A bucket is in the form of a frustum of a cone with a capacity of 12308.8 cm^{3} of water.The radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of the bucket and the area of the metal sheet used in its making. (Use 𝜋 = 3.14).

#### Solution

Given that,

the radii of the top and bottom circles o0f the frustum bucket are

r_{1} = 20 cm and r_{2} = 12 cm respectively.

Volume of the frustum cone = capacity of the bucket

Capacity of the bucket

Capacity of the bucket (V)`=1/3pi(r_1^2+r_1r_2+r_2^2)xxh`

`=1/3xx22/7(20^2+20xx12+12^2)xxh`

`=1/3xx22/7xx784xxh`

It is given that capacity of bucket is 12308.8 cm^{3}.

Hence, we have

`rArr 1/3xx22/7xx784xxh = 12308.8`

`rArr h=(12308.8xx7xx3)/(22xx784)`

⇒ h= 14.98 ≅ 15 cm

The slant height of the bucket

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

`=sqrt(15^2+(20-12)^2`

`=sqrt(225+64)`

= 17 cm

The surface area of the used metal sheet to make the bucket

`S = pi(r_1+r_2)xxl+pir_2^2`

`= 3.14(20+12)xx17+22/7xx12^2`

= 3.14 x 32 x 17 + 3.14 x 144

= 1708.16 + 452.16

= 2160.32 cm^{2}

Hence, the surface area of the metal sheet is 2160.32 cm^{2}.