2003 AMC 8 Problem 25

Question

In the figure, the area of square

WXYZ

is

25 \text{ cm}^2.

The four smaller squares have sides

1

cm long, either parallel to or coinciding with the sides of the large square. In

\triangle ABC,

AB = AC,

and when

\triangle ABC

is folded over side

\overline{BC},

point

A

coincides with

O,

the center of square

WXYZ.

What is the area of

\triangle ABC,

in square centimeters?

\dfrac{15}{4}

\dfrac{21}{4}

\dfrac{27}{4}

\dfrac{21}{2}

\dfrac{27}{2}

Let

M

be the midpoint of

\overline{BC}.

Folding makes the distance from

A

to

\overline{BC}

equal the distance from

O

to

\overline{BC}.

The side length of

WXYZ

is

5

cm, since

\sqrt{25} = 5.

We also know that the distance from

\overline{WZ}

to

\overline{BC}

is

2

since it is the sum of the side lengths of

2

unit squares.

Finally, the distance from

A

to

\overline{BC}

is the same as the distance from

\overline{BC}

to

O,

which is

2 + \dfrac{5}{2} = \dfrac{9}{2} \text{ cm}.

Now, we can find

BC,

which is

WZ - 2 = 5 - 2 = 3 \text{ cm}.

Therefore, the area of

\triangle ABC

is

\dfrac{1}{2} \cdot 3 \cdot \dfrac{9}{2} = \dfrac{27}{4} \text{ cm}^2.

Thus, C is the correct answer.

Problem 25 in other years