1980 AHSME Problems/Problem 7

Problem

Sides $AB$ , $BC$ , $CD$ and $DA$ of convex polygon $ABCD$ have lengths $3$ , $4$ , $12$ , and $13$ , respectively, and $\angle CBA$ is a right angle. The area of the quadrilateral is

$[asy] defaultpen(linewidth(0.7)+fontsize(10)); real r=degrees((12,5)), s=degrees((3,4)); pair D=origin, A=(13,0), C=D+12*dir(r), B=A+3*dir(180-(90-r+s)); draw(A--B--C--D--cycle); markscalefactor=0.05; draw(rightanglemark(A,B,C)); pair point=incenter(A,C,D); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$3$", A--B, dir(A--B)*dir(-90)); label("$4$", B--C, dir(B--C)*dir(-90)); label("$12$", C--D, dir(C--D)*dir(-90)); label("$13$", D--A, dir(D--A)*dir(-90));[/asy]$

$\text{(A)} \ 32 \qquad \text{(B)} \ 36 \qquad \text{(C)} \ 39 \qquad \text{(D)} \ 42 \qquad \text{(E)} \ 48$

Solution

Draw $\overline{AC}$ . Then $\triangle ABC$ is a $3-4-5$ right triangle and $\triangle ACD$ is a $5-12-13$ right triangle. Therefore, the area of $ABCD$ is $\frac{3\cdot4}{2}+\frac{5\cdot12}{2}=\boxed{(\textbf{B})\ 36}$ .

1980 AHSME (Problems • Answer Key • Resources)
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1980 AHSME Problems/Problem 7

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