2006 AIME I Problems/Problem 1: Difference between revisions

Revision as of 20:17, 4 July 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: $18+21+14+31=084$

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2006 AIME I Problems/Problem 1: Difference between revisions

Revision as of 20:17, 4 July 2006

Problem

Solution

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Revision as of 22:20, 30 June 2006 view source Xantos C. Guin (talk \| contribs) editor 580 edits →Solution ← Older edit	Revision as of 20:17, 4 July 2006 view source Joml88 (talk \| contribs) 1,619 edits m 2006 AIME I Problem 1 moved to 2006 AIME I Problems/Problem 1 Newer edit →
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