# Difference between revisions of "2014 AMC 12B Problems/Problem 9"

## Problem

Convex quadrilateral $ABCD$ has $AB=3$, $BC=4$, $CD=13$, $AD=12$, and $\angle ABC=90^{\circ}$, as shown. What is the area of the quadrilateral?

$[asy] pair A=(0,0), B=(-3,0), C=(-3,-4), D=(48/5,-36/5); draw(A--B--C--D--A); label("A",A,N); label("B",B,NW); label("C",C,SW); label("D",D,E); draw(rightanglemark(A,B,C,25)); [/asy]$

$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 58.5$

## Solution

Note that by the pythagorean theorem, $AC=5$. Also note that $\angle CAD$ is a right angle because $\triangle CAD$ is a right triangle. The area of the quadrilateral is the sum of the areas of $\triangle ABC$ and $\triangle CAD$ which is equal to $$\frac{3\times4}{2} + \frac{5\times12}{2} = 6 + 30 = \boxed{\textbf{(B)}\ 36}$$

~ pi_is_3.14