# Difference between revisions of "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 {A={\sqrt {s(s-a)(s-b)(s-c)}},} A = \sqrt{s(s-a)(s-b)(s-c)},. S is the semi-perimeter of the triangle, {s={\frac {a+b+c}{2}}.} s=\frac{a+b+c}{2}=30$. So$ABCD$is$30-6 = \boxed{\textbf{(B)}\ 24}.\$

Heron's formula states that the area of a triangle whose sides have lengths a, b, and c is

{\displaystyle A={\sqrt {s(s-a)(s-b)(s-c)}},} A = \sqrt{s(s-a)(s-b)(s-c)}, where s is the semi-perimeter of the triangle; that is,

{\displaystyle s={\frac {a+b+c}{2}}.} s=\frac{a+b+c}{2}.