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2006 AIME I Problems/Problem 1

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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:

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...

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 $\boxed{084}$ .

See also

2006 AIME I (Problems • Resources)
Preceded by First Question	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2006_AIME_I_Problems/Problem_1"

Category: Intermediate Geometry Problems

Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.

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