Question

Part C: Find the distance from B to E and from P to E. Show your work. (4 points)

Answer 1

Answer:

Part 1) $BE=266\frac{2}{3}\ ft$

Part 2) $PE=233\frac{1}{3}\ ft$

Step-by-step explanation:

Part  1)

Find the distance from B to E

To solve this problem I assume that CGRP is a parallelogram

so

Triangles CGB and PEB are similar by AA Similarity Theorem

Remember that

If two figures are similar, then the ratio of its corresponding sides is proportional and its corresponding angles are congruent

so

$\frac{BG}{BE}=\frac{CB}{PB}$

substitute the values

$\frac{400}{BE}=\frac{300}{200}$

solve for BE

$BE=400(200)/300$

$BE=\frac{800}{3}\ ft$

Convert to mixed number

$BE=\frac{800}{3}\ ft=\frac{798}{3}+\frac{2}{3}=266\frac{2}{3}\ ft$

Part 2)

Find the distance from P to E

$\frac{CG}{PE}=\frac{CB}{PB}$

substitute the values

$\frac{350}{PE}=\frac{300}{200}$

solve for PE

$PE=350(200)/300$

$PE=\frac{700}{3}\ ft$

Convert to mixed number

$PE=\frac{700}{3}\ ft=\frac{699}{3}+\frac{1}{3}=233\frac{1}{3}\ ft$