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

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== Solution ==
 
== Solution ==
<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>.
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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>.
  
 
''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 $AB$ and $CD$ of a given circle are perpendicular to each other and intersect at a right angle. Given that $BE = 16, DE = 4,$ and $AD = 5$, find $CE$.

Solution

$ADE$ is a right triangle with hypotenuse 5 and leg 4. Thus, by the Pythagorean Theorem, $AE = 3$ (or by just knowing your Pythagorean Triples). Applying the Power of a Point Theorem gives $AE\cdot BE = CE\cdot DE$, or $3\cdot 16 = x\cdot 4$. Solving gives $x = 12$.

Back to the Power of a Point Theorem.

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