2007 AMC 8 Problems/Problem 14

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Problem

The base of isosceles $\triangle ABC$ is $24$ and its area is $60$. What is the length of one of the congruent sides?

$\mathrm{(A)}\ 5 \qquad \mathrm{(B)}\ 8 \qquad \mathrm{(C)}\ 13 \qquad \mathrm{(D)}\ 14 \qquad \mathrm{(E)}\ 18$

Solution

The area of a triangle is shown by 1 half bh We set the base equal to 24 and the area equal to 60 and we get the height, or altitude, of the triangle to be 5 In this isosceles triangle, the height bisects the base a^2+b^2=c^2 we can solve for one of the legs of the triangle (it will be the the hypotenuse, $c$). a = 12 b = 5 c = 13 The answer is C:13

Video Solution by WhyMath

https://youtu.be/9sVdsKcpJ9U

~savannahsolver

See Also

2007 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AJHSME/AMC 8 Problems and Solutions

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