This exam was adminstered in August 2025.
August 2025 Geometry, Part II
Each correct answer is worth up to 2 credits. Partial credit can be given. Work must be shown or explained.
25. Triangle D'A'N' is the image of △DAN after a translation. Explain why △D'A'N' must be congruent to △DAN.
Answer:
A translation is a rigid transformation that preserves size and shape. Therefore △D'A'N' must be congruent to △DAN.
That's all you need to write.
26. The table below lists five metals and their densities.
A solid metal cube has an edge length of 5 cm and a mass of 982.5 grams.
Using the table above, determine and state the type of metal from which this cube is made.
Answer:
Density is equalt to the mass divided by the Volume. Find the Volume, then find the density. Check your answer against the table.
V = (5)3 = 125 cm3.
d = 982.5 / 125 = 7.86
According to the table, the cube must be made of Iron.
27. The endpoints of CAS are C(–3,1) and S(7,6). Determine and state the coordinates of point A
such that the ratio of CA:AS is 3:2.
[The use of the set of axes below is optional.]
Answer:
Graphing may make this easier to find. Or you can use a formula.
CA will be 3/5 of the length of CS, and AS will be 2/5 of the length of CS.
The difference of the x-coordinates is 10, and 3/5 (10) = 6. Add 6 to -3 and the x-coordinate of A is 3.
The difference of the y-coordinates is 5, and 3/5 (5) = 3. Add 3 to 1 and the x-coordinate of A is 4.
Point A is located at (3,4).
Another formula you use is the following:
(-6/5, 2/5) + (21/5, 18/5)
(15/5, 20/5)
(3, 4)
I only started teaching this a year ago because the formula showed up in material. It looks counterintuitive, but it works. I showed it to my students. Many scratched their heads, but a few liked it and started using it.
28. The ramp shown in the diagram below has an angle of elevation of 4.8°. The ramp is built to a
landing 0.6 m above the ground.
Determine and state the length of the ramp, to the nearest tenth of a meter.
Answer:
You have the height, which is opposite the angle, and you are looking for the hypotenuse. O and H means the SINE ratio.
sin (4.8°) = 0.6 / x
x = 0.6 / sin (4.8°) = 7.17...
The ramp has a length of about 7.2.
29. Angle KML is the vertex angle of isosceles triangle KLM below. Side LM is extended through
vertex M to point N.
If m∠K = 15°, determine and state m∠KMN.
Answer:
KML is an isosceles triangle with M as the vertex angle. We are given the size of angle K, which means that we know the size of angle N.
Because of the Remote Angle Theorem, we know that m∠KMN = 15 + 15 = 30 degrees.
If you forgot the Remote Angle Theorem, you can add 15 and 15 to get 30, then subtract 180 - 30 to get 150, which is the size of angle KML. Since KML is supplementary to KMN, then KMN must be 180 - 150 = 30.
30. In the diagram below of circle L, the area of the shaded sector KLM is 7.5π and LK = 5.
Determine and state the degree measure of angle KLM, the central angle of the shaded sector.
Answer:
The Area of a sector is equal to the Area of the entire circle times the fraction of the circle that the sector represents. That fraction is the central angle divided by 360.
(x / 360) π(5)2 = 7.5π
(x / 360) 25π = 7.5π
x = (360) (7.5π) / (25π)
x = 108
The measure of KLM is 108 degrees.
31. Using a compass and straightedge, construct the image of point A after a reflection over BC.
[Leave all construction marks.]
Answer: This is easier than drawing a perpendicular line. Make an arc with a length of BA, centered on point B. Make an arc with a length of CA, centered on Point C. The two arcs will meet at the reflection of point A.
End of Part II
How did you do?
Questions, comments and corrections welcome.
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