Question

The radii of ends of a frustum are 14 cm and 6 cm respectively and its height is 6 cm. Find its i) curved surface area ii) total surface area
iii ) volume

($\mathrm{\pi }$= 3.14)

Answer 1

Here, r₁ = 14 cm, r₂ = 6 cm and h = 6 cm.

Slant height of the frustum, l = $\sqrt{h^2+{\left(r_2-r_1\right)}^2}=\sqrt{6^2+{\left(14-6\right)}^2}=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}$ = 10 cm

i)
Curved surface area of frustrum
$$\begin{gathered} =\mathrm{\pi }\left(r_1+r_2\right)l \\ =3.14\times \left(14+6\right)\times 10 \\ =3.14\times 20\times 10 \\ =628{\mathrm{cm}}^2 \end{gathered}$$
ii)
Total surface area of frustrum
$$\begin{gathered} =\mathrm{\pi }\left(r_1+r_2\right)l+\mathrm{\pi }{r}_1^2+\mathrm{\pi }{r}_2^2 \\ =3.14\times \left(14+6\right)\times 10+3.14\times {14}^2+3.14\times 6^2 \\ =3.14\times 20\times 10+3.14\times 196+3.14\times 36 \\ =628+615.44+113.04 \\ =1356.48{\mathrm{cm}}^2 \end{gathered}$$
iii)
Volume of the frustum
$$\begin{gathered} =\frac{1}{3}\mathrm{\pi }h\left(r_1^2+r_1{r}_2+r_2^2\right) \\ =\frac{1}{3}\times 3.14\times 6\times \left({14}^2+14\times 6+6^2\right) \\ =3.14\times 2\times \left(196+84+36\right) \\ =6.28\times 316 \\ =1984.48{\mathrm{cm}}^3 \end{gathered}$$