center D and radius DA. This circle will pass through B; let E be the point where it intersects CD, and let F and G be the points where it intersects the circles centered at A and B, respectively. Produce FE until it intersects at H the circle centered at B; produce GE until it intersects at I the circle centered at A. Then the intersection K of the circles of radius a, centered at H and I, gives the fifth vertex of the pentagon.
How regular is this pentagon? If it were exactly regular, all sides should subtend an angle of 360°/5 = 72° at the center of the pentagon. Therefore, all angles at the vertices of the pentagon should be equal to 180°-72°= 108°(since each of them is the sum of two equal angles of [180°-72°]/2). In fact, we will see thatangleABH= 108°21'58" --- a little more than 108°, so that some other angle must be a little less than 108°. First note that angle FBG = 90°; then, since FG = 2a and GB = a, we have FB = a√3 . Also, angle DFE = 45°andangle DFC =angle DFA = 60° (F, A, C are collinear, as well as C, B, G and F, D, G ) , and thereforeangle AFE = 15°. Since angle BFE subtends an arc of circle FABG equal to that subtended by angle AFE, we also have angle BFE= 15°.
Question: What is the main purpose of the approximate construction method for a regular pentagon described in the text? Answer: The main purpose is to provide a quick and practical method for drawing a regular pentagon, suitable for artisans and stonemasons who valued simplicity over perfect accuracy.
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We now use the law of sines fortriangle FBH; this gives sin(angleBHF) =√3 sin15° (observe that since angle BHF < 90°, there is no ambiguity). From hereangle BHF can be found and then angle HBF; then, subtractingangle ABF = 30° from angle HBF, we find that angle ABH is approximately 108°22', a value close to that of Pedoe [2].
Since angle BAI is equal to angle ABH, it too is a little greater than 108°. Also,each of anglesBHK andAIK is a little greater than 107°, while angle HKI is a little greater than 109°... and still, this would be barely noticeable in our drawing.
Dürer does not warn the reader of his book that this construction is approximate. (It is, also, executed with a "rusty" compass, i.e., a compass with a fixed opening.) In fact, he also quotes, as we said, the exact construction of the pentagon.
"Dürer's interest in the construction of regular polygons is explained by the applications of geometry during the Middle Ages in Islamic and Gothic decorative design and, after the invention of guns, in the building of fortified towns. (It is curious that very few buildings in history have been built based on the pentagonal shape. The Pentagon, near Washington, DC, is a notable exception.)" [2]
Summarizing this survey of Renaissance polygon constructions, we may say that both Leonardo da Vinci and Albrecht Dürer were great mathematical amateurs. "Leonardo wrote a lot about polygons; but it was Dürer, more than Leonardo, who transmitted to us the popular medieval constructions." [2]
Question: What is the name of the mathematical concept used to find the measure of angle BHF? Answer: The law of sines
Question: Who transmitted popular medieval constructions to us? Answer: Albrecht Dürer, more than Leonardo da Vinci
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The angle bisector theorem is concerned with the relative lengths of the two segments that a triangle's side is divided into by a line that bisects the opposite angle. It equates their relative lengths to the relative lengths of the other two sides of the triangle.
Lengths
If the side lengths of a triangle are , the semiperimeter and A is the angle opposite side , then the length of the internal bisector of angle A is[2]
If the internal bisector of angle A in triangle ABC has length and if this bisector divides the side opposite A into segments of lengths m and n, then[2]
where b and c are the side lengths opposite vertices B and C; and the side opposite A is divided in the proportion b:c.
If the internal bisectors of angles A, B, and C have lengths and , then[3]
No two non-congruent triangles share the same set of three internal angle bisector lengths.[4][5]
Angle bisectors of a rhombus
Bisectors of the sides of a triangle
Medians
Each of the three medians of a triangle is a line segment going through one vertex and the midpoint of the opposite side, so it bisects that side (though not in general perpendicularly). The three medians intersect each other at the centroid of the triangle, which is its center of mass if it has uniform density; thus any line through a triangle's centroid and one of its vertices bisects the opposite side. The centroid is twice as close to the midpoint of any one side as it is to the opposite vertex.
Perpendicular bisectors
The interior perpendicular bisector of a side of a triangle is the segment, falling entirely on and inside the triangle, of the line that perpendicularly bisects that side. The three perpendicular bisectors of a triangle's three sides intersect at the circumcenter (the center of the circle through the three vertices). Thus any line through a triangle's circumcenter and perpendicular to a side bisects that side.
In an acute triangle the circumcenter divides the interior perpendicular bisectors of the two shortest sides in equal proportions. In an obtuse triangle the two shortest sides' perpendicular bisectors (extended beyond their opposite triangle sides to the circumcenter) are divided by their respective intersecting triangle sides in equal proportions.[6]
Area bisectors and perimeter bisectors of a triangle
There are an infinitude of lines that bisect the area of a triangle. Three of them are the medians of the triangle (which connect the sides' midpoints with the opposite vertices), and these are concurrent at the triangle's centroid; indeed, they are the only area bisectors that go through the centroid. Three other area bisectors are parallel to the triangle's sides; each of these intersects the other two sides so as to divide them into segments with the proportions .[7] These six lines are concurrent three at a time: in addition to the three medians being concurrent, any one median is concurrent with two of the side-parallel area bisectors.
Question: What is the relationship between the lengths of the segments created by the angle bisector and the other two sides of the triangle? Answer: Their relative lengths are equated.
Question: If the internal bisector of angle A in triangle ABC has length x and divides the side opposite A into segments of lengths m and n, what is the relationship between m, n, b, and c? Answer: m/n = b/c
Question: Where do the three perpendicular bisectors of a triangle's sides intersect? Answer: At the circumcenter
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Examples identities's examples
An identity that shows that the cosine of the difference of two angles is related to the cosines and sines of the angles themselves. This identity is given below (A and B are used in place of alpha and beta, respectively since HTML does not support Greek characters). — "Algebra II: Trigonometric Identities - Math for Morons Like Us",
Definition of identities from Webster's New World College Dictionary. Meaning of identities. Pronunciation of identities. Definition of the word identities. Origin of the word identities. — "identities - Definition of identities at ",
In Mathematics, the term identity has several different important meanings An identity is an equality that remains true regardless of the values of An important application is the integration of non-trigonometric functions: a common trick involves. — "List of trigonometric identities - Citizendia",
true precisely when a = ±b. The formulas or trigonometric identities introduced in For student's convenience, the identities presented in this lesson are sumarized in. — "LESSON 6: TRIGONOMETRIC IDENTITIES by Thomas E. Price Directory", math.uakron.edu
Israelis and Palestinians, India and Pakistan, governments and insurgents, Protestants and Catholics, whites and blacks, labor and management these are all examples of identities that have at some times and some places resulted in intractable conflicts. — "Identity Issues",
The Pythagorean Identities are listed below: (Note that the second Using the double angle identities, we can now derive half angle identities. — "Trigonometric identities - AoPSWiki",
Identities are statements that are always true, as opposed to equations which are only true under certain conditions. For example 3x + 2x = 5x is an identity which is always true whereas 3x = 15 is an equation (or more precisely, a conditional equation) which is only true if x = 5. — "14.2 - Trigonometric identities",
Those identities marked with ** must be memorized. Those identities marked with * should remaining identities may be easily derived from those marked with a ** or a *. All of the proofs appear in. the pages after all. — "C:\Documents and Settings\Kendr", webalg.math.tamu.edu
You have seen quite a few trigonometric identities in the past few pages. These identities mostly refer to one angle denoted t, but there are a few of them involving two angles, and for those, the other angle is denoted. — "Summary of trigonometric identities", clarku.edu
In mathematics, trigonometric identities are equalities that involve trigonometric functions and are true for every single value of the occurring variables. Geometrically, these are identities involving certain functions of one or more angles. — "List of trigonometric identities - Wikipedia, the free",
If an equation contains one or more variables and is valid for all replacement values of the variables for which both sides of the equation are defined, then the equation Try to apply the Pythagorean identities as much as possible. — "Trigonometry: Fundamental Identities - CliffsNotes",
Question: Which identities are marked for memorization in the provided text? Answer: Those marked with .
Question: What does the term "identity" refer to in the context of trigonometry? Answer: An identity is an equality that remains true regardless of the values of the variables involved.
Question: What is the condition for the identity sin^2(x) + cos^2(x) = 1 to be true? Answer: The identity is true for all values of x.
Question: What is the difference between an identity and an equation, according to the text? Answer: An identity is always true, while an equation is only true under certain conditions.
