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

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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