Question

The circumferences of circular faces of 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. `(π = 22/7)`

Circumference₁ = 2πr₁ = 132

`r_1 = 132/(2π) = square`

Circumference₂ = 2 = 2πr₂ = 88

`r_2 = 88/(2π) = square`

Slant height frustum = `(l) = sqrt(h^2 + (r_1 - r_2)^2`

= `sqrt(square^2 + square^2)`

= `square` cm

∴ Curved surface area of frustum = `π(r_1 + r_2)l = square  cm^2`

Answer 1

Given: 

Circumference₁ = 132 cm

Circumference₂ = 88 cm

height h = 24 cm

`π = 22/7`

Step-wise calculation:

1. `r_1 = ("Circumference"_1)/(2π)`

= `132/(2π)` 

= `132/(2 xx 22/7)` 

= `132 × 7/44`

= \[\boxed{21 \phantom{.}\text{cm}}\]

2. `r_2 = ("Circumference"_2)/(2π)` 

= `88/(2π)` 

= `88/(2 × 22/7)` 

= `88 xx 7/44` 

= \[\boxed{14 \phantom{.}\text{cm}}\]

3. Slant height `l = sqrt(h^2 + (r_1 - r_2)^2)` 

= \[\sqrt{\boxed{24}^2 + \boxed{21 - 14}^2}\]

= `sqrt(576 + 49)`

= `sqrt(625)`

= \[\boxed{25}\] cm

4. Curved surface area = `π(r_1 + r_2)l` 

= `22/7 xx (21 + 14) xx 25`

= `22/7 xx 35 xx 25`

= 22 × 125

= \[\boxed{2750}\] cm²

Curved surface area of the frustum = 2750 cm².