# The perimeters of the two circular ends of a frustum of a cone are 48 cm and 36 cm. If the height of the frustum is 11 cm. find its volume and curved surface area.

Given: perimeter of upper circle = 36 cm

Perimeter of lower circle = 48 cm

Height of frustum = h = 11 cm

Let r: radius of upper circle & R: radius of lower circle Total surface area of frustum = πr2 + πR2 + π(R + r)l cm2

Where l = slant height Now perimeter of circle = circumference of circle = 2π × radius

Perimeter of upper circle = 2πr

36 = 2 × 3.14 × r

r = 36/6.28 cm

r = 5.732 cm

Perimeter of lower circle = 2πR

48 = 2 × 3.14 × R

R = 48/6.28 cm

R = 7.643 cm l = 11.164 cm

Volume of frustum = (1/3) × 3.14 × 11 × (7.6432 + 5.7322 + 7.643 × 5.732) cm3

= (1/3) × 34.54 × (58.415 + 32.855 + 43.809) cm3

= 11.513 × 135.079 cm3

= 1555.164 cm3

Volume of frustum = 1555.164 cm3

Now we have asked curved surface area so we should subtract the top and bottom surface areas which are flat circles.

Surface area of top = πr2

Surface area of bottom = πR2

Curved surface area = total surface area - πr2 - πR2 cm2

= πr2 + πR2 + π(R + r)l - πr2 - πR2 cm2

= π(R + r)l cm2

= 3.14 × (7.643 + 5.732) × 11.164 cm2

= 3.14 × 13.375 × 11.164 cm2

= 468.86 cm2

curved surface area = 468.86 cm2

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