# 2017 AMC 8 Problems/Problem 18

## Problem

In the non-convex quadrilateral $ABCD$ shown below, $\angle BCD$ is a right angle, $AB=12$, $BC=4$, $CD=3$, and $AD=13$. What is the area of quadrilateral $ABCD$?

$[asy]draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); label("B", (0, 0), SW); label("A", (12, 0), ESE); label("C", (2.4, 3.6), SE); label("D", (0, 5), N);[/asy]$

$\textbf{(A) }12 \qquad \textbf{(B) }24 \qquad \textbf{(C) }26 \qquad \textbf{(D) }30 \qquad \textbf{(E) }36$

## Solution 1

We first connect point $B$ with point $D$.

$[asy]draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); draw((0,0)--(0,5)); label("B", (0, 0), SW); label("A", (12, 0), ESE); label("C", (2.4, 3.6), SE); label("D", (0, 5), N);[/asy]$

We can see that $\triangle BCD$ is a 3-4-5 right triangle. We can also see that $\triangle BDA$ is a right triangle, by the 5-12-13 Pythagorean triple. With these lengths, we can solve the problem. The area of $\triangle BDA$ is $\frac{5\cdot 12}{2}$, and the area of the smaller 3-4-5 triangle is $\frac{3\cdot 4}{2}$. Thus, the area of quadrialteral $ABCD$ is $30-6 = \boxed{\textbf{(B)}\ 24}.$

## Solution 2

$\triangle BCD$ is a 3-4-5 right triangle. Then we can use Heron's formula to resolve the problem. The area of $\triangle ABD$ whose sides have lengths 5-12-13 is \sqrt {s(s-5)(s-12)(s-13)}. S is the semi-perimeter of the triangle, $s=(5+12+13)/2= 15$. Then the area is \sqrt {15(15-5)(15-12)(15-13)}= 30$. So$ABCD$is$30-6 = \boxed{\textbf{(B)}\ 24}.\$