# 2006 AIME I Problems/Problem 1

## Problem

In quadrilateral $ABCD$, $\angle B$ is a right angle, diagonal $\overline{AC}$ is perpendicular to $\overline{CD}$, $AB=18$, $BC=21$, and $CD=14$. Find the perimeter of $ABCD$.

## Solution

From the problem statement, we construct the following diagram: $[asy] pointpen = black; pathpen = black + linewidth(0.65); pair C=(0,0), D=(0,-14),A=(-(961-196)^.5,0),B=IP(circle(C,21),circle(A,18)); D(MP("A",A,W)--MP("B",B,N)--MP("C",C,E)--MP("D",D,E)--A--C); D(rightanglemark(A,C,D,40)); D(rightanglemark(A,B,C,40)); [/asy]$

Using the Pythagorean Theorem: $(AD)^2 = (AC)^2 + (CD)^2$ $(AC)^2 = (AB)^2 + (BC)^2$

Substituting $(AB)^2 + (BC)^2$ for $(AC)^2$: $(AD)^2 = (AB)^2 + (BC)^2 + (CD)^2$

Plugging in the given information: $(AD)^2 = (18)^2 + (21)^2 + (14)^2$ $(AD)^2 = 961$ $(AD)= 31$

So the perimeter is $18+21+14+31=84$, and the answer is $\boxed{084}$.

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