2005 AMC 10A Problem 8

Question

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?

25

32

36

40

42

The four corner triangles, such as

\triangle ABH,

are congruent right triangles

Each corner triangle is right with hypotenuse

\sqrt{50}

and one leg

BE = 1;

the inner square’s side is the other leg minus

1

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.

Problem 8 in other years