2008 AMC 10A Problem 20

The diagonals split the trapezoid into four triangles around $K$. Find all four areas.

1. How the diagonals are cut. Because $AB$ and $DC$ are parallel, triangle $AKB$ and triangle $CKD$ have the same angles (they are similar, one is a scaled copy of the other). The scale is $AB : DC = 9 : 12 = 3 : 4$. So $K$ cuts each diagonal in the ratio $3 : 4$:
$$BK : KD = 3 : 4, \qquad AK : KC = 3 : 4.$$
2. Triangle $AKB$. Triangles $AKB$ and $AKD$ both have their top corner at $A$ and their bases $BK$ and $KD$ on the same line $BD$. So they have the same height, and their areas are in the ratio of the bases, $3 : 4$:
$$\text{area } AKB = \frac{3}{4} \times 24 = 18.$$
3. Triangle $CKD$. Triangles $AKD$ and $CKD$ both have their corner at $D$ and bases $AK$ and $KC$ on line $AC$. So their areas are in the ratio $3 : 4$:
$$\text{area } CKD = \frac{4}{3} \times 24 = 32.$$
4. Triangle $BKC$. Triangles $AKB$ and $BKC$ both have their corner at $B$ and bases $AK$ and $KC$. So
$$\text{area } BKC = \frac{4}{3} \times 18 = 24.$$
5. Add the four pieces:
$$18 + 24 + 32 + 24 = 98.$$

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

Two tools do everything: the triangles on the two parallel sides are scaled copies of each other, which tells you how $K$ cuts the diagonals; and triangles with the same height have areas in the same ratio as their bases. In every trapezoid, the two side triangles ($AKD$ and $BKC$) always have equal area, which is a good check.

A shortcut to remember: if the parallel sides are in ratio $a : b$, the four triangles have areas in ratio $a^2 : ab : ab : b^2$. Here that is $9 : 12 : 12 : 16$. Triangle $AKD$ is one of the $12$ parts and has area $24$, so one part is $2$. The whole trapezoid is $9 + 12 + 12 + 16 = 49$ parts, which is $49 \times 2 = 98$.

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 $\boxed{\textbf{(D)}\ 98}$.