2007 AMC 8 Problems/Problem 14: Difference between revisions

Revision as of 12:20, 9 December 2012

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

2007 AMC 8 (Problems • Answer Key • Resources)
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

@@ Line 24: / Line 24: @@
 <math>c = 13</math>
-The answer is <math>\boxed{C}</math>
+The answer is <math>\boxed{\textbf{(C)}\ 13}</math>
 ==See Also==
 {{AMC8 box|year=2007|num-b=13|num-a=15}}