Question

A cone has a height of `12 cm` and its base has a radius of `8 cm`. If the cone is horizontally cut into two segments `8 cm` from the base, what would the surface area of the bottom segment be?

Answer 1

`A = (640/9 + 256/9 sqrt(13)) pi ~~ 173.7 pi " "cm^2 ~~549.6 " "cm^2`

Explanation:

Given: cone of `h= 12 cm, b = 8 cm`. Cut `8cm` from the base. Find the surface area of the bottom segment.

The new shape is called a conical frustum . It has the surface area formula :

`A = A_"lateral" + A_"bases"`

`A_"lateral" = pi (r_1 + r_2) sqrt((r_1 - r_2)^2 + h_f^2)`

`A_"bases" = pi((r_1)^2 + (r_2)^2)`, where

`r_1 = "base radius = 8 cm", " "r_2 = "top radius" = ?,`

`h_f = "the height of the frustum" = 8 cm`

`h_"top cone"/h_"bottom come" = (12-8)/12 = 4/12`

`color(blue) "Find the radius of the top"` using proportions of the two cones:

`h_"top cone"/h_"bottom come" = r_2/r_1; " "4/12 = r_2/8`

Use the cross-product: `12 r_2 = 4*8 = 32`

`r_2 = 32/12 = 8/3 cm`

`A_"lateral" = pi(8 + 8/3) sqrt((8-8/3)^2 + 8^2)= 32/3 pi sqrt((16/3)^2 + 64)`

`A_"lateral" = 32/3 pi sqrt(256/9 + 576/9) = 32/3 pi sqrt(832)/3`

`A_"lateral" = 32/9 pi sqrt(16) sqrt(4)sqrt(13) = 32/9 pi *8 sqrt(13)`

`A_"lateral" = 256/9 sqrt(13) pi " "cm^2`

`A_"bases" = pi(8^2 + (8/3)^2) = 640/9 pi " "cm^2`

`A = (640/9 + 256/9 sqrt(13)) pi ~~ 173.7 pi " "cm^2 ~~549.6 " "cm^2`

Answer 2

`~~545.60"cm"^2"to the nearest 2 decimal places"`

Explanation:

`tan theta=12/8=1.5=56^@18'36''`

`"top radius"=cot 56^@18'36''=0.666666666 xx4=2.666666666cm`

`Lateral area= F=pi(r_1+r_2)sqrt((r_1-r_2)^2+h^2)`

`F=pi(8+2.666666666)sqrt((8-2.666666666)^2+8^2)`

`F=pi(10.667)sqrt((5.333^2+8^2)`

`F=33.511sqrt28.441+64`

`F=33.511sqrt92.441`

`F=33.511xx9.615`

`F=322.196"cm"^2`

`S=F+ pi (r_1^2+r_2^2)`

`S=322.196+pi(8^2+2.666666666^2)`

`S=322.196+pi(64+7.111)`

`S=322.196+pi(71.113)`

`S=322.196+223.402`

`S=545.598"cm"^2`

`S="surface area of bottom segment"`

`=545.60"cm"^2"to the nearest 2 decimal places"`