2011 AMC 10A Problem 11

Problem 11 of 25IntermediateGeometry

Square

EFGH

has one vertex on each side of square

ABCD.

Point

E

is on

\overline{AB}

with

AE=7\cdot EB.

What is the ratio of the area of

EFGH

to the area of

ABCD?

Show solution

Solution

Let

x = EB.

Then

AB = 8x.

Applying the Pythagorean Theorem to a side of

EFGH,

we get

\sqrt{(7x)^2 + x^2} = \sqrt{50x^2}

The desired ratio is then

\dfrac{\sqrt{50x^2}^2}{(8x)^2} = \dfrac{50}{64} = \dfrac{25}{32}.

Thus, B is the correct answer.