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Step 1: Recognize that a tangent to a circle makes a right angle with the radius at the point of tangency (hereinafter referred to as point T). Hence, the line segment between the center of the circle and the point (4,6) (hereinafter referred to as point A) is the hypotenuse of a right triangle where the legs are the radius at point T and the segement between point T and point A. Also note that your circle is centered at the origin, point O.
Step 2: Determine the length of segment TA. Use the distance formula to calculate the measure of OA. Determine the radius of the given circle by inspection of the equation. Use these values for the hypotenuse and one leg of the right triangle described in Step 1 to calculate the measure of TA.
Step 3: Notice that the segment TA is the radius of a circle centered at A that intersects the given circle at T. It is here that we realize that this problem will have two solutions because this new circle does, in fact, intersect the original circle at two points.
Step 4: Having the radius and center of this new circle, write the equation of this new circle.
Step 5: Solve the 2X2 quadratic system consisting of the equations of the two circles. It will be convenient to write both equations as functions of using the positive square root on the original circle equation and the negative square root on the new circle equation. The two roots of the resulting quadratic will be the -coordinates of points and . Finding the -coordinates is then trivial arithmetic.
Step 6: Using the two-point form of an equation of a line, derive the two desired equations -- one containing the segment and the other containing the segment
As you can see from the description of the solution, this is a complex problem with a labor intensive and difficult solution, and not something I care to do for free. Please write back for a quotation for a complete solution including a full explanation and graphical illustration, if desiredNumbers_Word_Problems/717262: The second of three numbers is five times the first. The third is three more than the second. If the second is decreased by twice the third, the answer is -21. Find the three numbers. 1 solutions Answer 440181 by solver91311(16897) on 2013-02-21 16:24:14 (Show Source):
Set the expression equal to -21 and then solve for . Then calculate , and
John Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
Miscellaneous_Word_Problems/717261: The amount P of pollution varies directly with the population N of people. City A has a population of
442,000 and produces 260,000 tons of pollutants. Find how many tons of pollution we should expect
Question: What is the length of segment OA in the text? Answer: The length of segment OA is not provided in the text.
Question: What is the two-point form of the equation of the line containing segment BC in the text? Answer: The two-point form of the equation of the line containing segment BC is not provided in the text.
Question: What are the coordinates of points B and C in the text? Answer: The coordinates of points B and C are not provided in the text.
Question: What is the quadratic equation resulting from the system of equations of the two circles in the text? Answer: The quadratic equation is not provided in the text.
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Ptolemy's Theorem
In Trigonometric Delights (Chapter 6), Eli Maor discusses this delightful theorem that is so useful in trigonometry. In case you cannot get a copy of his book, a proof of the theorem and some of its applications are given here. The theorem refers to a quadrilateral inscribed in a circle. As you know, three points determine a circle, so the fourth vertex of the quadrilateral is constrained, and the quadrilateral is not a general one. This constraint gives it the property that the product of the diagonals is the sum of the products of opposite sides
Refer to the diagram at the left. The sides of the quadrilateral are chords of the circle, so the angles that each subtends at points on the circumference (such as A, B, C, and D) are equal. The angles marked with single and triple arcs are equal for this reason. Ptolemy's proof uses the line BE drawn so that the angles marked with double arcs are equal. The point E divides the diagonal AC so that AE + EC = AC. The triangles ABE and DBC are similar, because their angles are equal. Since sides in similar triangles are proportional, AE/DC = AB/BD. The triangles CBE and DBA are also similar, so that EC/AD = BC/BD. These ratios are not easy to see; if you have difficulty, draw the triangles separately. From these two equalities, we have AE + EC = AB.DC/BD + AD.BC/BD = AC, from which the theorem follows on multiplication by BD. This theorem is not in Euclid; the first we hear of it is in the Almagest, but it may have been known to Apollonius much earlier.
If the quadrilateral is a rectangle, the Pythagorean theorem follows at once, because the opposite sides are the sides of right triangles, and the diagonals, which are diameters of the circle, are the hypotenuses. Sketch the diagram and verify this for yourself. If the circle is assumed to have unit diameter, then the chords are equal to the sines of the angles they subtend at the circumference. By taking special quadrilaterals, important trigonometric identities are obtained. For example, if one diagonal is a diameter, then it subtends right angles at the circumference, so that the sides of the quadrilateral are the sines and cosines of the two acute angles of the right triangles, and the other diagonal is the sine of the sum of the acute angles (draw a diagram!). Hence, we get sin(α+β) = sin α cos β
+ cos α sin β. The formula for the sine of the difference of two angles can be obtained in the same way by taking the other two vertices of the quadrilateral on the same side of the diameter, instead of on opposite sides.
Question: What angles are equal in the diagram due to the sides being chords of the circle? Answer: The angles subtended by each side at points on the circumference (like A, B, C, and D).
Question: Who is the first person we hear of in the text who discussed Ptolemy's Theorem? Answer: Ptolemy, in the Almagest.
Question: What does the line BE in Ptolemy's proof help to achieve? Answer: It helps to make the angles marked with double arcs equal.
Question: Is Ptolemy's Theorem about a general quadrilateral? Answer: No, it's about a quadrilateral inscribed in a circle.
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Area of a circle's perpendicular components
Area of a circle's perpendicular components
I asked a similar question about gravity, but I would just like to check the math of this first.
In the first picture I have a circle of radius r, There is a point p on the circle, and I am first trying to show that the equation below gives the area of the circle:
[itex]2\int^{90}_{0} \frac{\pi r^2cos^2\theta}{90}d\theta[/itex]
I got this from the area of a sector of a circle, but rather than from the center, go with the sector of a circle whose radius is a chord. It cuts some off, but as d[itex]\theta[/itex]→0 so does the error.
So that gives you the area of a circle figured up by little slices, the angle of each we know, it is [itex]\theta[/itex], or to be more correct, in each sector there is an angle [itex]\theta^{\ast}_{\imath}[/itex] that points to the center of the area, and as [itex]d\theta[/itex] approaches 0, [itex]\theta^{\ast}_\imath[/itex] approaches [itex]\theta[/itex]
Alright, so if that is all good, then I introduced a vector, whose magnitude was the area of the sector defined above, and whose direction is [itex]\theta[/itex], relative to a line going through the point and the center, for each slice. I could figure out how much of the area is parallel to a line going from the point P to the center of the circle with:
[itex]2A\cos \theta[/itex]
Substituting in for A we get:
[itex]\int^{90}_{0}\frac{\pi r^2\cos^3\theta}{45}d\theta[/itex]
*bump* Ah, I already feel this going unanswered...if someone could please check my math I would greatly appreciate it. And correction on the title, parallel is what I meant. Also, I dropped the [itex]d\theta[/itex] from the A= equation on the second page, but it should be there.
Question: What is the integral used to find the area of the circle by considering sectors with a radius that is a chord? Answer: The integral used is \(2\int^{90}_{0} \frac{\pi r^2cos^2\theta}{90}d\theta\).
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egg shaped. An ellipse can be drawn by positioning pins at two points, F1 and F2, and loosely tying a length of
string between these points, the foci. Using a pencil point to hold the string
taut, follow the string to form the shape shown in Figure 19.10. Notice that
the distance remains constant.
The ellipse also has a center. It is the point of intersection
of the major and minor axes. These are special line segments that intersect at
right angles in the ellipse.
The major axis runs through the foci and center; the minor axis is the
perpendicular bisector of the major axis.
A circle is a special kind of ellipse
in which the two focal points are coincident.
FIGURE 19.10 Important
features of an ellipse
Some examples of ellipses in technical fields are:
• The shape of the end section on a pipe that has been cut on a
slant.
• The projection of a circle in an isometric view of a part to
be machined.
• The orbits of the planets around the sun.
Circles and ellipses are examples of conic sections. The other conic sections are the
parabola and the hyperbola. Conic sections are the shapes obtained by passing a
plane through a right circular cone or cones at various angles.
EXERCISES
19.1 Define the terms:
19.2 What is an inscribed polygon?
19.3 What is a circumscribed polygon?
19.4How
many foci does an ellipse have?
19.5Draw
an ellipse whose major and minor axes are equal. What figure has been drawn?
Question: What are the two special line segments in an ellipse? Answer: The two special line segments in an ellipse are the major axis and the minor axis.
Question: What is the shape formed by an ellipse? Answer: An ellipse is a shape that is formed by positioning pins at two points, F1 and F2, and loosely tying a length of string between these points, the foci. Using a pencil point to hold the string taut, following the string forms the shape of an ellipse.
