Page 27


Problem 1


IF there is an isometry, then the distance from A to B will be equal to the distance from C to D. The distance from C to D is found easily to be



Calculating the distance from A to B is a little more difficult without a calculator, but it is something that you should be able to do. The distance from A to B is given by



Simplifying the stuff under the BIG radical



So the distance from A to B is



which is equal to the distance from C to D. So there IS and isometry. Finding it is another matter, but we will be getting there.



Problem 2




Problem 3


HYPOTENUSE-LEG IS A VERY IMPORTANT CONGRUENCE CRITERION FOR RIGHT TRIANGLES!

For right triangles, SSA IS a congruence criterion. Sketch two right triangles, clearly marking the right angles in each, as well as marking one leg congruent to the corresponding leg and marking the two hypotenuse as congruent. Well, if you know the hypotenuse and a leg, then you know the other leg by the Pythagorean Theorem. Essentially, Hypotenuse-Leg (HL) is just Side-Side-Side (SSS) in disguise.



Problem 4


If there are 12 pentagonal faces, and no edge is shared by another pentagon, there are (12)(5)=60 edges.
Notice that each pentagon has five vertices, and each vertex is shared by two pentagons, so there are 60/2=30 vertices.
Notice that each triangle has three edges, none being shared by another triangle, so there are 60/3=20 triangles.

In summary, there are 60 edges, 32 faces, and 30 edges. Compare these results to the icosahedron you did a couple of pages back.



Problem 5


(a) Square both sides of a+b=6. Then make the substitution ab=7.
(b) I will look for this on your page.



Problem 6


There does not seem to be an easy way to do this, but if you can find the area of this triangle, you can find the length of the altitude from A.

I can find the area of the triangle by using the parallelogram area formula and dividing by 2.

The vector from E to A is [7,1]. The vector from E to F is [3,-4]. The area of the parallelogram formed by these two vectors is |(7)(-4)-(1)(3)|=|-28-3|=31. The triangle area is then 31/2. Therefore, 31=(EF)(Altitude), so Altitude=31/5=6.2. The time for Avery to reach the river is 6.2 / 10 = 0.62 hours.

page_27_problem_6a.png



Problem 7


5x-2y=5(7)-2(9) which leads to 5x-2y=17



Problem 8


This is another variation of the "find a point that is some fraction of the way from this point to that point."

Notice that A, B, and C are collinear. Point A is one-third of the way from Q to R, so A' will be one-third of the way from M to K.
B is three-fifths of the way from P to R. B' will be three-fifths of the way from L to K.
C is one-fourth of the way from B to A. C' will be one-fourth of the way from B' to A'.

Construct this transformation in GeoGebra and check your answers.

page_27_problem_8a.png



Problem 9




Problem 10


Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now)



Problem 11


Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now)



Problem 12


Knowing that the area of a triangle is half the product of the side length and the altitude length drawn to that side, the three altitudes lengths are 168/13, 168/14, and 168/15.





Page 28


Problem 1


Use trig and slope triangle to confirm what is in the applet below.

Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now)



Problem 2


The line described by P_t is



The line described by Q_t is



Solve for the intersection point of the two lines. It's Algebra 1.



Problem 3


Divide this isosceles triangle in half, making a right triangle with sides 10, 2.5, and whatever the altitude of the isosceles triangle is. Find the area of this triangle, and double it.

In a formula, it looks like this:



Now...why??



Problem 4


Be certain to drag the points around. How do you know this is NOT an isometry? Is it possible to have the square's image align along the grid lines? Is it possible to make the image of the square a rectangle? Is it possible to make the image of the square a square? Is it possible to make the image a RHOMBUS? How does the area of the image compare to the area of the square? Is it possible to get a vertex of the image and the vertex of the square to coincide? I am trying to get you to PLAY with the applet, to figure things out about this transformation.

Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now)



Problem 5, 6, and 7

5

Since isometries do not change lengths of segments, if an isometry is applied to a right triangle, the image must be a right triangle as well, in that the lengths of the sides of both triangles are congruent, and if one is right, then they both are right.

6

The image of the origin MUST be the intersection of the two lines. The transformation is NOT an isometry, because the lines are not perpendicular.

7

If you read #6, then you know the answer.



Problem 8


The area of the triangle is 6 square units. The altitude drawn from (1,4) is 4 units long. The altitude drawn from the origin is 12/sqrt(20) units long. The altitude drawn from (3,0) is 12/sqrt(17) units long.



Problem 9


I believe they want you to do it algebraically.



Problem 10


The vector from P to B is [4,4]. To find point D such that the vector from D to M is also [4,4], you can take the coordinates of M and subtract [4,4]. D is then (1,-2).

Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now)



Problem 11


The net translation is [7,12].



Problem 12


The midpoint of the two points is (2,3). The vector from one point to the other is [6,-2]. The perpendicular vector is [2,6] or [1,3]. The equation of the line of reflection is x+3y=(2)+3(3) which is x+3y=11.



Problem 13


The area f the parallelogram is |(2)(-1)-(5)(7)| = 37.



Problem 14


I begin by drawing a triangle from the point to be reflected to the intercepts of the line. The area of this triangle is 3/2.

To find the distance from A to the line, I will write the area calculation in another way.



The distance from point A to its image will be twice this, or



Now I need to find a vector that is perpendicular to the line, and that is one unit long. A perpendicular vector is [2,3]. A perpendicular vector that is one unit long is



The image of point A is therefore found by



page_28_problem_14.png



Problem 15


page 28 problem 15
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