Problem:
The figure shown is the union of a circle and two semicircles of diameters a aa and b bb , all of whose centers are collinear. The ratio of the area of the shaded region to that of the unshaded region is
Answer Choices:
A. a b \sqrt{\dfrac{a}{b}}b a β β
B. a b \dfrac{a}{b}b a β
C. a 2 b 2 \dfrac{a^{2}}{b^{2}}b 2 a 2 β
D. a + b 2 b \dfrac{a+b}{2 b}2 b a + b β
E. a 2 + 2 a b b 2 + 2 a b \dfrac{a^{2}+2 a b}{b^{2}+2 a b}b 2 + 2 a b a 2 + 2 a b β
Solution:
The area of the shaded region is
Ο 2 ( ( a + b 2 ) 2 + ( a 2 ) 2 β ( b 2 ) 2 ) = Ο 2 a + b 2 ( a + b 2 + a β b 2 ) = Ο ( a + b ) a 4 \dfrac{\pi}{2}\left(\left(\dfrac{a+b}{2}\right)^{2}+\left(\dfrac{a}{2}\right)^{2}-\left(\dfrac{b}{2}\right)^{2}\right)=\dfrac{\pi}{2} \dfrac{a+b}{2}\left(\dfrac{a+b}{2}+\dfrac{a-b}{2}\right)=\dfrac{\pi(a+b) a}{4}
2 Ο β ( ( 2 a + b β ) 2 + ( 2 a β ) 2 β ( 2 b β ) 2 ) = 2 Ο β 2 a + b β ( 2 a + b β + 2 a β b β ) = 4 Ο ( a + b ) a β
and the area of the unshaded region is
Ο 2 ( ( a + b 2 ) 2 β ( a 2 ) 2 + ( b 2 ) 2 ) = Ο 2 a + b 2 ( a + b 2 + b β a 2 ) = Ο ( a + b ) b 4 \dfrac{\pi}{2}\left(\left(\dfrac{a+b}{2}\right)^{2}-\left(\dfrac{a}{2}\right)^{2}+\left(\dfrac{b}{2}\right)^{2}\right)=\dfrac{\pi}{2} \dfrac{a+b}{2}\left(\dfrac{a+b}{2}+\dfrac{b-a}{2}\right)=\dfrac{\pi(a+b) b}{4}
2 Ο β ( ( 2 a + b β ) 2 β ( 2 a β ) 2 + ( 2 b β ) 2 ) = 2 Ο β 2 a + b β ( 2 a + b β + 2 b β a β ) = 4 Ο ( a + b ) b β
Their ratio is a / b a / ba / b .