1973 AHSME Problems/Problem 25

Problem

A circular grass plot 12 feet in diameter is cut by a straight gravel path 3 feet wide, one edge of which passes through the center of the plot. The number of square feet in the remaining grass area is

$\textbf{(A)}\ 36\pi-34 \qquad \textbf{(B)}\ 30\pi - 15 \qquad \textbf{(C)}\ 36\pi - 33 \qquad$

$\textbf{(D)}\ 35\pi - 9\sqrt3 \qquad \textbf{(E)}\ 30\pi - 9\sqrt3$

Solution

$[asy] draw(circle((0,0),6)); draw((-6,0)--(6,0)); pair A=(-5.196,3),B=(5.196,3); draw(A--B--(0,0)--A); draw((0,0)--(0,3)); label("6",(-2.598,1.5),SW); label("3",(0,1.5),E); draw((-0.5,3)--(-0.5,2.5)--(0,2.5)); [/asy]$

The wanted area can be divided into a semicircle and a circular segment. The area of the semicircle is $\tfrac12 \cdot \pi \cdot 6^2 = 18\pi$ square feet. To find the area of the circular segment, note that by using 30-60-90 triangles (or Law of Cosines), the chord's length is $6\sqrt{3}$ ft, and the angle of the sector is $120^\circ$ . That means the area of the circular segment is $(\tfrac13 \cdot \pi \cdot 6^2) - (\tfrac12 \cdot 6\sqrt{3} \cdot 3) = 12\pi - 9\sqrt{3}$ square feet. Thus, the area of the remaining grassy area is $\boxed{\textbf{(E)}\ 30\pi - 9\sqrt3}$ square feet.

1973 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Problem 26
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1973 AHSME Problems/Problem 25

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