Question

The diagram below models the layout at a carnival where G, R, P, C, B, and E are various locations on the grounds. GRPC is a parallelogram.

Parallelogram GRPC with point B between C and P forming triangle GCB where GC equals 350 ft, CB equals 300 ft, and GB equals 400 ft, point E is outside parallelogram and segments BE and PE form triangle BPE where BP equals 200 ft.

Part A: Identify a pair of similar triangles. (2 points)

Part B: Explain how you know the triangles from Part A are similar. (4 points)

Part C: Find the distance from B to E and from P to E. Show your work. (4 points)

Answer 1

[AI-generated answer — Bot GPT 3.5]

Part A:

The pair of similar triangles are triangle GCB and triangle BPE.

Part B:

The triangles GCB and BPE are similar because they share an angle at point B, and the other angles are congruent due to being corresponding angles of the parallelogram GRPC. Additionally, both triangles have a right angle.

Part C:

To find the distance from B to E, we can use the Pythagorean theorem in triangle BPE:

BE^2 = BP^2 + PE^2
BE^2 = 200^2 + PE^2
BE^2 = 40000 + PE^2

Since BE = 400 ft, we have:

400^2 = 40000 + PE^2
160000 = 40000 + PE^2
PE^2 = 160000 - 40000
PE^2 = 120000
PE = √120000
PE = 346.41 ft

Therefore, the distance from B to E is 346.41 ft.

To find the distance from P to E, we can use the fact that PE is the height of parallelogram GRPC, so it is equal to the distance GC, which is 350 ft.

Therefore, the distance from P to E is 350 ft.