Find the volume of the frustum cone whose base and top radius is 12.4 ft and 4.5 ft respectively. The height of the cone is 1,200 ft. (Use $π$ = 3).

265,772 ft

^{3}

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255,772 ft

^{3}

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245,772 ft

^{3}

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275,772 ft

^{3}

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Solution

The correct option is D 275,772 ft $^{3}$
Volume of the frustum cone is $V = \frac{π h}{3} (R^{2} + R r + r^{2})$
$R = 12.4 f t, r = 4.5 f t, h = 1, 200 f t$
$V = (3 \times 1, 200) / (3) [{12.4}^{2} + 12.4 \times 4.5 + {4.5}^{2}]$
$= 1, 200 [153.76 + 55.8 + 20.25]$
$= 1, 200 [229.81]$
$=$ 275,772 ft $^{3}$

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Find the volume of the frustum cone whose base and top radius is 12.4 ft and 4.5 ft respectively. The height of the cone is 1,200 ft. (Use π = 3).

The correct option is D 275,772 ft3Volume of the frustum cone is V=πh3(R2+Rr+r2)R=12.4ft,r=4.5ft,h=1,200ftV=(3×1,200)/(3)[12.42+12.4×4.5+4.52] =1,200[153.76+55.8+20.25] =1,200[229.81] = 275,772 ft3

Find the volume of the frustum cone whose base and top radius is 12.4 ft and 4.5 ft respectively. The height of the cone is 1,200 ft. (Use $π$ = 3).