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
,
~megaboy6679
Video Solution by WhyMath
~savannahsolver
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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