Difference between revisions of "2013 AMC 12A Problems/Problem 1"

Revision as of 03:15, 7 February 2013

We are given that the area of $\triangle ABE$ is $40$ , and that $AB = 10$ .

The area of a triangle:

$A = \frac{bh}{2}$

Using $AB$ as the height of $\triangle ABE$ ,

$40 = \frac{10b}{2}$

and solving for b,

$b = 8$

@@ Line 1: / Line 1: @@
-We are given that the area of ∆ABE is 40, and that side AB is of length 10.
+We are given that the area of <math>\triangle ABE</math> is <math>40</math>, and that <math>AB = 10</math>.
-Side AB can be used as the height of right ∆ABE.
+The area of a triangle:
-Since A= (bh)/2,
+<math>A = \frac{bh}{2}</math>
-=(10h)/2
+Using <math>AB</math> as the height of <math>\triangle ABE</math>,
-and solving for h will give us 8, or E.
+<math>40 = \frac{10b}{2}</math>
+and solving for b,
+<math>b = 8</math>