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13 Jan 2021, 09:15
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D
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Difficulty:
65%
(hard)
Question Stats:
43% (01:20) correct
58%
(01:31)
wrong
based on 40
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A sector of a circle of radius 5 cm is recast into a right circular cone of height 4 cm. What is the volume of the resulting cone?
A. \(100 π\) cm^3
B. \(33 π\) cm^3
C. \(32 π\) cm^3
D. \(12 π\) cm^3
E. \(4 π\) cm^3
A. \(100 π\) cm^3
B. \(33 π\) cm^3
C. \(32 π\) cm^3
D. \(12 π\) cm^3
E. \(4 π\) cm^3
13 Jan 2021, 09:31
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Given, a section of circle is made to right circular cone.
So, r = radius of cone = sqrt (5^2-4^2) = sqrt(9) = 3
Volume of circular cone = 1/3 * Pie *r^2 *h = 1/3 * pie *3^2 *4 = 12*pie cm^3
I think D.
So, r = radius of cone = sqrt (5^2-4^2) = sqrt(9) = 3
Volume of circular cone = 1/3 * Pie *r^2 *h = 1/3 * pie *3^2 *4 = 12*pie cm^3
I think D.
13 Jan 2021, 22:30
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The radius\((r_{ci})\) of the circle will be the slant height\((l)\) of the cone when recasted, so the slant height \((l)\) of the cone \(=5 cm\)
Therefore, using Pythagoras theorem, the radius of the cone,\( r_{co} =\sqrt{l^2-h^2}=\sqrt{5^2-4^2}=\sqrt{9}=3\)
\(V_{cone}=\frac{1}{3} π r^2 h=\frac{1}{3} π 3^2 *4=12π\)
Ans D
Therefore, using Pythagoras theorem, the radius of the cone,\( r_{co} =\sqrt{l^2-h^2}=\sqrt{5^2-4^2}=\sqrt{9}=3\)
\(V_{cone}=\frac{1}{3} π r^2 h=\frac{1}{3} π 3^2 *4=12π\)
Ans D
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