Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

Difference between revisions of "1980 AHSME Problems/Problem 7"
(Shrunk diagram because it was huge.)
 
(One intermediate revision by the same user not shown)
Line 4: Line 4:
  
 
<asy>
 
<asy>
 +
size(150);
 
defaultpen(linewidth(0.7)+fontsize(10));
 
defaultpen(linewidth(0.7)+fontsize(10));
 
real r=degrees((12,5)), s=degrees((3,4));
 
real r=degrees((12,5)), s=degrees((3,4));
Line 24: Line 25:
 
== Solution ==
 
== Solution ==
  
−
Draw <math>\overline{AC}</math>. Then <math>\triangle ABC</math> is a <math>3-4-5</math> right triangle and <math>\triangle ACD</math> is a <math>5-12-13</math> right triangle. Therefore, the area of <math>ABCD</math> is <math> \frac{3\cdot4}{2}+\frac{5\cdot12}{2}=\boxed{(\textbf{B})\ 36} </math>.
+
Draw <math>\overline{AC}</math>. By the Pythagorean Theorem, <math>AC = \sqrt{AB^2 + BC^2} = \sqrt{3^2+4^2} = 5</math>.  Notice that <math>AD^2 = AC^2 + CD^2 = 169</math>, so by the converse of the Pythagorean Theorem, <math>\triangle ACD</math> is right. Thus the area of <math>ABCD</math> is <math>\frac{3\cdot4}{2}+\frac{5\cdot12}{2}=\boxed{(\textbf{B})\ 36} </math>.
  
 
== See also ==
 
== See also ==
 
{{AHSME box|year=1980|num-b=6|num-a=8}}
 
{{AHSME box|year=1980|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:27, 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 $\frac{3\cdot4}{2}+\frac{5\cdot12}{2}=\boxed{(\textbf{B})\ 36}$.

See also

1980 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1980_AHSME_Problems/Problem_7&oldid=255559"