# Difference between revisions of "1999 AMC 8 Problems/Problem 14"

## Problem

In trapezoid $ABCD$, the sides $AB$ and $CD$ are equal. The perimeter of $ABCD$ is

$[asy] draw((0,0)--(4,3)--(12,3)--(16,0)--cycle); draw((4,3)--(4,0),dashed); draw((3.2,0)--(3.2,.8)--(4,.8)); label("A",(0,0),SW); label("B",(4,3),NW); label("C",(12,3),NE); label("D",(16,0),SE); label("8",(8,3),N); label("16",(8,0),S); label("3",(4,1.5),E); [/asy]$

$\text{(A)}\ 27 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 34 \qquad \text{(E)}\ 48$

## Solution

$[asy] draw((0,0)--(4,3)--(12,3)--(16,0)--cycle); draw((4,3)--(4,0),dashed); draw((12,3)--(12,0),dashed); draw((3.2,0)--(3.2,.8)--(4,.8)); label("A",(0,0),SW); label("B",(4,3),NW); label("C",(12,3),NE); label("D",(16,0),SE); label("8",(8,3),N); label("8",(8,0),S); label("3",(4,1.5),E); label("4",(2,0),S); label("4",(14,0),S); label("5",(0,0)--(4,3),NW); label("5",(12,3)--(16,0),NE); [/asy]$

There is a rectangle present, with both horizontal bases being $8$ units in length. The excess units on the bottom base must then be $16-8=8$. The fact that $AB$ and $CD$ are equal in length indicate, by the Pythagorean Theorem, that these excess lengths are equal. There are two with a total length of $8$ units, so each is $4$ units. The triangle has a hypotenuse of $5$, because the triangles are $3-4-5$ right triangles. So, the sides of the trapezoid are $8$, $5$, $16$, and $5$. Adding those up gives us the perimeter, $8 + 5 + 16 + 5 = 13 + 21 = \boxed{\text{(D)}\ 34}$ units.