2005 AMC 10A Problem 12

Question

The figure shown is called a trefoil and is constructed by drawing circular sectors about sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length

2?

\dfrac{1}{3}\pi + \dfrac{\sqrt{3}}{2}

\dfrac{2}{3}\pi

\dfrac{2}{3}\pi + \dfrac{\sqrt{3}}{4}

\dfrac{2}{3}\pi + \dfrac{\sqrt{3}}{3}

\dfrac{2}{3}\pi + \dfrac{\sqrt{3}}{2}

Answer: B

Concepts:sector area decomposition equilateral triangle

Difficulty rating: 1460

Small hint:

The base of length

2

spans two radii, so each circular sector has radius

1

Big hint:

Rearrange the four equilateral triangles and four circular segments into four

60^\circ

sectors of radius

1

Solution:

Since the base

2

equals two radii, the radius is

1.

The trefoil is made of four equilateral triangles and four circular segments, which reassemble into four

60^\circ

sectors of a circle of radius

1.

Their total area is

4 \cdot \dfrac{60}{360}\pi (1)^2 = \dfrac{2}{3}\pi.

Thus, the correct answer is B.