# 2008 Mock ARML 2 Problems/Problem 1

## Problem

$ABCD$ is a convex quadrilateral such that $|AB| = 5$, $|BC| = 17$, $|CD| = 7$, and $|DA| = 25$. Given that $m\angle{ABC} + m\angle{BCD} = 270^{\circ}$, find the area of $ABCD$.

## Solution

$[asy] pointpen = black; pathpen = black + linewidth(0.7); pair A=(0,0), D=(25,0), F=(16,12), B=(3*A + F)/4, C=(8*D+7*F)/15; D(MP("A",A,SW)--MP("B",B,NW)--MP("C",C,NE)--MP("D",D,SE)--cycle); D(B--MP("E",F,N)--C,linetype("4 4")); D(rightanglemark(A,F,D,30),linetype("4 4")); [/asy]$

Note that $m\angle{BAD} + m\angle{ADC} = 360 - 270 = 90^{\circ}$. Thus, if we let $E$ be the intersection of the extensions of $\overline{AB}$ and $\overline{CD}$, it follows that $\triangle EAD$ is a right triangle. Immediately we notice that $\triangle ECB$ is a $8-15-17,\,\triangle$ and that $\triangle EAD$ is a $15-20-25\, \triangle$; otherwise we can determine these lengths through the Pythagorean Theorem.

The answer is $[ABCD] = [EAD] - [ECB] = \frac{1}{2}(15)(20) - \frac{1}{2}(8)(15) = \boxed{90}$.