# Difference between revisions of "2021 JMPSC Sprint Problems/Problem 16"

## Problem

$ABCD$ is a concave quadrilateral with $AB = 12$, $BC = 16$, $AD = CD = 26$, and $\angle ABC=90^\circ$. Find the area of $ABCD$.

## Solution

Notice that $[ABCD] = [ADC] - [ABC]$ and $AC = \sqrt{12^2 + 16^2} = 20$ by the Pythagorean Thereom. We then have that the area of triangle of $ADC$ is $\frac{20 \cdot \sqrt{26^2 - 10^2}}{2} = 240$, and the area of triangle $ABC$ is $\frac{12 \cdot 16}{2} = 96$, so the area of quadrilateral $ABCD$ is $240 - 96 = 144$.

~Mathdreams

## Solution 2

$$[ACD] = \frac{24 \cdot 20}{2}=240$$ $$[ABC] = \frac{12 \cdot 16}{2}=96$$ Therefore, $[ABCD] = 240-96=144$

- kante314 -