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

Answer:

The similar triangles are $\triangle BPE$ and $\triangle BCG$

$BE = 321$ and $PE =286$

Step-by-step explanation:

Given

See attachment for the required figure

Solving (a): The similar triangles

The similar triangles are $\triangle BPE$ and $\triangle BCG$

Solving (b): Why they are similar

Both triangles are similar because $\triangle BCG$ is dilated (i.e. enlarged) and then reflected to give $\triangle BPE$.

Solving (c): Calculate BE and PE

The following are equivalent ratios

$BP: BC = BE : BG$

and

$BP: BC = PE : CG$

Solving for BE, we have:

$BP: BC = BE : BG$

Substitute the known values

$250:350 = BE:450$

Express as fraction

$\frac{250}{350} = \frac{BE}{450}$

Multiply both sides by 450

$450 * \frac{250}{350} = BE$

$321 = BE$

$BE = 321$ -- approximated

Solving for PE, we have:

$BP: BC = PE : CG$

Substitute known values

$250:350 = PE:400$

Express as fraction

$\frac{250}{350} = \frac{PE}{400}$

Multiply both sides by 400

$400 * \frac{250}{350} = PE$

$286 = PE$

$PE =286$