2007 AMC 8 Problems/Problem 14

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 $\frac{1}{2}bh$ . 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, so by using the Pythagorean Theorem, $a^2+b^2=c^2$ , 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 $\boxed{\textbf{(C)}\ 13}$

Video Solution by WhyMath

https://youtu.be/9sVdsKcpJ9U

~savannahsolver

2007 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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All AJHSME/AMC 8 Problems and Solutions

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2007 AMC 8 Problems/Problem 14

Contents

Problem

Solution

Video Solution by WhyMath

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