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

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<math> 18+21+14+31=084 </math>
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<math> 18+21+14+31=84 </math>
  
 
== See also ==
 
== See also ==

Revision as of 22:33, 25 October 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

From the problem statement, we construct the following diagram:

Aime06i.1.PNG

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: $18+21+14+31=84$

See also

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