2008 AMC 10A Problems/Problem 20

Problem

Trapezoid $ABCD$ has bases $\overline{AB}$ and $\overline{CD}$ and diagonals intersecting at $K$ . Suppose that $AB = 9$ , $DC = 12$ , and the area of $\triangle AKD$ is $24$ . What is the area of trapezoid $ABCD$ ?

We now introduce the concept of area ratios: given two triangles that share the same height, the ratio of the areas is equal to the ratio of their bases. Since $\triangle AKB, \triangle AKD$ share a common altitude to $\overline{BD}$ , it follows that (we let $[\triangle \ldots]$ denote the area of the triangle) $\frac{[\triangle AKB]}{[\triangle AKD]} = \frac{KB}{KD} = \frac{3}{4}$ , so $[\triangle AKB] = \frac{3}{4}(24) = 18$ . Similarly, we find $[\triangle DKC] = \frac{4}{3}(24) = 32$ and $[\triangle BKC] = 24$ .