2005 AMC 12A Problem 7

Problem 7 of 25EasierGeometry

Square

EFGH

is inside square

ABCD

so that each side of

EFGH

can be extended to pass through a vertex of

ABCD.

Square

ABCD

has side length

\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

By the symmetry of the figure, triangles

ABH,

BCE,

CDF,

and

DAG

are congruent right triangles. Hence

\begin{aligned} &BH = CE = \sqrt{BC^2 - BE^2} \\ &= \sqrt{50 - 1} = 7. \end{aligned}

Since

E

lies between

B

and

H,

the side of the inner square is

EH = BH - BE = 7 - 1 = 6.

Therefore the area of

EFGH

is

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.