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 $\boxed{\textbf{(C)}\ 13}$

See Also

2007 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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