Question

The circumferences of circular faces of a frustum are 132 cm and 88 cm and its height is 24 cm. To find the curved surface area of the frustum complete the following activity.($\mathrm{\pi }=\frac{22}{7}$)

circumference₁ = 2$\mathrm{\pi }$r₁ = 132

r₁ = $\frac{132}{\mathrm{\pi }}=\boxed{134}$

circumference₂ =2$\mathrm{\pi }$r₂= 88

r₂= $\frac{88}{2\mathrm{\pi }}=\boxed{134}$

slant height of frustum, l = $$\begin{gathered} \sqrt{h^2+{\left(r_1-r_2\right)}^2} \\ \sqrt{{\boxed{1223}}^2+\boxed{1234}{}^2} \\ =\boxed{1234}\mathrm{cm} \end{gathered}$$

curved surface area of the frustum = $\mathrm{\pi }$ (r₁ + r₂ ) l

= $$\begin{gathered} \mathrm{\pi }\times \boxed{123}\times \boxed{132} \\ =\boxed{123}\mathrm{sq}.\mathrm{cm}. \end{gathered}$$

Answer 1

Circumference₁ = 2$\mathrm{\pi }$r₁ = 132

r₁ = $\frac{132}{2\mathrm{\pi }}=\frac{132\times 7}{2\times 22}=\boxed{21\mathrm{cm}}$

Circumference₂ = 2$\mathrm{\pi }$r₂ = 88

r₂ = $\frac{88}{2\mathrm{\pi }}=\frac{88\times 7}{2\times 22}=\boxed{14\mathrm{cm}}$

Slant height of frustum,
$$\begin{gathered} l=\sqrt{h^2+{\left(r_1-r_2\right)}^2} \\ =\sqrt{{24}^2+{\left(21-14\right)}^2} \\ =\sqrt{{\boxed{24}}^2+\boxed{7}{}^2} \\ =\sqrt{576+49} \\ =\sqrt{625} \\ =\boxed{25}\mathrm{cm} \end{gathered}$$
Curved surface area of the frustum = $\mathrm{\pi }$ (r₁ + r₂ ) l
$$\begin{gathered} =\mathrm{\pi }\times \left(21+14\right)\times 25 \\ =\mathrm{\pi }\times \boxed{35}\times \boxed{25} \\ =\frac{22}{7}\times 35\times 25 \\ =\boxed{2750}\mathrm{sq}.\mathrm{cm}. \end{gathered}$$