find two numbers that have square roots between 326 and 327, are divisible by 100, and are a multiple of 6

step-by-step explanation:

points a and b are equidistant from the desired line, so the line you find by using them will go through point r.

using points other than a and b will give the desired line only if they are equidistant from r. if they are randomly located, the perpendicular line you create will not necessarily go through point r, hence will not be a solution to the problem.

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