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Ma3 Shape, space and measures
Coordinates
Teacher's notes
Next steps
Extend the problem: for example, given almost all of the coordinates and the name of the
resulting shape, find the missing coordinates.
Measuring
Teacher's notes
Measures distance between two dots to the nearest 0.1 cm.
Measures angles to the nearest degree.
Knows vocabulary related to angle, for example acute, obtuse.
Next steps
Estimate first before measuring angles, as a way of checking which scale to use on the
protractor.
Discuss sensible degrees of accuracy for measuring different lengths, for example the
appropriate degree of accuracy for measuring the length of the playground.
What the teacher knows about Peter's attainment in Ma3
Peter recognises irregular 2-D shapes such as a pentagon or octagon. He recognises and
names most triangles and quadrilaterals, such as isosceles triangle, trapezium and
parallelogram; he also understands their properties: for example, a square is a special rectangle
and has four lines of symmetry; a trapezium has only one pair of parallel sides; a scalene
triangle has no sides equal in length. He finds it difficult to draw 2-D shapes in different
orientations or to visualise 3-D shapes from nets. He understands mathematical terms such as
horizontal, vertical, congruent (to describe the same shape and size), parallel, regular, irregular.
Peter reflects simple shapes in horizontal and vertical mirror lines even when the shape is not
touching or parallel to the mirror line, but is not yet confident with oblique mirror lines, unless the
shape is touching the line. He can draw a shape after a rotation of 180°, with the help of tracing
paper.
In measures, Peter uses standard units of length, mass, capacity and time, choosing which
ones are suitable for a task. He measures lengths to the nearest millimetre and reads scales
between labelled divisions, for example, he reads 700 g on a scale going up in 50 g increments
labelled every 500 g. He measures angles to the nearest degree, understanding whether they
are acute or obtuse, and reads the time on an analogue clock. He calculates time durations
going over the hour on a 12-hour clock: for example, he knows that 5:20 pm is 1 h 40 min after
3:40 pm. He calculates perimeters of rectangles and finds areas by counting squares and half
squares.
Question: What is the smallest unit Peter uses to measure lengths? Answer: Millimetre
Question: What is one of the properties Peter understands about a trapezium? Answer: A trapezium has only one pair of parallel sides.
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Geometry Help I need help to find the coordinates of the circumcenter of each triangle. Isosceles triangle CDE with vertices C(0, 6), D(0, –6), and E(12, 0)
Thursday, November 15, 2012 at 8:25pm
geometry Write a paragraph proof for the following given- AD bisects CB and AD is perpendicular to CB. Prove- triangle ACD = triangle ABD
Thursday, November 15, 2012 at 4:13pm
geometry question proof needed Medians AX and BY of Triangle ABC are perpendicular at point G. Prove that AB=CG.
Thursday, November 15, 2012 at 10:39am
geometry Calculate the area of a circular patio that has a radius r = 3 yd. Use Ð= 3.14
Thursday, November 15, 2012 at 10:08am
geometry Write a flowchart proof: given BD bisects <abc and <1 is congruent to <3 and prove angle 2 is congruent to <3
Wednesday, November 14, 2012 at 9:37pm
geometry Prove that if the area of a rectangular patio abcd is 25sq units and one lengths is 5 units then abcd is a square
Wednesday, November 14, 2012 at 6:17pm
geometry Rick's 76ft times 50ft house sits on a plot of land as shown in the diagram below. If 1 bag of lawn food covers 5000 ft squared , how many bags should Rick buy to cover the lawn (shaded)?
Wednesday, November 14, 2012 at 1:05pm...
Wednesday, November 14, 2012 at 4:56am
Geometry Find the coordinates of the circumcenter of triangle DEF D(6,0) E(0,6) F(-6,0)
Tuesday, November 13, 2012 at 7:59pm
Geometry Given: Segment CE bisects <BCD; <A is congruent to <B Prove: Segment CE ll to segment AB -I used the exterior angle theorem to set the four angles equal to each other, but i don't know how to move on from there. The converse of something (maybe alternate interior ...
Thursday, November 8, 2012 at 5:51am
Geometry In a garden, a birdbath 2ft 6 in. tall casts an 18-in. shadow at the same time an oak tree casts a 90-ft shadow. How tall is the oak tree?
Wednesday, November 7, 2012 at 9:09pm
Geometry When a point is on a segment and it is equidistant from endpoints of the segment, what is it?
Question: What is the name of the point that is equidistant from the endpoints of a segment? Answer: The midpoint.
Question: What are the coordinates of the circumcenter of isosceles triangle CDE with vertices C(0, 6), D(0, –6), and E(12, 0)? Answer: The circumcenter of this triangle is at (6, 0).
Question: What is the total area of Rick's lawn (shaded) in square feet? Answer: The lawn is 76ft * 50ft = 3800 sq ft.
Question: What is the height of the oak tree in the given problem? Answer: The height of the oak tree is 45 feet.
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You can put this solution on YOUR website! There are two very important facts that you must know to do these types of questions.
1. In ANY triangle, the measurements of the three angles must add up to 180.
2. In a RIGHT triangle, one of the angles measures 90 degrees.
We have a right triangle with an angle measure of 35. The fact that the triangle is right tells us that one of the angles is 90. We need to find the measure of the third angle, which we will call x. Since the measures of the angles must add up to 180, we have
35 + 90 + x = 180
and we must solve for x. Simplifying a bit on the left hand side gives
125 + x = 180
Now subtract 125 from both sides to get
x = 180 - 125 = 55
so the third angle is 55 degrees. I hope this helps!
Question: What is the measure of the third angle (x) in the given right triangle? Answer: 55 degrees
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Points
Points are the basis of all Geometry. There are so many things you can do with the little buggers that the possibilities are endless. Points are zero-dimensional. That basically means that they have no height, length, or width. They are just there.
There are four main definitions of a point. They are the dot, the exact location, the ordered pair, and the node. A point has four definitions because, over the years, many different mathematicians have come up with their own ideas as to what a point should be. Since their ideas were all equally true, the point was given four main definitions instead of a single definition. In fact, the point is considered undefined for that reason (among others). When being written out, points are always represented by a capitol letter. If a point is on a line, it is often represented by the same letter.
The first definition of a point is the dot. This was probably the first kind of point ever thought up. You see, a dot has size - it has a definite, measureable, length and width. Probably the best example of a dot today would be the pixel. Yup, that's right - a pixel. Those tiny spots of color that make up your computer screen. A matrix is a rectangular array made up of lots of pixels, so that's what your computer screen is. As you most likely know, the more pixel in a computer or TV screen, the better the resolution.
The second definition of a point is an exact location. The exact location is the perfect example of the normal, zero-dimensional point. No matter how much you zoom in, there will always be another point in between two others. This definition of a point was discovered sometime between 550 B.C. and 150 A.D. One example of where these are used in real life is in measuring distances, especially between two cities. Some cities are more than a mile across, so mapmakers have to pick one exact location in the city to measure from. One use of the exact location ties in with the the next definition, the ordered pair. The number line, or coordinatized line, is a line where every point is represented by a number and vice versa.
Question: What does it mean for a point to have no dimensions? Answer: It means a point has no height, length, or width
Question: What are the four main definitions of a point? Answer: The dot, the exact location, the ordered pair, and the node
Question: What is an example of a dot in modern technology? Answer: A pixel
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Functions and Analysis
The Introduction of Twist (The Skew) in the Mathematics
The article define a mathematic entity called twist, which generates, in this way, notion of straight line.
Straight line becom thus a twist of eccentricity e = 0, and broken line (zigzag line) is a twist of s = ± 1.
Question: What is the relationship between a straight line and a broken line in this context? Answer: A straight line is a special case of a broken line with an eccentricity of 0.
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Ah, thanks a bunch Seer, and if it's not too much trouble, can anybody define point and line symmetry?
Symmetry, at least bilateral symmetry, which is what you seem to have in mind, is defined as being equal on two parts. Point symmetry is when a pair of points matches perfectly at 90 degree, 180 degree or acute or obtuse angles.
As for point and line symmetry, a square is good. All 90 degree angles, each side equal, each side therefore symmetric. Or, alternately, one could go with an even-numbered polygon, such as a hexagon, octagon, decagon, or so on. Each line must be at a specific angle, and so it can be divided equally. Tricky as hell, but it can be done.
As best as I can do. Take some with a grain of salt, it's been years since I studied geometry
Question: What is bilateral symmetry? Answer: Bilateral symmetry is when an object can be divided into two equal parts by a line, with each part being a mirror image of the other.
