A cylindrical quarter has a \(\frac{15}{16}\) inch diameter and a \(\frac{1}{16}\) inch height. What would be the number of inches in the height of a coin whose volume is exactly four times that of the given quarter and whose diameter equals \(1 \frac{1}{8}\) inches? Express your answer as a common fraction.
+128470 The old radius is ..... (1/2)(15/16) = 15/32 inches
The old volume is ..... pi * ( 15/32)^2 * (1/16) in ^3
And 4 times this volume is pi * (15/32)^2 * (1/16) * 4 in^3 = pi * (15/32)^2 *(1/4) in^3
The new diameter is 1 + 1/8 in = 9/8 inches.....so....the new radius is 1/2 of this = (9/8) (1/2) = 9/16 inches
So we have that
pi * (15/32)^2 * (1/4) = pi * (9/16)^2 * height [ divide out pi ]
(15/32)^2 / 4 = (9/16)^2 * height [ divide both sides by (9/16)^2 ]
(15/32)^2 / [ 4 * (9/16)^2 ] = height = 25 /144 in
