2003 AMC 8 Problems/Problem 25

Problem

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?

$[asy] defaultpen(fontsize(8)); size(225); pair Z=origin, W=(0,10), X=(10,10), Y=(10,0), O=(5,5), B=(-4,8), C=(-4,2), A=(-13,5); draw((-4,0)--Y--X--(-4,10)--cycle); draw((0,-2)--(0,12)--(-2,12)--(-2,8)--B--A--C--(-2,2)--(-2,-2)--cycle); dot(O); label("$A$", A, NW); label("$O$", O, NE); label("$B$", B, SW); label("$C$", C, NW); label("$W$",W , NE); label("$X$", X, N); label("$Y$", Y, S); label("$Z$", Z, SE); [/asy]$

$\textbf{(A)}\ \frac{15}4\qquad\textbf{(B)}\ \frac{21}4\qquad\textbf{(C)}\ \frac{27}4\qquad\textbf{(D)}\ \frac{21}2\qquad\textbf{(E)}\ \frac{27}2$

Solution

The side lengths of square $\text{WXYZ}$ must be 5 cm, since the area is $25 {cm}^2$ . First, you should determine the height of $\triangle{ABC}$ . The distance from $\text{O}$ to line $\text{WZ}$ must be 2.5 cm, since line $\text{WX}$ = 5 cm, and the distance from $\text{O}$ to $\text{Z}$ is half of that. The distance from line $\text{WZ}$ to line $\text{BC}$ must be 2, since the side lengths of the small squares are 1, and there are two squares from line $\text{WZ}$ to line $\text{BC}$ . So, the height of $\triangle{ABC}$ must be 4.5, which is 2.5 + 2. The length of $\text{BC}$ can be determined by subtracting 2 from 5, since the length of $\text{WZ}$ is 5, and the two squares in the corners give us 2 together. This gives us the base for $\triangle{ABC}$ , which is 3. Then, we multiply 4.5 by 3 and divide by 2, to get an answer of ${(C)}\ \frac{27}{4}$ .

2003 AMC 8 (Problems • Answer Key • Resources)
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2003 AMC 8 Problems/Problem 25

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