Find the volume of the frustum cone with $r = 2$ m, $R = 4$ m and $h = 10$ m. (Use $π = \frac{22}{7}$ ).

2640 m

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2640 m^{2}

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293 m^{3}

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2640 c m^{3}

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Solution

The correct option is C $293 m^{3}$
Volume of frustum of cone $=$ $\frac{1}{3} π (R^{2} + r^{2} + R r) \times h$

$=$ $\frac{1}{3} \times \frac{22}{7} \times 10 (4^{2} + 2^{2} + 4 \times 2)$

$=$ $\frac{220}{21} (16 + 4 + 8)$

$=$ $\frac{220}{21} \times 28 = \frac{880}{3} = 293 m^{3}$

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Find the volume of the frustum cone with r=2m, R=4m and h=10m. (Use π=227).

The correct option is C 293m3Volume of frustum of cone = 13π(R2+r2+Rr)×h = 13×227×10(42+22+4×2) = 22021(16+4+8) = 22021×28=8803=293m3

Find the volume of the frustum cone with $r = 2$ m, $R = 4$ m and $h = 10$ m. (Use $π = \frac{22}{7}$ ).

The correct option is C $293 m^{3}$
Volume of frustum of cone $=$ $\frac{1}{3} π (R^{2} + r^{2} + R r) \times h$

$=$ $\frac{1}{3} \times \frac{22}{7} \times 10 (4^{2} + 2^{2} + 4 \times 2)$

$=$ $\frac{220}{21} (16 + 4 + 8)$

$=$ $\frac{220}{21} \times 28 = \frac{880}{3} = 293 m^{3}$