This exam was adminstered in August 2024 .

### August 2024 Geometry, Part II

Each correct answer is worth up to 2 credits. Partial credit can be given. Work must be shown or explained.

**25.** * In the diagram below, △SUV ~ △TRE.
*

*If SU = 5, UV = 7, TR = 14, and TE = 21, determine and state the length of SV.*

**Answer: **

Because the triangles are similar, then the corresponding sides will be proportional. If we compare the corresponding sides TR and SU, then we will have the scale factor for the dilation from triangle SUV to TRE.

TR/SU = 14/5, which is the scale factor.

Therefore (14/5) SV = 21, which is the length of TE.

Using inverse operations, SV = 21 (5/14) = 15/2 = 7.5

SV = 7.5

You could have created this proportion and solved:

This would be solved using the same operations.

**26.** * Using a compass and straightedge, construct the line of reflection that maps △ABC
onto its image, △DEF. [Leave all construction marks.]
*

**Answer: **

The line of reflection will be the perpendicular bisector of the line segment drawn from CF.

Using a straightedge to draw CF. Then create a perpendicular bisector, which is two arcs and a straight line.

First, draw the blue line connecting points C and F.

Second, make a circle centered around C and a circle *the same size* centered on F. They need to be big enough to overlap in two places. *Note: my circles overlap, but they should've been a little bigger to give me more room to work with.*

Finally, draw the line of reflection through the two points of intersection in the two circles.

**27.** *Triangle MAX has vertices with coordinates M(-5,-2), A(1,4), and X(4,1).
Determine and state the area of △MAX.
[The use of the set of axes below is optional.]
*

**Answer: **

Plotting the points and drawing the triangle will not get you any credit, but it will make it easier to visualize the problem.

Once you have the triangle plotted, draw a rectangle around the triangle so that M is a corner of the rectangle and A and X are on the sides of the rectnagle. You know have four rectangles.

You can figure out the size of the three outer triangles using 1/2 * base * height and counting the boxes. Find the area of the rectangle, and subtract the area of the three outer triangles. Whatever is left is the area of MAX.

The area of the rectangle is Length * Width = 9 * 6 = 54.

The area of the three outer triangles are 1/2(6)(6) = 18, 1/2(3)(3) = 4.5, and 1/2(9)(3) = 13.5.

Therefore, the area of MAX = 54 - 18 - 4.5 - 13.5 = **18**.

You could find the base and the height of the triangle. Note: this is a two-point question, so the answer shouldn't be complicated.

To find the any of a triangle, you need to find the length of the base and the length of the altitude (or height). Those two lines need to be perpendicular to each other. You can use the distance formula to find these.

As soon as the image is graphed, we can see that MA has a slope of 1, it has a rise of 1 and a run of 1, and 1/1 = 1. We can also see that AX has a slope of -1 because it was a rise of -1 and a run of 1, and -1/1 = -1. Since (1)(-1) = -1, then MA is perpedicular to AX, so their lengths can be used as the base and the height.

Use the distance formula or the Pythagorean Theorem to find the lengths of these segments.

MA = &sqrt;(6^{2} + 6^{2}) = &sqrt;(72). AX = &sqrt;(3^{2} + 3^{2}) = &sqrt;(18).

The area of the triangle is 1/2 * base * altitude = 1/2 &sqrt;(72) &sqrt;(18) = 1/2 &sqrt;(1296) = 1/2 (36) = 18.

**28.** *A person observes a kite at an angle of elevation of 32° from a line of sight that begins 4 feet above
the ground, as modeled in the diagram below. At the moment of observation, the kite is 70 feet
above the ground.
*

Determine and state the horizontal distance, x, between the person and the point on the ground directly below the kite, to the nearest foot.

**Answer: **

You have the opposite side and you are looking for the adjacent side, so you need to use the TANGENT ratio. The

*Trick*to this question is that you have to remember to subtract 4 from the 70 feet above the ground because the kite is only 66 feet above the observer.

Tan 32 = 66 / x

x = 66 / (Tan 32) = 105.622...

The distance along the ground is 106 feet, to the nearest foot.

**29.** *In △AGL below, N and E are the midpoints of AG and AL, respectively, NE is drawn.
*

*If NE = 15 and GL = 3x - 12, determine and state the value of x.
*

**Answer: **

The midsegment (the line connecting two midpoints) is equal to half of the length of the third side of the triangle. You can write an equation that says that double the size of NE is equal to the length of GL.

2(15) = 3x - 12

3x - 12 = 30

3x = 42

x = 14.

**30.** * In the diagram below, △TAN is the image of △SUN after a reflection over NZ.
*

*Use the properties of rigid motions to explain why △TAN ≅ △SUN.*

**Answer: **

In a rigid transformation, the image is the same size and shape as the pre-image. Since distance is preserved the corresponding sides must be congruent. Therefore by SSS, the triangles must be congruent.

Note that in a question like this, I'm never sure exactly how much of an explanation that they are looking for. They basically want a definition of rigid transformation but it has to address this problem specifically.

**31.** * A pyramid with a square base is made of solid glass. The pyramid has a base with a side length of
5.7 cm and a height of 7 cm. The density of the glass is 2.4 grams per cubic centimeter.
*

*Determine and state, to the nearest gram, the mass of the pyramid.
*

**Answer: ** You need to find the Volume of the pyramid and then using the density and the Volume, find the mass of the pyramid.

The Volume of a pyramid is 1/3 * the Area of the base * the height. The base is a square, so it's area is (5.7)^{2}, or 32.49. Therefore the Volume is 1/3 * (32.49) * (7) = 75.81.

Density is equal to mass divided by the Volume (d = m/V), so the mass is equal to Density times Volume (m = Vd). Therefore, the mass = (75.81) * (2.4) = 181.944, which is 182 grams to the nearest gram.

**End of Part II**

How did you do?

Questions, comments and corrections welcome.

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