Difference between revisions of "1973 AHSME Problems/Problem 25"

Latest revision as of 13:03, 20 February 2020

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
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

@@ Line 30: / Line 30: @@
 ==See Also==
-{{AHSME 35p box|year=1973|num-b=24|num-a=26}}
+{{AHSME 30p box|year=1973|num-b=24|num-a=26}}
 [[Category:Introductory Geometry Problems]]

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Difference between revisions of "1973 AHSME Problems/Problem 25"

Latest revision as of 13:03, 20 February 2020

Problem

Solution

See Also