# 1952 AHSME Problems/Problem 24

## Problem

In the figure, it is given that angle $C = 90^{\circ}$, $\overline{AD} = \overline{DB}$, $DE \perp AB$, $\overline{AB} = 20$, and $\overline{AC} = 12$. The area of quadrilateral $ADEC$ is: $[asy] unitsize(7); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair A,B,C,D,E; A=(0,0); B=(20,0); C=(36/5,48/5); D=(10,0); E=(10,75/10); draw(A--B--C--cycle); draw(D--E); label("A",A,SW); label("B",B,SE); label("C",C,N); label("D",D,S); label("E",E,NE); draw(rightanglemark(B,D,E,30)); [/asy]$ $\textbf{(A)}\ 75\qquad\textbf{(B)}\ 58\frac{1}{2}\qquad\textbf{(C)}\ 48\qquad\textbf{(D)}\ 37\frac{1}{2}\qquad\textbf{(E)}\ \text{none of these}$

## Solution $[ADEC]=[BCA]-[BDE]$

Note that, as right triangles sharing $\angle B$, $\triangle BDE \sim \triangle BCA$. $\overline{BD}=\frac{20}{2}=10$

Because the sides of $\triangle BCA$ are in the ratio $3:4:5$, $\overline{BC}=16$.

The sides of our triangles are in the ratio $\frac{10}{16}=\frac{5}{8}$, and the ratio of their areas is $\left(\frac{5}{8}\right)^2=\frac{25}{64}$. $[BCA]=\frac{12 \cdot 16}{2}=96$, and $[BDE]=\frac{96 \cdot 25}{64}=\frac{75}{2}$. $[ADEC]=\frac{192-75}{2}=\boxed{\textbf{(B)}\ 58\frac{1}{2}}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 