2003 AMC 8 Problems/Problem 21

Problem

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

Solution

Using the formula for the area of a trapezoid, we have $164=8(\frac{BC+AD}{2})$ . Thus $BC+AD=41$ . Drop perpendiculars from $B$ to $AD$ and from $C$ to $AD$ and let them hit $AD$ at $E$ and $F$ respectively. Note that each of these perpendiculars has length $8$ . From the Pythagorean Theorem, $AE=6$ and $DF=15$ thus $AD=BC+21$ . Substituting back into our original equation we have $BC+BC+21=41$ thus $BC=10\Rightarrow \boxed{B}$

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2003 AMC 8 Problems/Problem 21

Problem

Solution