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

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

Revision as of 04:15, 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,

$b = 8$