Question

In the diagram below, triangle ABC is an isoceles right triangle that overlaps square ADEF. EF=4 and AC=8. What is the ratio of the area of the quadrilateral EDBC to the area of Pentagon AFECB? Express your answer as a common fraction.

Answer 1

EDBC is a trapezoid

We know AC=8 and since the sides of an isosceles right triangle are in ratio

$1 : 1 : \sqrt 2$

we conclude

$AB = BC = AC/\sqrt{2} = 8/sqrt{2} = 4 \sqrt 2$

We have a square so AD=ED=EF=4.

DB is the height of the trapezoid,

$DB = AB - AD = 4 \sqrt 2 - 4$

So the area of the trapezoid is

$t = \frac 1 2 (b_1+b_2)h = \frac 1 2 (4 \sqrt{2} + 4)(4 \sqrt 2 - 4) = 8(\sqrt 2 + 1)(\sqrt 2 -1) = 2(2 - 1) = 8$

AFECB is the sum of two isosceles right triangles AFE + ABC so has area

$p = \frac 1 2 (4)^2 + \frac 1 2 (4 \sqrt{2})^2 = \frac 1 2 (16 + 32) =24$

That's a ratio of 8:24:8 or 1:3

That must mean that ADE and AFE each have area 8 as well.

Answer: 1/3