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 375 ft, CB equals 325 ft, and GB equals 425 ft, point E is outside parallelogram and segments BE and PE form triangle BPE where BP equals 225 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

Triangle BGC and triangle BPE are similar by the Side-Angle-Side (SAS) similarity theorem. The distance from B to E is approximately 550 ft and the distance from P to E is approximately 225 ft.

Step-by-step explanation:

In this diagram, the pair of similar triangles can be identified as triangle BGC and triangle BPE.

To explain why these triangles are similar, we can use the Side-Angle-Side (SAS) similarity theorem. Triangle BGC and triangle BPE share two important pieces of information: BP is proportional to BC, and angle GBC is congruent to angle PBE. These shared attributes satisfy the SAS similarity theorem and prove that the triangles are similar.

To find the distance from B to E, we can use the concept of proportions. Since triangle BGC and triangle BPE are similar, we can set up a proportion using their corresponding sides: BC/BE = BG/BP. Plugging in the given values, we have 325/BE = 375/225. Cross-multiplying and solving for BE, we find that the distance from B to E is approximately 550 ft.

To find the distance from P to E, we can use the same proportion as before: BG/BP = GC/PE. Plugging in the given values, we have 375/225 = 375/PE. Cross-multiplying and solving for PE, we find that the distance from P to E is approximately 225 ft.