2008 Mock ARML 2 Problems/Problem 1
Problem
is a convex quadrilateral such that
,
,
, and
. Given that
, find the area of
.
Solution
![[asy] pointpen = black; pathpen = black + linewidth(0.7); pair A=(0,0), D=(25,0), F=(16,12), B=(3*A + F)/4, C=(8*D+7*F)/15; D(MP("A",A,SW)--MP("B",B,NW)--MP("C",C,NE)--MP("D",D,SE)--cycle); D(B--MP("E",F,N)--C,linetype("4 4")); D(rightanglemark(A,F,D,30),linetype("4 4")); [/asy]](http://latex.artofproblemsolving.com/b/8/d/b8d0874fa95cec121998ed22159efe50097b8b81.png)
Note that . Thus, if we let
be the intersection of the extensions of
and
, it follows that
is a right triangle. Immediately we notice that
is a
and that
is a
; otherwise we can determine these lengths through the Pythagorean Theorem.
The answer is .
See also
2008 Mock ARML 2 (Problems, Source) | ||
Preceded by First question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 |