Find the volume of the frustum cone whose base and top radius is 20 ft and 10 ft respectively. The height of the cone is 300 ft. (Use $π$ = 3).

200,000 ft

^{3}

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220,000 ft

^{3}

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210,000 ft

^{3}

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230,000 ft

^{3}

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Solution

The correct option is B 210,000 ft $^{3}$
Volume of the frustum cone is V = $\frac{π h}{3} (R^{2} + R r + r^{2})$
$R = 20 f t, r = 10 f t, h = 300 f t$
$V = (3 \times 300) / (3) [20^{2} + 20 \times 10 + 10^{2}]$
$= 300 [400 + 200 + 100]$
$= 300 [700]$
$= 210, 000 f t^{3}$

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Find the volume of the frustum cone whose base and top radius is 20 ft and 10 ft respectively. The height of the cone is 300 ft. (Use π = 3).

The correct option is B 210,000 ft3Volume of the frustum cone is V = πh3(R2+Rr+r2)R=20ft,r=10ft,h=300ftV=(3×300)/(3)[202+20×10+102] =300[400+200+100] =300[700] =210,000ft3

Find the volume of the frustum cone whose base and top radius is 20 ft and 10 ft respectively. The height of the cone is 300 ft. (Use $π$ = 3).

The correct option is B 210,000 ft $^{3}$
Volume of the frustum cone is V = $\frac{π h}{3} (R^{2} + R r + r^{2})$
$R = 20 f t, r = 10 f t, h = 300 f t$
$V = (3 \times 300) / (3) [20^{2} + 20 \times 10 + 10^{2}]$
$= 300 [400 + 200 + 100]$
$= 300 [700]$
$= 210, 000 f t^{3}$