This exam was adminstered in June 2026.
June 2026 Geometry, Part II
Each correct answer is worth up to 2 credits. Partial credit can be given. Work must be shown or explained.
25. On the set of axes below, △DAN ≅ △MIK.
Describe a sequence of rigid motions that maps △DAN onto △MIK.
Answer:
Make sure your transformations go in the correct direction, in this case from the top left to the lower right.
You should see that the orientation has changed in that the triangle is facing in the opposite direction. This indicates that a reflection has taken place. After that, a tranlation will move the figure to its image.
A reflection over the y-axis moves the triangle to Quadrant I with point A' at (1,8). Point I is at (1,-3) with is 11 units down and 0 units left or right.
The reflection over the y-axis (ry-axis) is followed by a translation 11 units down. (T0,-11).
This sequence could be done in the reverse order and achieved the same result.
26. In parallelogram LEAD below, C is on DE and S is on DA such that SC ⊥ DE.
If m∠LED = 21°, determine and state the measure of ∠CSA.
Answer:
Fill in what you know: LED = 21, so CDA must be 21 degrees as well, and angle DCS = 90 because it's a right angle.
This means that by the Remote Angle Theorem, angle CSA = 90 + 21 = 111 degrees.
If you forgot that theorem, you could have found that CSD = 180 - 90 - 21 = 69, and then CSA = 180 - 69 = 111.
27. A section of tree trunk from a white pine tree can be modeled by a cylinder. The diameter of the trunk is 1.5 feet, and the height of this section is 8 feet. If white pine weighs 25 pounds per cubic foot, determine and state the weight of this section of the white pine tree, to the nearest pound.
Answer:
You need to find the volume of the cylinder first. Once you have the volume, you can use that and the density to the mass (weight) using the formula m = VD.
The volume formula uses radius, but we were given diameter. Cut that in half. r = 1.5/2 = .75.
V = π r2 h = π (7.5)2 (8) = 14.137...
Leave the number in the calculator, or write down at least three decimal places. DO NOT ROUND IN THE MIDDLE OF A PROBLEM.
m = VD = 14.137 (25) = 353.429...
The weight is 353 to the nearest pound.
28. In △ADC below, points B and F are on AC and AD , respectively, such that AB = 40, BC = 30, and BF is drawn.
If AF = 64, determine and state the length of FD that would prove BF || CD.
Answer:
If BF || CD then triangles ABF and ACD are similar and their corresponding sides are proportional. Conversely, if the corresponding sides are proportional, then the triangles are similar and those two sides must be parallel.
Solve the proportion AB / BC = AF / FD
40 / 30 = 64 / FD
40 FD = (30)(64)
FD = (30)(64)/(40) = 48
It is also easy to see that QUAD could be broken up into four congruent triangles, for which the area could be found.
29. In △CLM below, m∠C = 33°, CL = 8, and CM = 15.
Determine and state, to the nearest tenth, the area of △CLM.
Answer:
You have two options here. The first is the use the Law of Sines: A = 1/2 (a)(b) sin C. The second is the use the sine function to find the height of the triangle, and then use A = 1/2 b h. This is the same exact thing but doing it in two steps might make it seem easier to you.
The Law of Sines allows you to find the area of a triangle without knowing the height if you know the lengths of two sides and the size of the angles included between them.
A = 1/2 (8) (15) sin (33) = 32.678...
The area is 32.7.
If you tried to find the height, you could use sin (33) = h / 8, and then h = 8 sin (33) = 4.357...
Then A = 1/2 (15)(4.357...) = 32.678
Which is the same answer.
30. In the diagram of △ABC below, use a compass and straightedge to construct the angle bisector of ∠ABC. [Leave all construction marks.]
Answer:
Starting at point B, make an arc that intersects the two sides of the angle. (I made a circle because it's easier with my paint program, but it isn't necessary.) Make two more arcs at the points where the first arcs met the sides. These points will intersect on the bisector of the angle. Draw a line from the intersection of these two arcs to the vertex of the angle.
31. The diagonals of parallelogram GRAM intersect at P. If RP = 12 and GA = 24, explain why GRAM is a rectangle.
Answer: The diagonals of a parallelogram biset each other. Therefore, if RP = 12 than RM = 24, which makes it congruent to GA.
If the diagonals of a parallelogram are congruent, then the parallelogram must be a rectangle.
End of Part II
How did you do?
Questions, comments and corrections welcome.
More to come.
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