2008 AMC 10A Problem 20

Because $AB \parallel DC$, the alternate interior angles at $K$ make $\triangle AKB \sim \triangle CKD$ with ratio $\frac{AB}{DC} = \frac{9}{12} = \frac34$. Hence the diagonals are cut in the ratio $3:4$:
$$\frac{BK}{KD} = \frac{AK}{KC} = \frac{3}{4}.$$

Now use shared altitudes. Triangles $AKB$ and $AKD$ share the vertex $A$ and have bases $BK$ and $KD$ on the same line, so
$$[AKB] = \frac{3}{4}[AKD] = \frac{3}{4}\cdot 24 = 18.$$
Similarly $[CKB] = \frac34 [CKD]$, and $[CKD] = \left(\frac43\right)^2 [AKB] = \frac{16}{9}\cdot 18 = 32$, giving $[CKB] = 24$.

(Indeed $[BKC] = [AKD]$ always: triangles $ABD$ and $ABC$ share base $AB$ and height, and both contain $AKB$.)

Adding the four pieces:
$$18 + 24 + 24 + 32 = 98.$$

The answer is $\boxed{\textbf{(D)}\ 98}$.

The diagonals of a trapezoid create two similar triangles on the bases (ratio $a:b$) and two equal-area triangles on the legs. With the crossing ratio $a:b$ known, every one of the four triangles is a fixed multiple of any other via "same altitude, bases in ratio $a:b$." Memorable form: if the base triangles have areas $a^2 t$ and $b^2 t$, the leg triangles each have area $abt$, and the trapezoid is $(a+b)^2 t$. Here $t = 2$ and $(3+4)^2 \cdot 2 = 98$.

• Scaling the area of AKD by the ratio 3:4 or (3:4)^2 directly; only the triangles on the two bases are similar, and AKD is not one of them.
• Assuming the two leg triangles have different areas, or forgetting to include $[AKD] = 24$ itself in the final sum.