Question

The radii of the ends of 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

Answer 1

Let '$\ell$' & R be the radii of the bottom and top circular ends of a frustum of a cone, respectively and 'h' be height of cone.

Volume of the frustum of the cone $=\frac{1}{3}\times \frac{22}{7}\times 45({28}^2+7^2+28\times 7)c{m}^3$

$=\frac{22}{7}\times 15(784+49+196)c{m}^3$

$=\frac{22}{7}\times 15\times 1029c{m}^3=22\times 15\times 147c{m}^3$

$=330\times 147c{m}^3$

Now, $l=\sqrt{h^2+(R-r^2)}=\sqrt{(45{)}^2+(28-7{)}^2}=\sqrt{2025+441}=\sqrt{2466}$

$=\sqrt{2466}=49.65cm$

curved surface area of the frustum of the cone

$=\pi (R+r)\ell =[\frac{22}{7}\ast (28+7)\ast 49.65]c{m}^2$

$=[22\times 5\times 49.65]c{m}^2$

$5461.5c{m}^2$

total surface area of frustum of the cone

$\pi (R+r)\ell +\pi R^2+\pi r^2$

$=[5461.5+\frac{22}{7}\times (28{)}^2+\frac{22}{7}\times \left(7{)}^2\right]c{m}^2$

$=[5261.5+\frac{22}{7}\times (28{)}^2+\left(7{)}^2\right]c{m}^2$

$=[5461.5+22\times 1+9]c{m}^2$

$=8079.5c{m}^2$