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Question 2:
Perform induction on the number of line segments within a weakly
simple polygon. Suppose that you form the convex hull of the given
segments. Together with the segments which have one endpoint touching
the hull, you get a weakly simple polygon. This polygon contains
some number of segments which don't intersect anything. If you
manage to attach one of these segments to the boundary of your
polygon, you have reduced the number of purely interior segments,
so by induction you can triangulate. I'm sure you can see that
its not difficult to do this. The base case is also quite easy
(one interior segment).
Question 3:
This is explained in section 5.7 of the textbook (Voronoi diagrams
- Connection to convex hulls , p.182, 2nd edition)
Question 4:
a) Focus on one line, and WLOG let it be vertical. There are
n-1 intersections on it, so n-2
segments between intersection points. The two lines that form
any two consecutive intersection points meet to form a triangle
that has a base on our vertical line. So if each of these triangles
is empty, we're done. Now look at how any two pairs of lines might
interact. Lets say we have pair of lines A,B
that form a triangle T to the right
of the vertical line. If another pair of lines C,D
has its triangle T2 to the
left, then the lines might extend to cross through T.
They would just form a new triangle within T.
So can still "charge" a triangle to the base of T.
If C and D meet on the right
side, we could have just one cross through T,
which is ok, as above. Or, both could cross through T:
case 1: if they actually intersect within T,
then we get 2 new triangles inside T.
One of the new triangles will belong to the base of T,
and the other to the base of T2.
You can draw an example to see how to consistently assign the
new triangles... meaning, in this case only one of A,B
crosses through T2 so it
is uniquely assigned one of the two newly formed triangles.
case 2: if C and D
dont intersect in T, then there is a
new triangle in T (and outside of T2),
and a new triangle in T2
(and outside of T). So for every triangle
T that we start out with, we can go
through all other pairs of adjacent lines and conclude that we
will always get a new triangle inside T,
that can be charged to the base of T.
This gives us n-2 triangles.
b) Assume that the statement is true for n lines.
When you add a new one, it cuts through n+1 faces.
For any face F that is cut into two new ones,
L and R, its not hard to show
that the new contribution to the sum is increased only if the
cut is very uneven. The difference is at most some linear amount
Question: What is the main topic of the text? Answer: The text discusses the process of triangulation for weakly simple polygons and the interaction of lines in Voronoi diagrams.
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Intersecting Secant-Tangent Theorem
If a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.
In the circle, is a tangent and
is a secant. They intersect at point U .
So, .
Example :
In the circle shown, if UX = 8 and XY = 10, then find the length of UV .
Question: What happens if the exterior point is on the circle? Answer: The theorem does not apply, as there would be no tangent or secant segments
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Copy onto cards the following descriptions, one per card: one right angle, one obtuse angle, and two acute angles. Thoroughly mix the three cards. Label the sections of each of two spinners "3," "4," "5," "6," "7," "8." The numbers on these spinners represent the lengths in centimeters of bases of trapezoids. Label equal sections of a third spinner "3," "4," "5." These numbers represent the heights of trapezoids.
Each group member spins a spinner. Then each group member selects a card and draws a trapezoid with the height and bases shown on the spinners and the angle(s) described on the card. Finally, each group member computes the area of his or her trapezoid and shares his or her work with the group.
Repeat the activity and then, as a group, write what you have observed about the areas and shapes of trapezoids.
Question: How many sections should be labeled on the third spinner? Answer: 3
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As they say, be careful what you wish for. The references you want may be found at:
In particular, paper 1 and paper 2. Paper 1 gives the basic concepts such as that a linear approximation at the reference point defines most of the keywords and then some projection is applied to that. Paper 2 is all the details of celestial coordinates and projections and it is big and full of equations, etc.
You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum You cannot vote in polls in this forum
Question: What is the purpose of a linear approximation in the context of this text? Answer: It defines most of the keywords
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Triangles/343272: Is it possible for four lines in a plane to have exactly zero points of intersection? One point? Two points? Three points? Four points? Five points? Six points? Draw a figure to support each of your answer. 1 solutions Answer 245704 by Fombitz(13828) on 2010-09-15 08:56:08 (Show Source):
You can put this solution on YOUR website! Look at the possible outcomes.
There are 8 of them.
H H H
H H T
H T H
H T T
T H H
T H T
T T H
T T T
Count the ones that have at least 2 heads.
There are four of them.
The probability is then 4 possible outcomes out of 8 total outcomes.
You can put this solution on YOUR website! Plug the ordered pair into the equation.
If it satisfies the equation, it is a solution.
If it doesn't, it's not.
Try (2,11)
False, (2,11) is not a solution.
.
.
.
Try (-2,1)
True, (-2,1) is a solution.
Question: What is the equation used to determine if an ordered pair is a solution? Answer: The equation is not explicitly stated in the text.
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Ellipse
Ellipse is one of the easiest topics in the Conic Sections of Co-ordinate Geometry in Mathematics.
"Ellipse" is defined as the locus of a point which moves such that the ratio of its distance (Eccentricity) from a fixed point (Locus) and a fixed line (Directrix) is less than one i.e. a point moves such that its distance from a fixed point is always less than the distance from a fixed line, we get a different types of curve for one value of eccentricity, which are similar for all values of eccentricity less than one. Thus curve looks like a circle but it is not exactly a circle. Rather it is more like the edges of an egg. And if we plot the movement of the Earth and other planets around the Sun, it is the same curve satisfying the above condition of eccentricity less than one. This beautiful curve has been named as "Ellipse".
In this chapter we will discuss in detail the nature/properties of this beautiful and important curve. As you will see, the curve is symmetrical about two axes. We will study the standard form of ellipse where the X and y-axes will be taken as these axes. The main emphasis in this chapter should be on learning the properties ofellipse. The judgement of using parametric co-ordinates, which can reduce the complexity of the problem, should also be learnt.
Ellipse is important from the perspective of scoring high in IIT JEE as it fetches 1-2 questions in most of the engineering examination Ellipseand
Question: What is the ratio of the distance of a point on an ellipse from a fixed point (locus) and a fixed line (directrix)? Answer: The ratio is less than one.
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What is geometry? A dictionary might say it is the study of shapes and configurations. An even better dictionary might say it is the study of figures in a space of a given number of dimensions and of a given type. A comic might say life is pointless without geometry (and you may or may not get it). But I want you to think of geometry as a game. It's a game that starts with a few rules that we must accept, like passing "GO" brings us $200, or five cards of the same suit are called a flush. Just like any other game, if you play it the same way long enough, even math can get boring. So, mathematicians have been twisting those rules and adding to the game for centuries. For instance, when German mathematician Bernhard Riemann asked the question, "What if parallel lines could intersect?" he started a new branch of geometry that led to Albert Einstein's Theory of Relativity. Think of geometry as a game whose rules you need to learn, and then, all I ask is that you play the game.
Question: What is geometry defined as in the text? Answer: The text defines geometry as a game that starts with a few rules that we must accept.
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first one i dont see a diagram.....GMAT/MBA Expert
they ask for a length, so statement 2 is not going to be useful - it gives us an angle only, and you can have a triangle of any size, teeny-tiny to huge, with those angles.
So, ACE.
AD is 6. <BAD is x. <BCD is 2x. <BCD is 2x. They want to know side BC.
<BDC is what's called an exterior angle of triangle BAD. You create this by taking any leg and just extending it out in a straight line. That exterior angle is always equal to the sum of the two opposite interior angles - in this case, angles BAD and ABD. So BAD + ABD = BDC. Fill in what you know. x + ABD = 2x. Solve. ABD = x.
So if ABD = x, it's the same angle as BAD. This means the sides opposite these angles are the same also. We know AD is 6, so BD is also 6.
Now look at the triangle on the right. We also have two identical angles there, so the two opposite sides are the same. One of those sides is BD (which we already know is 6) and the other side is BC, which we want to find. BC is also 6.
So statement 1 is sufficient. Answer is A.
_________________ Please note: I do not use the Private Messaging system! I will not see any PMs that you send to me!!
To determine how the angles affect each other, PLUG IN.
As you plug in, be sure to satisfy the conditions in the problem as well as the rules of geometry.
The drawings above show two valid combinations of angles.
In each case, AD = BD = BC.
Question rephrased: What is the length of AD or BD?
Statement 1: AD = 6.
Sufficient.
Statement 2: x=36.
No information about the any of the lengths.
Insufficient.
The correct answer is ALast edited by GMATGuruNY on Mon Feb 13, 2012 7:25 pm; edited 1 time in totalEven I reckon the answer is Option C that is both the statements are required. What is the OA.
The same reasoning was applied to figure 2, in which I plugged in x=25.
