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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)

Sagot :

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Answer:

A. The triangles that are similar would be ΔGCB and ΔPEB due to Angle, Angle, Angle similarity theorem.

B. The proof of the fact that ΔGCB and ΔPEB are similar pairs of triangles is as follow;

∠CGB ≅ ∠PEB (Alternate Interior Angles)

∠BPE ≅ ∠BCG (Alternate Interior Angles)

∠GBC ≅ ∠EBP (Vertical Angles)

C. To find the distance from B to E and from P to E, we will first find PE and then BE by proportion;

225/325 = PE/375

PE = 260 ft

BE/425 = 225/325

BE = 294 ft

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