Grade ten. Quadratic equations. Best answer will be brainliest

this is a good question! i'll do some research and come back when i have the answer!

part a: triangles gcb and epb

part b. sides cg and rp are parallel by definition of parallelogram.

angles cgb and e are congruent by being alternate interior angles of parallel lines cut by a transversal. angles gbc and ebp are congruent because they are vertical angles. by aa similarity, the triangles are similar.

part c. since the triangles are similar, the ratios of the lengths of corresponding sides are equal.

calculation of be:

gb/be = cb/bp

(400 ft)/be = 300/200

(400 ft)/be = 3/2

3be = 2 * 400 ft

3be = 800 ft

be = 800/3 ft

calculation of pe:

gc/pe = cb/bp

(350 ft)/pe = 300/200

(350 ft)/pe = 3/2

3pe = 2 * 350 ft

3pe = 700 ft

pe = 700/3 ft