The result was the same: AB = BD, implying that AB = BD = BC
Question: What is the measure of angle ABD?
Answer: x
Question: What is the length of side BC?
Answer: 6
Question: What is the measure of angle BDC?
Answer: 2x
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While I had been teaching math in the classroom prior to our geometry
unit, this unit was the first unit that I introduced and carried through
completely. Obviously, I wanted to adhere to the curriculum of the
school, but I was interested in working within this curriculum to
design a more hands-on experience. The unit which follows is thus
closely aligned with the math textbook that serves as the math curriculum
for fourth grade. Each lesson follows the instructional goals laid
out in each lesson in the textbook and I use the homework assignments,
quizzes and tests from the text, so that I could ensure that I was
meeting the goals of the curriculum.
Three-Week Objectives:
Students will be able to categorize polygons by number of sides, side
length, angles, and if they have parallel or perpendicular sides.
Students will be able to describe the differences and similarities
between various polygonal forms, using the terminology introduced in
the unit.
Students will be able to define and find perimeter of any polygon;
find the area and volume of rectangles and rectangular prisms; and be
able to approximate the area and volume of other shapes.
Lesson 1: What is Geometry?
Objective:
Students will be able to write down a meaningful definition of geometry.
Procedure:
Introductory Activity:
Ask the students: "Who's heard of 'geometry'?"
(Knowing that the students have seen geometry in other classes at the
Swarthmore-Rutledge Elementary School, I assume that most of the students
will feel comfortable raising their hands.) Next, ask students: "What
do you think geometry is? What things are part of geometry?" Allow
several students to volunteer answers, and write students' initial
ideas on the board/overhead. Then, pulling ideas from the students'
brainstorms, write down a more formal definition of geometry on the
board/overhead.
At this time, have all the students take out a piece of loose-leaf
and fold it vertically (hot-dog fold). Have students label it with their
name and the heading "Geometry Dictionary." Have them label
the two columns "word" and "definition." The students
will record the definition of geometry in this dictionary. Students
will use this dictionary throughout the unit as an introduction to taking
and using notes.Examples
of Student Work on Dictionary Every word that students
put in their dictionary will also be included on a word wall in the
classroom. (These words will be introduced in bold-face throughout the
lesson plans).
Why Is Geometry Important?
Ask students to think about why geometry is important in the "real
world." Write down any ideas that surface in the discussion, making
sure students touch on the topics of art, architecture and engineering.
Lesson 2: Exploring Polygons
Objectives:
Students will be able to identify whether a given shape is a polygon
using the properties of polygons.
Students will be able to identify and name polygons that are triangles,
quadrilaterals, pentagons, hexagons, and octagons.
Question: What are the three main objectives for the students in the three-week geometry unit? Answer: 1) Categorize polygons, 2) Describe differences and similarities between polygonal forms, 3) Define and calculate perimeter, area, and volume of various shapes
Question: Was this the first unit on geometry that the teacher introduced and completed? Answer: Yes
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Latitudes are imaginary lines running parallel to the equator and longitudes are imaginary lines running parallel to the prime meridian and perpendicular to the equator. Latitude and longitude are used to reference a specific point on Earth.
Lines of latitude and longitude are equally spaced and their values are the angular distance of a point from equator and prime meridian.
Latitude and longitude are geographical terms used for locating a place on earth. Latitude is the distance around the earth starting from the prime merdidian. Longitude is the distance up or down starting from the equator.
Question: Which lines are perpendicular to the equator and run parallel to the prime meridian? Answer: Longitudes
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And while I'm at it please could you explain how, if you know 1 side of a right angled triangle e.g. 5 (ie. commonly memorised 5,12,13) and you know it's the shortest side, why you cannot automatically know the other sides will be 12 and 13?
Bottom two paragraphs: Let's say the sides can be represented in terms of x, 2x, and x*sqrt3. Now, if they tell us that the side opposite the 60 is 27, and we know that this 27 represents x*sqrt3, then we can find x from there -- x*sqrt3 = 27, so x = 27/sqrt3 The GMAT doesn't like radicals in denominators, so we multiply by sqrt3/sqrt3 (It's equal to 1, so it doesn't change the value) and we get 27sqrt3 / 3, which is 9sqrt3. (I think this answers statement 1, though I believe you may have meant 27sqrt3 rather than 29sqrt3)
The second triangle (bottom paragraph) works the same way. The side opposite the 60 degree angle is sqrt48, and is also x*sqrt3 in our "pattern" triangle. x*sqrt3 = sqrt48 Divide both sides by sqrt3 and we have x = sqrt48 / sqrt3 Remember that sqrt(a) / sqrt(b) = sqrt(a/b), so x = sqrt (48/3) which is sqrt(16) or 4. This tells us that the value of x is 4, so the hypotenuse, which is 2x in our "pattern" is 8.
As for your final question - If we know a triangle has one side equal to 5 and it's the shortest side, there's no way we can determine the other two sides without knowing more about the triangle. If we know it's a right triangle, it MIGHT be a 5-12-13 triangle, but there are many many triangles, even right triangles, that can be formed with 5 as the shortest side. Take 5 and 6 as the perpendicular sides, for example -- the third side can be found using the Pythagorean Theorem, and is sqrt(61). If, on the other hand, we know that the angles in the triangle are all the same as the angles in a 5-12-13 triangle (in the case of similar triangles), and we know that the shortest side is 5, then we DO know the other two sides are 12 and 13. We must know a fair amount about the triangle to determine this
Question: What is the difference between knowing just one side of a right-angled triangle and knowing it's a 5-12-13 triangle? Answer: Knowing it's a 5-12-13 triangle provides enough information to determine the other two sides, while knowing just one side does not.
Question: Can you determine the other two sides of a right-angled triangle if you know it's a 5-12-13 triangle? Answer: Yes, they would be 12 and 13.
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Given a line and a point not on the line, there exist(s) ____________ through the given point and parallel to the given line.
a) exactly one line
(Euclidean)
b) no lines
(spherical)
c) infinitely many lines
(hyperbolic)
Euclid's fifth postulate is ____________.
a) true
(Euclidean)
b) false
(spherical)
c) false
(hyperbolic)
The sum of the interior angles of a triangle ______ 180 degrees.
a) =
(Euclidean)
b) >
(spherical)
c) <
(hyperbolic)
The non-Euclidean geometries developed along two different historical threads. The first thread started with the search to understand the movement of stars and planets in the apparently hemispherical sky. For example, Euclid (flourished c. 300 bc) wrote about spherical geometry in his astronomical work Phaenomena. In addition to looking to the heavens, the ancients attempted to understand the shape of the Earth and to use this understanding to solve problems in navigation over long distances (and later for large-scale surveying). These activities are aspects of spherical geometry.
If a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, will meet on that side on which the angles are less than the two right angles.
For 2,000 years following Euclid, mathematicians attempted either to prove the postulate as a theorem (based on the other postulates) or to modify it in various ways. (Seegeometry: Non-Euclidean geometries.) These attempts culminated when the Russian Nikolay Lobachevsky (1829) and the Hungarian János Bolyai (1831) independently published a description of a geometry that, except for the parallel postulate, satisfied all of Euclid's postulates and common notions. It is this geometry that is called hyperbolic geometry.
From early times, people noticed that the shortest distance between two points on Earth were great circle routes. For example, the Greek astronomer Ptolemy wrote in Geography (c.ad 150):
It has been demonstrated by mathematics that the surface of the land and water is in its entirety a sphere…and that any plane which passes through the centre makes at its surface, that is, at the surface of the Earth and of the sky, great circles.
Question: What is the shape of the Earth according to Ptolemy? Answer: a sphere
Question: Which ancient activities are aspects of spherical geometry? Answer: The search to understand the movement of stars and planets, understanding the shape of the Earth, and navigation over long distances.
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Great circles are the "straight lines" of spherical geometry. This is a consequence of the properties of a sphere, in which the shortest distances on the surface are great circle routes. Such curves are said to be "intrinsically" straight. (Note, however, that intrinsically straight and shortest are not necessarily identical, as shown in the figure.) Three intersecting great circle arcs form a spherical triangle (see ); while a spherical triangle must be distorted to fit on another sphere with a different radius, the difference is only one of scale. In differential geometry, spherical geometry is described as the geometry of a surface with constant positive curvature.
There are many ways of projecting a portion of a sphere, such as the surface of the Earth, onto a plane. These are known as maps or charts and they must necessarily distort distances and either area or angles. Cartographers' need for various qualities in map projections gave an early impetus to the study of spherical geometry.
