# 2011 AMC 8 Problems/Problem 20

## Problem

Quadrilateral $ABCD$ is a trapezoid, $AD = 15$, $AB = 50$, $BC = 20$, and the altitude is $12$. What is the area of the trapezoid? $[asy] pair A,B,C,D; A=(3,20); B=(35,20); C=(47,0); D=(0,0); draw(A--B--C--D--cycle); dot((0,0)); dot((3,20)); dot((35,20)); dot((47,0)); label("A",A,N); label("B",B,N); label("C",C,S); label("D",D,S); draw((19,20)--(19,0)); dot((19,20)); dot((19,0)); draw((19,3)--(22,3)--(22,0)); label("12",(21,10),E); label("50",(19,22),N); label("15",(1,10),W); label("20",(41,12),E);[/asy]$ $\textbf{(A) }600\qquad\textbf{(B) }650\qquad\textbf{(C) }700\qquad\textbf{(D) }750\qquad\textbf{(E) }800$

## Solution $[asy] unitsize(1.5mm); defaultpen(linewidth(.9pt)+fontsize(10pt)); dotfactor=3; pair A,B,C,D,X,Y; A=(9,12); B=(59,12); C=(75,0); D=(0,0); X=(9,0); Y=(59,0); draw(A--B--C--D--cycle); draw(A--X); draw(B--Y); pair[] ps={A,B,C,D,X,Y}; dot(ps); label("A",A,NW); label("B",B,NE); label("C",C,SE); label("D",D,SW); label("X",X,SE); label("Y",Y,S); label("a",D--X,S); label("b",Y--C,S); label("15",D--A,NW); label("50",B--A,N); label("20",B--C,NE); label("12",X--A,E); label("12",Y--B,W); [/asy]$

If you draw altitudes from $A$ and $B$ to $CD,$ the trapezoid will be divided into two right triangles and a rectangle. You can find the values of $a$ and $b$ with the Pythagorean theorem. $$a=\sqrt{15^2-12^2}=\sqrt{81}=9$$ $$b=\sqrt{20^2-12^2}=\sqrt{256}=16$$ $ABYX$ is a rectangle so $XY=AB=50.$ $$CD=a+XY+b=9+50+16=75$$

The area of the trapezoid is $$12\cdot \frac{(50+75)}{2} = 6(125) = \boxed{\textbf{(D)}\ 750}$$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 