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 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)

Answer 1

The segment GB that extends through  parallelogram GRPC form triangles  

ΔGBC and ΔBPE.

Part A: The pair of similar triangles are ΔGBC and ΔBPE

Part B: Two angles in ΔGBC are congruent to two angles in ΔBPE, therefore, by Angle-Angle, AA similarity postulate, ΔGBC and ΔBPE are similar

Part C: The distance from B to E is $\underline {321\frac{3}{7} \ feet}$

The distance from P to E is $\underline {285\frac{5}{7} \ feet}$

Reasons:

Part A: The diagram of G, R, P, C, B, and E is attached

The pair of similar triangles in the figure are ΔGBC and ΔBPE

Part B: The reason why triangles ΔGBC and ΔBPE are similar are;

∠GBC in ΔGBC is congruent to ∠EBP in ΔBPE by vertical angle theorem

∠GBC ≅ ∠EBP

∠BPE in ΔBPE is congruent to ∠GCB in ΔGBC by alternate interior angles

formed between two parallel lines $\overline{CG}$ and $\overline{PR }$ that are intersected by a

common transversal $\overline{CP}$.

∠BPE ≅ ∠GCB

Therefore, ΔGBC and ΔBPE are similar by Angle-Angle, AA, similarity

postulate.

Part C: By similar triangles, we have;

$\dfrac{\overline{BP}}{\overline{BC}} = \mathbf{\dfrac{\overline{BE}}{\overline{BG}}}$

Therefore;

$\dfrac{250}{350} = \dfrac{\overline{BE}}{450}$

$\overline{BE} = \dfrac{250}{350} \times 450 = 321\frac{3}{7}$

The distance from B to E = $321\frac{3}{7}$ ft.

Similarly, we have;

$\mathbf{\dfrac{\overline{PE}}{\overline{GC}}} = \dfrac{\overline{BP}}{\overline{BC}}$

Therefore;

$\dfrac{\overline{PE}}{400} = \dfrac{250}{350}$

${\overline{PE}= 400 \times \dfrac{250}{350} = 285\frac{5}{7}$

The distance from P to E = $285\frac{5}{7}$ ft.

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