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

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(Solution)
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== Solution ==
 
== Solution ==
 +
Using the Pythagorean Theorem:
  
 +
<math> (AD)^2 = (AC)^2 + (CD)^2 </math>
 +
 +
<math> (AC)^2 = (AB)^2 + (BC)^2 </math>
 +
 +
Substituting <math>(AB)^2 + (BC)^2 </math> for <math> (AC)^2 </math>:
 +
 +
<math> (AD)^2 = (AB)^2 + (BC)^2 + (CD)^2 </math>
 +
 +
Plugging in the given information:
 +
 +
<math> (AD)^2 = (18)^2 + (21)^2 + (14)^2 </math>
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<math> (AD)^2 = 961 </math>
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<math> (AD)= 31 </math>
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So the perimeter is:
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<math> 18+21+14+31=084 </math>
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--[[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 15:10, 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: $18+21+14+31=084$

--Xantos C. Guin 15:10, 30 June 2006 (EDT)

See also

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