# Difference between revisions of "Power of a Point Theorem/Introductory Problem 4"

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− | <math> ADE </math> is a right triangle with hypotenuse 5 and leg 4. Thus, by the [[Pythagorean Theorem]], <math> AE = 3 </math> (or by just knowing your [[Pythagorean Triple]]s). Applying the Power of a Point Theorem gives <math> AE\cdot BE = CE\cdot DE </math> or <math> 3\cdot 16 = x\cdot 4 </math>. Solving gives <math> x = 12 </math>. | + | <math> ADE </math> is a right triangle with hypotenuse 5 and leg 4. Thus, by the [[Pythagorean Theorem]], <math> AE = 3 </math> (or by just knowing your [[Pythagorean Triple]]s). Applying the Power of a Point Theorem gives <math> AE\cdot BE = CE\cdot DE </math>, or <math> 3\cdot 16 = x\cdot 4 </math>. Solving gives <math> x = 12 </math>. |

''Back to the [[Power of a Point Theorem]].'' | ''Back to the [[Power of a Point Theorem]].'' |

## Revision as of 13:41, 3 July 2006

## Problem

(ARML) Chords and of a given circle are perpendicular to each other and intersect at a right angle. Given that and , find .

## Solution

is a right triangle with hypotenuse 5 and leg 4. Thus, by the Pythagorean Theorem, (or by just knowing your Pythagorean Triples). Applying the Power of a Point Theorem gives , or . Solving gives .

*Back to the Power of a Point Theorem.*