Elliptic geometry is the term used to indicate an axiomatic formalization of spherical geometry in which each pair of antipodal points is treated as a single point. An intrinsic analytic view of spherical geometry was developed in the 19th century by the German mathematician Bernhard Riemann; usually called the Riemann sphere (seefigure), it is studied in university courses on complex analysis. Some texts call this (and therefore spherical geometry) Riemannian geometry, but this term more correctly applies to a part of differential geometry that gives a way of intrinsically describing any
Question: Who developed an intrinsic analytic view of spherical geometry in the 19th century? Answer: The German mathematician Bernhard Riemann.
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What exactly do you mean by 'the shortest distance'? What are you trying to doThe shortest distance will either be a horizontal or vertical line (in which case any of a range of points can be used as the endpoints of the line), or a line connecting two of the corners.
If your x coordinates overlap, then the distance is vertical and you need only find the minimum distance between the minimum y of one rectangle and the maximum of the other. (If the y's overlap too, the distance should properly be 0.)
If your y coordinates overlap but x coords do not, the distance is horizontal and you need only find the minimum distance between the x's.
If neither overlap, find the minimum x distance and minimum y distance anyway, and use Pythagoras to get a diagonal.
Question: If the y-coordinates of the rectangles overlap but the x-coordinates do not, what should you do to find the shortest distance? Answer: Find the minimum distance between the x-coordinates.
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Question:
I am going to try out for the ARML math team in a few weeks and I have
noticed a type of problem which seems to appear on the tryouts with some
regularity. The problem will require the evaluation of an expression with
trigonometry functions such as (sin 25)(sin 35)(sin 85). We are not allowed
to use calculators and have a time limit. Here is another example:
(tan 54)(cos 54 + cos 162 + cos 270) All angles are in degrees. Can you
suggest any method for solving problems like these?
Good knowledge of the identities is crucial, but as you can see there is a
need for a certain amount of resourcefulness that can only come with
practice. (I can tell you that it took me an embarrassingly long time to
figure these out!) Good luck to you!
Question: Which of the following is NOT a trigonometric function mentioned in the text? A) sin B) cos C) tan D) log Answer: D) log
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I want to calculate the angle a line makes with the positive x-axis in a clockwise direction. It's a lot like a bearing except instead of North, I want the angle it makes with positive x-axis. The image illustrates what I'm after.
Below is the code I wrote to achieve this. It works fine but I am just wondering if there's a way to reduce all these if statements.
Question: What is the user's concern about their code? Answer: They want to know if there's a way to reduce all the if statements.
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Wendy - The details are nice, but the general form looks squashed. I'm sure you know, but you need to analyze carefully for the time being; with more practice, you will apply it intuitively :thumbsup:.
All lines of a set of parallel lines vanish to the same point.
Here are the 4 box still life's. Two copies as I had to take them at an angle with my camera .
Thank you, Diane
wendyc1276
12-04-2011, 06:40 PM
Arnoud - Can I ask what program you are using that you can add the lines out further than the drawing? I'd like to use that technique to check my work for this class and future works. I do have photoshop, but I don't know how easy that is to add the straight lines like you do. Thanks!
jrl11528
12-04-2011, 07:35 PM
Wendy,
Lacking anything better, I just checked my lesson with Powerpoint. You can only move the projection lines up down or tilt in increments, but it helped a lot. They are not exact, but at least point out the gross errors. I am making some adjustments to my outlines now and will be posting them in a few minutes. (I'm really struggling with this for some reason, so I am going to post the outlines for feedback before I spend more time with the details on the boxes.)
Wendy -
I do have photoshop, but I don't know how easy that is to add the straight lines Straight lines ( is easy enough in Photoshop. Do it in a separate layer, so you can safely erase and try again.
:wave:
arnoud3272
12-05-2011, 03:32 AM
Diane - Well done :clap:. There is one obvious error - we can see the left side of that box, so definitely the right side must converge to the left.
Jim - Good start :clap:. When the VP is near enough you did very well. For VP's far off the page, it is enough that it looks reaonable that the edges converge to a point on the EL (not so for architects, but their CAD software solves it for them :lol:) Your guess is mostly correct, except for the top box.
Arnoud, Thank you for the comment on the boxes. I have corrected and attached the work again.
Also, I have attached the Exercise 4 three items, except for the chair. Thanks for all you are doing. Diane
crafor
12-06-2011, 07:09 PM
Question: What does Wendy think about the general form of the lines? Answer: Wendy thinks the general form of the lines looks squashed.
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To calculate the distance between two atoms:Click/Enter these
atoms. Clicking them will input their names into the dialog
box. Click Execute to calculate the distance.
Bond Angle Structure > Bond angle.
To measure a bond angle, three atoms need to be selected so
that the angle can be measured between them. A dialog box opens.
Click on the atoms of interest or enter their names in each
text field. Select Execute to calculate the angle.
Dihedral
Angle Structure > Dihedral Angle
Just as described in Measure Distance and Bond
Angle, open the dialog box and click/enter the atoms
of interest. Dihedral angles require 4 atoms of interest.
Question: Which of the following is NOT a requirement to calculate a dihedral angle? A) 3 atoms B) 4 atoms C) 5 atoms Answer: A) 3 atoms
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Polar coordinates
With this calculator you can convert cartesian coordinates of a point
into polar coordinates and vice versa. The cartesian coordinates of a point
are the value of the abscissa x an the ordinate y.
The polar coordinates are the radius r for the distance between the point
and the pole (the origin of the cartesian coordinate system) and the angle
Θ (or azimut) for the angle (anti-clockwise) between the axis
with the angle 0° (corresponding to the abscissa in the cartesian coordinate system)
and the point.
Usage:
Type the cartesian coordinates or the polar coordinates of a point into the
corresponding fields. After a click with the mouse on any free space of the window
or the "calculate"-button the calculation is performed.
The fields with input data get a light-green background, fields with calculated
values are coloured pink. The calculation is performed with the pair of coordinates
changed last.
Move the mouse over a unit to read its full name.
Click on the "reset"-button to reset the calculation.
Example:
What are the polar coordinates of the point with the cartesian coordinates x=3 und y=5 ?
After a click with the mouse on any free space of the window or the
"calculate"-button you can read the result.
The radius (the distance between the origin and the point) is 5.831.
The angle between the connection origin to point and the abscissa-axis is 59.036°.
This value corresponds to 1.0304 radians or 0.32798*π
Question: What is the angle in radians for the point with Cartesian coordinates (x=3, y=5)? Answer: 1.0304 radians
Question: What is the result of converting the Cartesian coordinates (x=3, y=5) to polar coordinates in terms of radius? Answer: 5.831
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as we have seen, all lines that pass through the center of projection
are represented as a point, the representation of this line
is the intersection of line 9 with the picture plane. This intersection
is D1 for line 9 is parallel to the diagonals that converge
at that point. Now consider the triangle O VD1. Because it is
a right triangle with one 45° angle, it is isosceles; and
because the length of O V is d, the length of D1V is also d.
QED.
Box 2.3
In two-point perspective, to determine the center of projection
of a picture, it must represent at least two distinct objects
each with two conjugate vanishing points so that we have four
distinct vanishing points QED.
14 This problem was analyzed in great depth by
Jules de la Gournerie (1814-83) in his monumental Traité de Perspective
Linéaire (1884). Methods such as his have been applied to a substantial
number of works of Renaissance art. For a recent bibliography, see Welliver
(1973) and
15 Although Janson (1967, p. 88, footnote 25) argues convincingly
that Sanpaolesi's reconstruction contains errors, I have chosen to reproduce
his rather than Janson's because its elevation shows the location of the figures,
whereas Janson's shows only the architecture. (See also Battisti, 1971.)
16 The distance points are known as a conjugate pair of vanishing
points. For future reference, we may define this term: The perspective images
of any two lines pass through their respective vanishing points. If the lines
to be represented intersect, and if the angle of their intersection is a right
angle, their respective vanishing points are said to form a conjugate pair
that project at a right angle at the eye. It is, in fact, this right angle
property that allows us to use pairs of conjugate vanishing points to define
the correct viewing distance for the perspective construction.
Question: What is the property that allows us to use pairs of conjugate vanishing points to define the correct viewing distance for perspective construction? Answer: The right angle property
Question: What is the name of Jules de la Gournerie's monumental work on linear perspective? Answer: Traité de Perspective Linéaire
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We'll each eat through the Radius,
To the center where we get a Lady and Tramp kiss.
The distance from one side, through the middle to the other side
Is the Diameter, so never say die. Perimeter, distance around a circle is 2 pi R. Area is pi R squared,
How much space is in there, do you care?
