2001 CEMC Pascal Problems/Problem 9

Problem

The perimeter of $\Delta ABC$ is

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$\text{ (A) }\ 23 \qquad\text{ (B) }\ 40 \qquad\text{ (C) }\ 42 \qquad\text{ (D) }\ 46 \qquad\text{ (E) }\ 60$

The perimeter of a shape is just the sum of its sides, meaning that we have to find the length of $BC$ .

Since the triangle is a right triangle, we can use the pythagorean theorem to find $BC$ :

$BC^2 = 8^2 + 15^2 = 64 + 225 = 289$

$BC = \sqrt{289} = 17$

The perimeter is then $AB + AC + BC = 8 + 15 + 17 = 23 + 17 = \boxed {\textbf {(B) } 40}$

~anabel.disher

Like in solution 1, we can get the length of $BC$ . However, to do this, we can remember that $8$ , $15$ , and $17$ form a pythagorean triple.

This means $BC = 17$ , which leads to the same answer of $\boxed {\textbf {(B) } 40}$ .

~anabel.disher