2007 AMC 8 Problems/Problem 14: Difference between revisions

Revision as of 16:10, 26 March 2010

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, $c$ ).

$c = 13$

The answer is $\boxed{C}$

@@ Line 20: / Line 20: @@
 we can solve for one of the legs of the triangle (it will be the the hypotenuse, <math>c</math>).
-&12^2+5^2=c^2<math>
+<math>c = 13</math>
-</math>c = 13<math>
+The answer is <math>\boxed{C}</math>
-The answer is </math>\boxed{C}$