Find the volume of a frustum cone, whose base and upper area of a circle is 20 and 80 $m^{2}$ . The height of the cone is 45.5 m.

6,270

m^{3}

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6,170

m^{3}

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2,123.33

m^{3}

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6,470

m^{3}

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Solution

The correct option is D 2,123.33 $m^{3}$
Volume of a frustum cone = $\frac{1}{3} (A_{1} + A_{2} + \sqrt{A_{1} A_{2}}) h$
$= \frac{1}{3} (20 + 80 + \sqrt{20 \times 80}) \times 45.5$
$= \frac{1}{3} (100 + \sqrt{1600}) \times 45.5$
$= \frac{1}{3} (100 + 40) \times 45.5$
$= \frac{1}{3} \times 140 \times 45.5$
$=$ 2,123.33 $m^{3}$

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