2009 Grade 8 CEMC Gauss Problems/Problem 11

Problem

The perimeter of $\Delta ABC$ is $32$ . If $\angle ABC = \angle ACB$ and $BC = 12$ , the length of $AB$ is

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$\text{ (A) }\ 11 \qquad\text{ (B) }\ 10 \qquad\text{ (C) }\ 9 \qquad\text{ (D) }\ 12 \qquad\text{ (E) }\ 8$

Since $\angle ABC = \angle ACB$ , the triangle is an isosceles triangle, where $AB = AC$ . That means that we can let $x$ represent $AB$ and $BC$ .

The perimeter is the sum of the side lengths of a polygon, meaning we can set up an equation:

$x + x + 12 = 32$

$2x + 12 = 32$

$2x = 20$

$x = \boxed {\textbf {(B) } 10}$

~anabel.disher

We can test answer choices, and see whether or not the side length results in the perimeter being too high or too low.

We can first use $10$ , and we get:

$10 + 10 + 12 = 20 + 12 = 32$

This happens to be exactly the perimeter of the triangle, so the answer is $\boxed {\textbf {(B) } 10}$

~anabel.disher