The base of isosceles $\triangle ABC$ is $24$ and its area is $60$. What is the length of one
of the congruent sides?
$\mathrm{(A)}\ 5 \qquad \mathrm{(B)}\ 8 \qquad \mathrm{(C)}\ 13 \qquad \mathrm{(D)}\ 14 \qquad \mathrm{(E)}\ 18$
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 $\boxed{\textbf{(C)}\ 13}$