Difference between revisions of "2006 AIME I Problems/Problem 1"

Revision as of 15:56, 29 December 2012

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 $\boxed{84}$ .

@@ Line 27: / Line 27: @@
 <div style="text-align:center"><math> (AD)= 31 </math></div>
-So the perimeter is <math> 18+21+14+31=84 </math>, and the answer is <math>\boxed{084}</math>.
+So the perimeter is <math> 18+21+14+31=84 </math>, and the answer is <math>\boxed{84}</math>.
 == See also ==

2006 AIME I (Problems • Answer Key • Resources)
Preceded by First Question	Followed by Problem 2
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Difference between revisions of "2006 AIME I Problems/Problem 1"

Revision as of 15:56, 29 December 2012

Problem

Solution

See also