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 400 ft, cb equals 350 ft, and gb equals 450 ft, point e is outside parallelogram and segments be and pe form triangle bpe where bp equals 250 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)

Answers and Step-by-step explanations:

A.

One pair of similar triangles are ΔGBC and ΔEBP.

B.

∠GBC and ∠EBP are vertical angles, so by definition, they are congruent. We also know that since GRPC is a parallelogram, then opposite sides are parallel, so GC ║ RE. Then, since GE is a transversal of these two lines, ∠CGB and ∠PEB are alternate interior angles. By definition, they are congruent.

Thus, by the Angle-Angle Similarity Theorem, ΔGBC ~ ΔEBP.

C.

Similar triangles have the property that corresponding sides have the same ratio. Here, GB corresponds to EB, BC corresponds with BP, and GC corresponds to EP.

We know that BC / BP = 350 / 250. So, this ratio is equivalent to GB / EB and GC / EP:

GB/EB = GC/EP = 350/250

GB = 450 and GC = 400, so:

450/EB = 400/EP = 350/250

Solve for EB and EP separately:

450/EB = 350/250

350EB = 112500

EB = 2250/7 ft

400/EP = 350/250

350EP = 100000

EP = 2000/7 ft