2008 AMC 10A Problem 17

Below is the professionally curated solution for Problem 17 of the 2008 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 10A solutions, or check the answer key.

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Concepts:area decomposition sector equilateral triangle

Difficulty rating: 1680

17.

An equilateral triangle has side length $6.$ What is the area of the region containing all points that are outside the triangle and not more than $3$ units from a point of the triangle?

$36 + 24\sqrt{3}$

$54 + 9\pi$

$54 + 18\sqrt{3} + 6\pi$

$\left(2\sqrt{3} + 3\right)^2 \pi$

$9\left(\sqrt{3} + 1\right)^2 \pi$

Solution:

Along each of the three sides is a $6 \times 3$ rectangle, contributing $3 \cdot 6 \cdot 3 = 54.$

At each vertex is a $120^\circ$ sector of radius $3;$ the three together form a full circle of area $\pi \cdot 3^2 = 9\pi.$

The total area is $54 + 9\pi.$

Thus, the correct answer is B.

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