Question

the top and bottom radius of frusturm are 7m and 14m and height of frusturm is 10 m find the volume and total surface area of frusturm.​

Answer 1

$\bf \to \: { \underline{given \div }}$

$\sf \to \: top \: and \: bottom \: of \: radius \: of \: frustum \: = 7m \: and \: 14m. \\ \\ \sf \to \: height \: of \: the \: frustum \: = 10m.$

$\bf \to \: { \underline{to \: find \div }} \\ \\ \sf \to \: 1) = volume \: of \: frustum \\ \\ \sf \to \: 2) = total \: surface \: area \: of \: frustum.$

$\bf \to \: \orange{{ \underline{step - by - step \: \: explanation}}}$

$\sf \to \:top \: of \: radius \: = r_{1} \: = 7m. \\ \\ \sf \to \: bottom \: of \: radius \: = r_{2} \: = 14m \\ \\ \sf \to \: height \: = 10m \\ \\ \sf \to \: { \underline{volume \: of \: frustum \: of \: cone}} \\ \\ \sf \to \: \dfrac{1}{3}\pi \: h \: ( r_{1} {}^{2} + r_{1} r_{2} \: + r_{2} {}^{2} ) \\ \\ \sf \to \: \frac{1}{3} \times \frac{22}{7} \times 10 (7 \times 7 + 7 \times 14 + 14 \times 14) \\ \\ \sf \to \: \frac{1}{3} \times \frac{22}{7} \times 10(49 + 98 + 196) \\ \\ \sf \to \: \frac{1}{3} \times \frac{22}{7} \times 10 \times 343 \\ \\ \sf \to \: \frac{1}{3} \times \frac{22}{ \cancel{7}} \times 10 \times \cancel{343} \\ \\ \sf \to \: \frac{22 \times 10 \times 49}{3} = 3593.3m {}^{3} \\ \\ \sf \to \: \green{{ \underline{volume \: of \: frustum \: = 3593.3 {m}^{3} }}}$

$\sf \to \: slant \: height \: of \: frustum \: of \: cone \\ \\ \sf \to \: \sqrt{( r_{2} \: - r_{1}) {}^{2} + {h}^{2} } \\ \\ \sf \to \: \sqrt{(14 - 7) {}^{2} + 10 {}^{2} } \\ \\ \sf \to \: \sqrt{49 + 100} = \sqrt{149} = 12.2m \\ \\ \sf \to \: { \underline{total \: surface \: area \: of \: frustum}} \\ \\ \sf \to \: \pi \: l \: ( r_{1} \: + r_{2} \: ) + \pi \: ( r_{1} {}^{2} + r_{2} {}^{2} ) \\ \\ \sf \to \: \frac{22}{7} \times 12.2(14 + 7) + \frac{22}{7}( {7}^{2} + {14}^{2} ) \\ \\ \sf \to \: \frac{22}{7} \times 12.2 \times 21 + \frac{22}{7} \times (49 + 196) \\ \\ \sf \to \: 804.468 + 349.86 = 1154.328m {}^{2} \\ \\ \sf \to \: \orange{{ \underline{total \: surface \: area \: of \: frustum \: = 1154.328 {m}^{2} }}}$