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?

$\dfrac{15}{4}$

$\dfrac{21}{4}$

$\dfrac{27}{4}$

$\dfrac{21}{2}$

$\dfrac{27}{2}$

Solution:

We get that the side lengths of $WXYZ$ are $5$ 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.

2003 AMC 8 Problem 25

Solution:

Problem 25 in Other Years