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

(Created page with "==Problem== Sides <math>AB,BC,CD</math> and <math>DA</math> of convex polygon <math>ABCD</math> have lengths 3,4,12, and 13, respectively, and <math>\measuredangle CBA</math> is...")
 
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==Problem==
 
==Problem==
  
Sides <math>AB,BC,CD</math> and <math>DA</math> of convex polygon <math>ABCD</math> have lengths 3,4,12, and 13, respectively, and <math>\measuredangle CBA</math> is a right angle. The area of the quadrilateral is
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Sides <math>AB,BC,CD</math> and <math>DA</math> of convex polygon <math>ABCD</math> have lengths 3, 4, 12, and 13, respectively, and <math>\angle CBA</math> is a right angle. The area of the quadrilateral is
  
 
<asy>
 
<asy>
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<math>\text{(A)} \ 32 \qquad \text{(B)} \ 36 \qquad \text{(C)} \ 39 \qquad \text{(D)} \ 42 \qquad \text{(E)} \ 48</math>
 
<math>\text{(A)} \ 32 \qquad \text{(B)} \ 36 \qquad \text{(C)} \ 39 \qquad \text{(D)} \ 42 \qquad \text{(E)} \ 48</math>
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== Solution ==
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Connect C and A, and we have a 3-4-5 right triangle and 5-12-13 right triangle. The area of both is <math> \frac{3\cdot4}{2}+\frac{5\cdot12}{2}=36\Rightarrow\boxed{(B)} </math>.
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== See also ==
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{{AHSME box|year=1980|num-b=6|num-a=8}}

Revision as of 20:08, 31 March 2013

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

Connect C and A, and we have a 3-4-5 right triangle and 5-12-13 right triangle. The area of both is $\frac{3\cdot4}{2}+\frac{5\cdot12}{2}=36\Rightarrow\boxed{(B)}$.

See also

1980 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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All AHSME Problems and Solutions