Well I sure hope you do,
Or I'll be eating pancakes, and you'll be eating Froot-Loops.
It's all mathematics. We add like addicts.
We subtract like taxes, we multiply like rabbits.
We divide like axes. It's all mathematics
polygon - a closed shape with straight sides
regular polygon - a polygon where all sides and all angles are equal
triangle - a shape or polygon with three sides
quadrilateral - a shape or polygon with four sides
pentagon - a shape or polygon with five sides
hexagon - a shape or polygon with six sides
heptagon - shape or polygon with seven sides
octagon - a shape or polygon with eight sides
nonagon - a shape or polygon with nine sides
decagon - a shape or polygon with ten sides
decimal - a fraction written with a "decimal point," like .
5 rather than 1/2
decimeter - one tenth of a meter
congruent - equal, generally relating to two geometric figures that are identical in size or shape
equidistant - equally distant from two points
adjacent - directly next to
perpendicular - two lines that meet at a 90 degree angle
parallel - two lines that will never cross
plane - a two dimensional shape that extends infinitely in all directions
circle - a round shape where all points are equidistant from the center
radius - the distance from the center of a cirle to its circumference
diameter - a line that passes through the center of a circle and cuts it in half
perimeter - the distance around the outside of a shape. The perimeter, or circumference, of a circle =2 pi r
area - the amount of space within a two dimensional shape. The area of a circle = pir R squared.
What is the word for any shape with a bunch of sides?
Polygon
What is special about a regular polygon?
All sides are the same length.
How many sides does a heptagon have?
Seven
An octopus has the same number of legs as what shape?
Octagon
What does "congruent" mean?
Equal
When two lines will never cross, they're _____.
Parallel
What dimension is a plane?
2-D
In what shape is every point equidistant to the center?
A circle
The length from the center of a circle to one end is called the _____.
Radius
What do we call the distance around the circle?
Perimeter
Question: If a circle has a diameter of 10 units, what is its radius? Answer: 5 units
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rangian Points'm not sure what you use to conceptualize things, but basically, we're just looking for orbits that keep everything in the same orientation.
So, If M1 is many orders of magnitude larger than M2, L1 & L2 are practically on M2. If M1 = M2, L1 is midway between them, and L2 is somewhere giving it a one (year) orbit around the center of mass (coincidentally L1). L3 for small M2 is the same orbit as M2. For M1=M2 is a bit further out, being the same distance from M1 as L2 is from M2.They exist, but they are unstable, according to the wiki article.Yes, of course.They would be symmetric, with respect to the barycenter, of course.
If the distance between M1 and M2 is 2R (distance to the barycenter, the center of the orbit, is R), then I get L2 and L3 at a distance of .7789R, where 1.7789 is the solution to x = 1/(x+1)^2 + 1/(x-1)^2, or x^5 - 2x^3 - 2x^2 + x +2 = 0
Question: What is the distance of Lagrange points L2 and L3 from the barycenter when the distance between M1 and M2 is 2R? Answer: They are at a distance of 0.7789R.
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The name geometry comes from two Greek words meaning "earth" and
"to measure".
The ancient Egyptians used geometry to measure their fields and find the
boundaries of their land.
Euclidean geometry was organized in about 300 B.C. by a Greek mathematician named
Euclid. He arranged mathematical propositions into thirteen books called The
Elements. These books contained not only geometry but also algebra and advanced
arithmetic. The basic ideas presented by Euclid in these books have not been changed
through the years.
Question: Which of the following is NOT a subject covered in "The Elements"? A) Geometry B) Algebra C) Calculus Answer: C) Calculus
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As the difference of affine points, x
and y are vectors in the euclidean
image plane. Perpendicularity is
preserved when (Sx)^T(Sy) = 0, and
aspect ratio is preserved if
[(Sx)^T(Sx)]/[(Sy)^T(Sy)] = w²/h².
Question: What condition preserves the aspect ratio of these vectors? Answer: [(Sx)^T(Sx)]/[(Sy)^T(Sy)] = w²/h²
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i need calculus help asap
AloofGhost257 asked
1) One side of a right triangle is known to be 36 cm long and the opposite angle is measured as 30°, with a possible error of ±1°.
(a) Use differentials to estimate the error in computing the length of the hypotenuse. (Round your answer to two decimal places.)
Answer in CM
2) (b) What is the percentage error? (Round your answer to the nearest integer.)
Question: What is the percentage error in the calculation of the hypotenuse? Answer: Approximately 2%
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Now, let's get two-dimensional here. We'll start with the easy case, which is when the points line up. In that case, we can use the same rule, right? For instance, let's look at (4,3) and (10,3). How far apart are they? Same as before—6. We can just count, or we can just subtract, because the y-coordinates are the same. (Show them this visually!) Similarly, suppose we take (-2,5) and (-2,-8). Since the xx-coordinates are the same, we can just count again, or just subtract the yy-coordinates, and get a distance of 13.
Now, what if neither coordinate is the same? Then it's a bit trickier. But we're not going to use any magic "distance formula"—if you ever memorized one, throw it out. All we need is what we've already done. Let's look at (-2,1) and (4,9). (Draw it!) To find that distance, we're going to find the distance across and the distance up. So draw in this other point at (4,1). Now, draw a triangle, with the distance we want over here, and the distance across here, and the distance up here. These two sides are easy, because they are just what we have already been doing, right? So this is 6 and this is 8. So how do we find this third side, which is the distance we wanted? Right, the Pythagorean Theorem! So it comes out as 10.
The moral of the story is—whenever you need to find a distance, use the Pythagorean Theorem.
OK, one more thing before you start the assignment. That was the distance between two points.How about the distance from a point to a line? For instance, what is the distance from you to the nearest street? The answer, of course, is—it depends on where on the street you want to get. But when we say, distance from you to the street, we mean the shortest distance. (Do a few drawings to make sure they get the idea of shortest distance from a point to a line. If the line is vertical or horizontal, then we are back to just counting. If it's diagonal, life gets much more complicated, and we're not going to get into it. Except I sometimes assign, as an extra credit assignment, "find the distance from the arbitrary point (xx,yy) to the arbitrary liney=mx+by=mx+b. It's ugly and difficult, but I usually have one or two kids take me up on it. For the rest of the class, just promise to stick with horizontal and vertical lines, and counting
Question: What is the distance between the points (-2,5) and (-2,-8)? Answer: 13
Question: How is the distance between points with the same x-coordinate calculated? Answer: By subtracting the y-coordinates
Question: Which theorem is used to calculate the distance between the points (-2,1) and (4,9)? Answer: Pythagorean Theorem
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Mathematical Properties of the Golden Ratio
The Golden Ratio, Φ, is an irrational number that has the following unique properties:
Taking the reciprocal of Φ and adding one yields Φ. phi=1/phi+1, or Φ=1/Φ+1.
Φ squared equals itself plus one. In other words, Phi^2 =Phi+1, or Φ^2=Φ+1. These characteristics are indeed very interesting; it is the only number in the world has such properties.
If we convert the equation from 2. into the equation Φ^2-Φ-1=0 which is in the format ax^2 + bx + c = 0, so we can solve using the quadratic formula, x= (-b ± √(b^2 - 4ac))/(2a). Doing this we get x = (1 ± √5)/2.
Together, these two solutions are known as Phi (1.618033989) and phi (0.618033989). Phi and phi are reciprocals.
Φ vs. Π
Φ and Π (pi) have this in common: where Π is the ratio of the circumference to its diameter, Φ is the ratio of the length to the width of a perfect rectangle.
Question: In the quadratic equation Φ^2 - Φ - 1 = 0, what is the value of 'a'? Answer: 1
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To identifying poles, find two great circles that intersect with the desired pole point. Find the zone directions of these great circles by finding two planes in them (not planes that are diametrically opposed, as these contain all possible planes), and take the cross product of those plane directions. Then take the cross product of the directions of these planes and identify the direction of the pole point.
It is also possible to identify poles by using vector addition. e.g.:
Note: This animation requires Adobe Flash Player 8 and later, which can be downloaded here.
Question: What is the first method mentioned to identify poles? Answer: Finding two great circles that intersect with the desired pole point.
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Autograph Activity - Angles in the Same Segment
An Autograph Player activity to illustrate The Angles in the Same Segment circle theorem. You can use this activity on the interactive whiteboard, or for your students to investigate on their own. Autograph does not need to be installed to use this activity (so your students can even use it at home) More…, but you will need to install the free Autograph Player add-on (instructions given after clicking the link). For a full list of Autograph Activities, please click on the other web-link.
