Question

The radii of the ends of a frustum of a cone 45 cm high are 28 cm and 7 cm . Find its volume, the curved surface area and the total surface area.

$(take \: \: \pi \: = \frac{22}{7} )$

Answer 1

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Answer 2

solutions Here ,

LET ‘r’ and ‘R’ be the radii of the bottom and top circular ends of a frustum of a cone, respectively and ‘h’ be the height of the cone.

=> volume of the frustum of the cone

=>1/3πh(R²+r²+Rr)
$= > [\frac{1}{3} \times \frac{22}{7} \times 45( {28}^{2} + {7}^{2} + 28 \times 7)] {cm}^{3} \\ \\ = > [\frac{22}{7} \times 15(784 + 49 + 196)] {cm}^{3} \\ \\ = > [\frac{22}{7} \times 15 \times 1029] {cm}^{3} \\ \\ = > [22 \times 15 \times 147 ]\: {cm}^{3} \\ \\ = > (330 \times 147) {cm}^{3} \\ \\ = >$
Now,
$l = \sqrt{( {h)}^{2} + ( {R - r)}^{2} } \\ \\ = > \sqrt{ {(45)}^{2} + {(28 - 7)}^{2} } \\ \\ = > \sqrt{2025 + 441 } \\ \\ = > \sqrt{2466} \\ \\ = > 49.65 \: cm$
curved surface area of the frustum of the cone.
$= > \pi(R+ r)l \\ \\ = > [\frac{22}{7} \times (28 + 7) \times 49.65] \: {cm}^{2} \\ \\ = >[ 22 \times 5 \times 49.65] \: {cm}^{2} \\ \\ = > 5461.5 \: {cm}^{2}$
Total surface area of the frustum of the cone
$= > \pi(R + r)l + \pi {R}^{2} + \pi {r}^{2} \\ \\ = > [5461.5 + \frac{22}{7} \times {(28)}^{2} + \frac{22}{7} \times {(7)}^{2} ]{cm}^{2} \\ \\ = > [5461.5 + \frac{22}{7} \times ( {28)}^{2} + ( {7)}^{2}] \: {cm}^{2} \\ \\ = > [5461.5 + \frac{22}{7} \times (784 + 49) ]\: {cm}^{2} \\ \\ = >[ 5461.5 + \frac{22}{7} \times 833 ]\: {cm}^{2} \\ \\ = >[ 5461.5 + 22 \times 119] \: {cm}^{2} \\ \\ = > [5461.5 + 2618] \: {cm}^{2} \\ \\ = > 8079.5 \: {cm}^{2} \: \: \: answer$
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