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

(Solution)
(Solution)
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So the perimeter is:  
 
So the perimeter is:  
 
<math> 18+21+14+31=084 </math>
 
<math> 18+21+14+31=084 </math>
 
--[[User:Xantos C. Guin|Xantos C. Guin]] 15:10, 30 June 2006 (EDT)
 
  
 
== See also ==
 
== See also ==
 
* [[2006 AIME I Problems]]
 
* [[2006 AIME I Problems]]

Revision as of 23:20, 30 June 2006

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

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: $18+21+14+31=084$

See also

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