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2006 AIME I Problems/Problem 1: Difference between revisions

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== See also ==
== See also ==
* [[2006 AIME I Problems/Problem 2 | Next problem]]
* [[2006 AIME I Problems]]
* [[2006 AIME I Problems]]
* [[Geometry]]
* [[Geometry]]


[[Category:Intermediate Geometry Problems]]
[[Category:Intermediate Geometry Problems]]

Revision as of 19:45, 11 March 2007

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:

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

The answer is 084.

See also

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