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

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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> | ||

+ | |||

+ | <math> (AD)^2 = 961 </math> | ||

+ | |||

+ | <math> (AD)= 31 </math> | ||

+ | |||

+ | So the perimeter is: | ||

+ | <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 15:10, 30 June 2006

## Problem

In quadrilateral is a right angle, diagonal is perpendicular to and Find the perimeter of

## Solution

Using the Pythagorean Theorem:

Substituting for :

Plugging in the given information:

So the perimeter is:

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