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:

a) A pair of similar triangles is triangles GBC and EBP

b) At least two angles in triangle GBC are congruent to two angles in triangle EBP, so according to the AA Similarity Postulate, triangles GBC and EBP are similar.

$\begin{gathered} \angle GBC\cong\angle EBP \\ \angle G\cong\angle E \end{gathered}$

c) Distance from B to E is 294.2 ft

Distance from P to E is 259.6 ft

Explanation:

Given a diagram that models the layout at a carnival where G, R, P, C, B, and E are various locations on the grounds and that GRPC is a parallelogram;

a) A pair of similar triangles is triangles GBC and EBP

b) The AA Similarity Postulate states that if two angles in one triangle are congruent to two angles in another triangle, therefore the two triangles are said to be similar.

Looking at the given diagram, we can see that angles GBC and EBP are vertically opposite angles and vertically opposite angles are always equal, so we have that;

$\angle GBC\cong\angle EBP$

Also, we can notice from the diagram that angles G and E are alternate interior angles and they are always equal, so we have that;

$\angle G\cong\angle E$

We can see from the above that at least two angles in triangle GBC are congruent to two angles in triangle EBP, so according to AA Similarity Postulate, triangles GBC and EBP are similar.

c) The distance from B to E can be found by setting up proportions as seen below;

$\begin{gathered} \frac{BE}{BG}=\frac{BP}{BC} \\ \frac{BE}{425}=\frac{225}{325} \\ BE=\frac{225\cdot425}{325} \\ BE=\frac{95625}{325} \\ BE=294.2ft \end{gathered}$

And the distance from P to E can be found by setting up proportions as seen below;

$\begin{gathered} \frac{PE}{CG}=\frac{PB}{CB} \\ \frac{PE}{375}=\frac{225}{325} \\ PE=\frac{375\cdot225}{325} \\ PE=\frac{84375}{325} \\ PE=259.6ft \end{gathered}$