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The perimeters of the ends of a frustum of a right circular cone are 44 cm and 33 cm. If the height of the frustum be 16 cm, find its volume, the slant surface and the total surface.

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If you open math book you will find a lot of mathematical questions but
youshould have solve this question.
It is from class 10 rd sharma chapter 16 mensuration .

In this question we have to find the volume,the slant surface and
total surface from the given informations that the perimeters of
the ends of a frustum of a right circular cone are 44 cm and 33 cm.
If the height of the frustum be 16 cm.

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1. Given,

Perimeter of the upper end = 44 cm

2 Ï€ r1 = 44

2(22/7) r1 = 44

r1 = 7 cm

Perimeter of the lower end = 33 cm

2 Ï€ r2 = 33

2(22/7) r2 = 33

r2 = 21/4 cm

Now,

Let the slant height of the frustum of a right circular cone be L

L = 16.1 cm

So, the curved surface area of the frustum cone = Ï€(r1 + r2)l

= Ï€(7 + 5.25)16.1

Curved surface area of the frustum cone = 619.65 cm3

Next,

The volume of the frustum cone = 1/3 Ï€(r22 + r12 + r1 r2 )h

= 1/3 Ï€(72 + 5.252 + (7) (5.25) ) x 16

= 1898.56 cm3

Thus, volume of the cone = 1898.56 cm3

Finally, the total surface area of the frustum cone

= Ï€(r1 + r2) x L + Ï€ r12 + Ï€ r22

= Ï€(7 + 5.25) Ã— 16.1 + Ï€72 + Ï€5.252

= Ï€(7 + 5.25) Ã— 16.1 + Ï€(72 + 5.252) = 860.27 cm2

Therefore, the total surface area of the frustum cone is 860.27 cm2

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