2007 AMC 8 Problems/Problem 14
Contents
[hide]Problem
The base of isosceles is and its area is . What is the length of one of the congruent sides?
Solution 1
The area of a triangle is shown by . We set the base equal to , and the area equal to , and we get the triangle's height, or altitude, to be . In this isosceles triangle, the height bisects the base, so by using the Pythagorean Theorem, , we can solve for one of the legs of the triangle (it will be the hypotenuse, ). , , . The answer is
Solution 2 (Heron's Formula)
According to Heron's Formula, setting side as , we have where is the triangle's semiperimeter (i.e. ). Since the triangle is isosceles, , so we can rewrite as . Substituting and solving the equation and taking the positive solution for ,
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Video Solution by WhyMath
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Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=omFpSGMWhFc
Video Solution by AliceWang
See Also
2007 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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