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 1

We construct the following diagram: $[asy] pathpen = black; pair C=(0,0),D=(0,-14),A=(-sqrt(765),0),B=IP(circle(C,21),circle(A,18)); D(MP("A",A,W)--MP("B",B,N)--MP("C",C,E)--MP("D",D,E)--A--C); D(rightanglemark(A,C,D,40)); D(rightanglemark(A,B,C,40)); [/asy]$ Using the Pythagorean Theorem, we get the following two equations: $AD^2 = AC^2 + CD^2$ $AC^2 = AB^2 + BC^2$ Substituting $AB^2 + BC^2$ for $AC^2$ gives us: $AD^2 = AB^2 + BC^2 + CD^2$ Plugging in the given information: $AD^2 = 18^2 + 21^2 + 14^2 = 961 \implies AD = 31$ So the perimeter is $AB+BC+CD+AD = 18+21+14+31 = \boxed{084}$ .

2006 AIME I (Problems • Answer Key • Resources)
Preceded by First Question	Followed by Problem 2
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2006 AIME I Problems/Problem 1

Problem

Solution 1

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