2003 AMC 8 Problems/Problem 25


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]

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


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 $\boxed{\textbf{(C)}\ \frac{27}{4}}$.

See Also

2003 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last Problem
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS