2005 AMC 12A Problem 7

Question

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?

25

32

36

40

42

By symmetry the four corner triangles are congruent right triangles

In right triangle

BCE,

the leg

CE = BH = \sqrt{BC^2 - BE^2}

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.

Problem 7 in other years