2005 AMC 10A Problem 8

Problem 8 of 25EasierGeometry

In the figure, the length of side

AB

of square

ABCD

is

\sqrt{50},

E

is between

B

and

H,

and

BE = 1.

What is the area of the inner square

EFGH?

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Solution

The triangles

ABH,

BCE,

CDF,

and

DAG

are congruent right triangles. In

\triangle BCE

the hypotenuse is

BC = \sqrt{50}

and

BE = 1,

so

CE = \sqrt{50 - 1} = 7.

Since

BH = CE = 7

and

E

lies on

BH

with

BE = 1,

the inner square’s side is

EH = 7 - 1 = 6,

giving area

6^2 = 36.

Thus, the correct answer is C.

More practice

Concepts: square (geometry) · Pythagorean Theorem · congruence (geometry)

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.