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 1

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 triangle's height, or altitude, 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 hypotenuse, $c$ ). $a = 12$ , $b = 5$ , $c = 13$ . The answer is $\boxed{\textbf{(C)}\ 13}$ .

Solution 2 (Heron's Formula)

According to Heron's Formula, setting side $a$ as $24$ , we have $\sqrt{s(s-24)(s-b)(s-c)}=60$ where $s$ is the triangle's semiperimeter (i.e. $\frac{a+b+c}{2}$ ). Since the triangle is isosceles, $b=c$ , so we can rewrite $s$ as $\frac{24+2b}{2}=12+b$ . Substituting and solving the equation and taking the positive solution for $b$ , $\sqrt{(12+b)(-12+b)(12)(12)}=60$ $\sqrt{144(144-b^2)}=60$ $144(144-b^2)=3600$ $-b^2=-169$ $b=\boxed{\textbf{(C)}\ 13}$

~megaboy6679

Video Solution by WhyMath

https://youtu.be/9sVdsKcpJ9U

~savannahsolver

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=omFpSGMWhFc

Video Solution by AliceWang

https://youtu.be/U8v4XVPXr18

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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