Difference between revisions of "1980 AHSME Problems/Problem 7"
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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>\ | + | Sides <math>AB</math>, <math>BC</math>, <math>CD</math> and <math>DA</math> of convex polygon <math>ABCD</math> have lengths <math>3</math>, <math>4</math>, <math>12</math>, and <math>13</math>, respectively, and <math>\angle CBA</math> is a right angle. The area of the quadrilateral is |
<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 21: | Line 22: | ||
<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> | ||
| + | |||
| + | == Solution == | ||
| + | |||
| + | 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 == | ||
| + | {{AHSME box|year=1980|num-b=6|num-a=8}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 16:27, 25 July 2025
Problem
Sides 
, 
, 
and 
of convex polygon 
have lengths 
, 
, 
, and 
, respectively, and 
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]](http://latex.artofproblemsolving.com/8/5/9/85952cf09f6d010354fdab35f8a1e246200a1e51.png)
Solution
Draw 
. By the Pythagorean Theorem, 
. Notice that 
, so by the converse of the Pythagorean Theorem, 
is right. Thus the area of is 
.
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. 
