Difference between revisions of "1980 AHSME Problems/Problem 7"

Latest revision as of 17:27, 25 July 2025

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

$[asy] size(150); defaultpen(linewidth(0.7)+fontsize(10)); real r=degrees((12,5)), s=degrees((3,4)); pair D=origin, A=(13,0), C=D+12*dir(r), B=A+3*dir(180-(90-r+s)); draw(A--B--C--D--cycle); markscalefactor=0.05; draw(rightanglemark(A,B,C)); pair point=incenter(A,C,D); label("$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 $\overline{AC}$ . By the Pythagorean Theorem, $AC = \sqrt{AB^2 + BC^2} = \sqrt{3^2+4^2} = 5$ . Notice that $AD^2 = AC^2 + CD^2 = 169$ , so by the converse of the Pythagorean Theorem, $\triangle ACD$ is right. Thus the area of $ABCD$ is $\frac{3\cdot4}{2}+\frac{5\cdot12}{2}=\boxed{(\textbf{B})\ 36}$ .

1980 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AHSME Problems and Solutions

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Difference between revisions of "1980 AHSME Problems/Problem 7"

Latest revision as of 17:27, 25 July 2025

Problem

Solution

See also

@@ Line 4: / Line 4: @@
 <asy>
+size(150);
 defaultpen(linewidth(0.7)+fontsize(10));
 real r=degrees((12,5)), s=degrees((3,4));