Question: What is the purpose of this activity? Answer: The purpose of this activity is to illustrate the Angles in the Same Segment circle theorem.
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For every whole number n greater than or equal to 3, it is possible to have a regular polygon with n sides. So far we've seen the equilateral triangle ( n = 3), the square ( n = 4), the regular pentagon ( n = 5), the regular hexagon ( n = 6), and the regular octagon ( n = 8). There can exist a regular polygon with 1000 sides (this might be called a "regular kilogon"), 1,000,000 sides (a "regular megagon"), or 1,000,000,000 sides (a "regular gigagon"). These last three would look pretty much like circles to the casual observer.
General, Many-sided Polygons
Once the restrictions are removed concerning the relationship among the sides of a polygon having four sides or more, the potential for variety increases without limit. Sides can have all different lengths, and the measure of each interior angle can range anywhere from 0° (0 rad) to 360° (2 π rad), non-inclusive.
Figure 4-4 shows some examples of general, many-sided polygons. The object at the top left is a non-convex octagon whose sides happen to all have the same length. The interior angles, however, differ in measure. The other two objects are irregular and non-convex. All three share the essential characteristics of a plane polygon:
Fig. 4-4 . General, many-sided polygons. The object with the shaded interior is the subject of Problem 2.
The vertices all lie in a single plane
No two sides cross
No two vertices coincide
No three vertices lie on a single straight line
All the sides are line segments of finite length
Polygons, Five Sides and Up Practice Problems
PROBLEM 1
What is the measure of each interior angle of a regular hexagon?
SOLUTION 1
Draw a horizontal line segment to start. All the other sides must be duplicates of this one, but rotated with respect to the first line segment by whole-number multiples of a certain angle. This rotation angle from side to side is 360° divided by 6 (a full rotation divided by the number of sides), or 60°. Imagine the lines on which two adjacent sides lie. Look back at Fig. 4-2. These lines subtend a 60° angle with respect to each other, if you look at the acute angle. But if you look at the obtuse angle, it is 120°. This obtuse angle is an interior angle of the hexagon. Therefore, each interior angle of a regular hexagon measures 120°.
Fig. 4-2 . A regular hexagon. Each side is s units long, and each interior angle has measure z . The extensions of sides (dashed lines) are the subject of Problem 1.
PROBLEM 2
Question: What would a regular polygon with 1,000,000 sides look like to a casual observer? Answer: It would look pretty much like a circle.
Question: What is the measure of each interior angle of a regular hexagon? Answer: 120 degrees.
Question: Which of the following is NOT a characteristic of a plane polygon? (a) All vertices lie in a single plane (b) No two sides cross (c) No two vertices coincide (d) All three vertices lie on a single straight line Answer: (d) All three vertices lie on a single straight line.
Question: Can a regular polygon have 2 sides? Answer: No, a regular polygon must have at least 3 sides.
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important thing where i got stuck... i found this on another site..its really help full.
I feel we are not supposed to take the literal meaning of the term "cord" {lineseg that lies inside the cirlce with its ends lieing on the circle"} [color=#FF4040]as its meaning but as a piece of wire/some material that goes around the 2 circles[/color].
so......... The answer is (B).
Since the tangents form squares with the radii.
the total length as tangents = 2*80+2*60 = 280
and the total lenght wound around the circles = 270deg/360deg * 2*pi*80 =120pi
Question: What is the total length of the "cord" wound around the circles? Answer: 120π units
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The compass can be opened arbitrarily wide, but (unlike some real compasses) it has no markings on it. A compass or pair of compasses is a Technical drawing instrument that can be used for inscribing Circles or arcs They can also be used as It can only be opened to widths that have already been constructed, and it collapses when not used for drawing.
The straightedge is infinitely long, but it has no markings on it and has only one edge, unlike ordinary rulers. It can only be used to draw a line segment between two points or to extend an existing line.
Each construction must be exact. "Eyeballing" it (essentially looking at the construction and guessing at its accuracy, or using some form of measurement, such as the units of measure on a ruler) and getting close does not count as a solution.
Stated this way, compass and straightedge constructions appear to be a parlour game, rather than a serious practical problem; but the purpose of the restriction is to ensure that constructions can be proven to be exactly correct. A parlor game is a group Game played indoors During the Victorian era in Great Britain and in the United States, these games were extremely One of the chief purposes of Greek mathematics was to find exact constructions for various lengths; for example, the side of a pentagon inscribed in a given circle. Regular pentagons The term pentagon is commonly used to mean a regular convex pentagon, where all sides are equal and all interior angles are equal (to The Greeks did not find constructions for three problems:
Squaring the circle: Drawing a square the same area as a given circle. Squaring the circle is a problem proposed by ancient Geometers.
Doubling the cube: Drawing a cube with twice the volume of a given cube. Doubling the cube (also known as The Delian Problem) is one of the three most famous geometric problems unsolvable by Compass and straightedge construction
Trisecting the angle: Dividing a given angle into three smaller angles all of the same size. The problem of trisecting the angle is a classic problem of Compass and straightedge constructions of ancient Greek mathematics.
For 2000 years people tried to find constructions within the limits set above, and failed. All three have now been proven under mathematical rules to be impossible generally — some angles, for example, can in fact be trisected, but many (in particular the innocent looking π/3) cannot. The radian is a unit of plane Angle, equal to 180/ π degrees, or about 57
The basic constructions
The basic constructions
All compass and straightedge constructions consist of repeated application of five basic constructions using the points, lines and circles that have already been constructed. These are:
Creating the line through two existing points
Creating the circle through one point with centre another point
Creating the point which is the intersection of two existing, non-parallel lines
Creating the one or two points in the intersection of a line and a circle (if they intersect)
Creating the one or two points in the intersection of two circles (if they intersect)
Question: What is the purpose of the restriction on compass and straightedge constructions? Answer: To ensure that constructions can be proven to be exactly correct.
Question: How many basic constructions are there in total? Answer: 5.
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Archimedes and Apollonius gave constructions involving the use of a markable ruler. Archimedes of Syracuse ( Greek:) ( c. 287 BC – c 212 BC was a Greek mathematician, Physicist, Engineer This would permit them, for example, to take a line segment, two lines (or circles), and a point; and then draw a line which passes through the given point and intersects both lines, and such that the distance between the points of intersection equals the given segment. This the Greeks called neusis ("inclination", "tendency" or "verging"), because the new line tends to the point. The neusis is a geometric construction method that was used in antiquity by Greek mathematicians
This construction extends geometry beyond the reach of Euclid's Elements. Euclid has no axiom, and can prove no theorem, that such verging lines even exist, so he cannot use them for constructions. In this expanded geometry, any distance whose ratio to an existing distance is the solution of a cubic or a quartic equation is constructible. This article discusses cubic equations in one variable For a discussion of cubic equations in two variables see Elliptic curve.In Mathematics, a quartic equation is one which can be expressed as a Quartic function equalling zero It follows that, if markable rulers and neusis are permitted, the trisection of the angle (see Archimedes' trisection) and the duplication of the cube can be achieved; the quadrature of the circle is still impossible. Some regular polygons, like the heptagon, become constructible; and John H. Conway gives constructions for several of them;[2] but the 11-sided polygon, the hendecagon, is still impossible, and infinitely many others. Construction A regular heptagon is not constructible with Compass and straightedge but is constructible with a marked Ruler and compassJohn Horton Conway (born December 26, 1937, Liverpool, England) is a prolific mathematician active in the theory of finite groups Use in coinage The Canadian dollar coin the Loonie, is patterned on a regular Hendecagonal prism, as is the Indian two-rupee coin
When only angle trisector is permitted, there is a complete description of all regular polygons which can be constructed, including above mentioned regular heptagon, triskaidecagon (13-gon) and enneadecagon (19-gon). Construction A regular heptagon is not constructible with Compass and straightedge but is constructible with a marked Ruler and compassIn Geometry, a triskaidecagon (or tridecagon is a Polygon with 13 sides and angles[3] It is open whether there are infinitely primes p for which a regular p-gon is constructible with ruler, compass and an angle trisector.
Origami
Question: What can be achieved with the use of markable rulers and neusis that couldn't be done with Euclid's Elements? Answer: The trisection of the angle, the duplication of the cube, and the construction of certain regular polygons like the heptagon
Question: What is the shape of the Canadian dollar coin, also known as the 'Loonie'? Answer: A regular hendecagonal prism
Question: Who were the mathematicians that introduced the concept of neusis? Answer: Archimedes and Apollonius
Question: Is the quadrature of the circle still impossible with these new tools? Answer: Yes, it is still impossible
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