2005 AMC 10A Problems/Problem 12

Problem

The figure shown is called a trefoil and is constructed by drawing circular sectors about the sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length $2$ ?

$\mathrm{(A) \ } \frac{1}{3}\pi+\frac{\sqrt{3}}{2}\qquad \mathrm{(B) \ } \frac{2}{3}\pi\qquad \mathrm{(C) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{4}\qquad \mathrm{(D) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{3}\qquad \mathrm{(E) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{2}$

Solution

The area of the trefoil is equal to the area of a small equilateral triangle plus the area of four $60^\circ$ sectors with a radius of $\frac{2}{2}=1$ minus the area of a small equilateral triangle.

This is equivalent to the area of four $60^\circ$ sectors with a radius of $1$ .

So the answer is:

$4\cdot\frac{60}{360}\cdot\pi\cdot1^2 = \frac{4}{6}\cdot\pi = \frac{2}{3}\pi \Rightarrow B$

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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2005 AMC 10A Problems/Problem 12

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Solution

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