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.
