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

#### Solution

Let the depth of the bucket is h cm. The radii of the top and bottom circles of the frustum bucket are r_{1} =20cm and r_{2} =12cm respectively.

The volume/capacity of the bucket is

`"V" = (1)/(3)π("r"_1^2 + "r"_1"r"_2 + "r"_2^2) xx "h"`

= `(1)/(3)π(20^2 + 20 xx 12 + 12^2) xx "h"`

= `(1)/(3) xx (22)/(7) xx 784 xx "h"`

= `1/3xx22xx112xx"h" "cm"^3`

Given that the capacity of the bucket is 12308.8 Cubic cm. Thus, we have

`(1)/(3) xx 22 xx 112 xx "h" = 12308.8`

⇒ `"h" = (12308.8 xx 3)/(22 xx 112)`

⇒ h = 15

Hence, the height of the bucket is 15 cm

The slant height of the bucket is

`"l" = sqrt(("r"_1 -"r"_2)^2 + "h"^2`

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

= `sqrt(289)`

= 17cm

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

`"S"_1 = π("r"_1 +"r"_2) xx l + π"r"_2^2`

= π x (20 + 12) x 17 + π x 12^{2}

= π x 32 x 17 + 144π

= 2160.32cm^{2}

Hence Surface area of the metal = 2160.32cm^{2}