Question 68

The radii of the two circular faces of the frustum of a cone of height 14 cm are 5 cm and 2 cm. What is its volume in cm$$^3$$ ($$\pi = \frac{22}{7}$$)

Solution

Volume of frustum of cone $$\frac{1}{3}\times\ \pi\ \times\ H\times\ \left\{\left(R1^2+R2^2\right)+\left(R1\times\ R2\right)\right\}$$

                                    =$$\frac{1}{3}\times\ \frac{22}{7}\ \times\ 14\times\ \left\{\left(5^2+2^2\right)+\left(5\times\ 2\right)\right\}$$

                                     =$$\frac{44}{3}\times\ \left\{\left(25+4\right)+10\right\}$$

                                    =$$\frac{44}{3}\times\ 39=572$$


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