Question

The area of trapezoid $ABCD$ is $164\text{ cm}^2.$ The altitude is $8$ cm, $AB$ is $10$ cm, and $CD$ is $17$ cm. What is $BC,$ in centimeters?

$9$

$10$

$12$

$15$

$20$

Drop perpendiculars from $B$ and $C$ to $\overline{AD}$ , meeting it at $E$ and $F$ .

In the left and right right triangles, $AE=\sqrt{10^2-8^2}=6$ and $FD=\sqrt{17^2-8^2}=15$ .

The two side triangles have areas $\frac12\cdot6\cdot8=24$ and $\frac12\cdot15\cdot8=60$ . The middle rectangle has area $8\cdot BC$ .

Thus $164=24+60+8BC$ , so $8BC=80$ and $BC=10$ .

Thus, B is the correct answer.