# Difference between revisions of "2017 AMC 8 Problems/Problem 18"

## Problem 18

In the non-convex quadrilateral $ABCD$ shown below, $\angle BCD$ is a right angle, $AB=12$, $BC=4$, $CD=3$, and $AD=13$. $[asy]draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); label("B", (0, 0), SW); label("A", (12, 0), ESE); label("C", (2.4, 3.6), SE); label("D", (0, 5), N);[/asy]$ What is the area of quadrilateral $ABCD$?

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

## Solution

We can see a Pythagorean triple's two longer lengths: 12, 13. So BD should be 5. This is certainly the case because $3^2 + 4^2 = 5^2$, which is $BD$. Thus the area of triangle ABD is $30$. So $30 - 6 = 24$, or $\text{B)}$ $24$.