1952 AMC 12 第 24 题

Place $AB$ on the $x$-axis and find the coordinates of $C$ from the two side lengths

The perpendicular through midpoint $D$ meets $CB$ at $E$; then use the shoelace formula on $A,D,E,C$

Since $AB=20,$ $AC=12,$ and angle $C$ is right, $BC=16.$ Put $A=(0,0)$ and $B=(20,0).$ Solving $$\begin{aligned} x_C^2+y_C^2&=144,\\ (x_C-20)^2+y_C^2&=256 \end{aligned}$$ gives $C=(\frac{36}{5},\frac{48}{5}).$ Also $D=(10,0).$

The line $CB$ meets $x=10$ at $E=(10,\frac{15}{2}).$ The shoelace formula for $A,D,E,C$ now gives $$\begin{aligned} [ADEC] &=\frac12\left( 10\cdot\frac{15}{2} +10\cdot\frac{48}{5}\right.\\ &\qquad\left.{} -\frac{15}{2}\cdot\frac{36}{5}\right)\\ &=\frac{117}{2}=58\frac12. \end{aligned}$$

Thus, the correct answer is B.