Difference between revisions of "1980 AHSME Problems/Problem 7"
Claudiafeng (talk | contribs) |
|||
| Line 28: | Line 28: | ||
== 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}} | ||
Revision as of 11:47, 5 July 2013
Problem
Sides 
and 
of convex polygon 
have lengths 3, 4, 12, and 13, respectively, and 
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]](http://latex.artofproblemsolving.com/0/3/4/03406705488fb5be15f3ce0fda54f0e61223736d.png)
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 
.
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. 
