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# A solid is in the form of a right circular cylinder, with a hemisphere at one end and a cone at the other end. The radius of the common base is 3.5 cm and the height of the cylindrical and conical portions are 10 cm and 6 cm, respectively. Find the total surface area of the solid. (Use π = 22/7).

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I would like to give a question which
i could not solve.Rd sharma class 10
maths exercise-16.2 mensuration.

In this question we have to find the total surface area of
the solid.from a solid is in the form of a right circular
cylinder, with a hemisphereat one end and a cone at the
other end. The radius of the common baseis 3.5 cm and the
height of the cylindrical and conical portions are
10 cm and 6 cm, respectively.

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1. Given,

Radius of the common base (r) = 3.5 cm

Height of the cylindrical part (h) = 10 cm

Height of the conical part (H) = 6 cm

Let, ‘l’ be the slant height of the cone

Then, we know that

l2 = r2 + H2

l2 = 3.52 + 62 = 12.25 + 36 = 48.25

l = 6.95 cm

So, the curved surface area of the cone (S1) = πrl

S1 = π(3.5)(6.95)

S1 = 76.38 cm2

And, the curved surface area of the hemisphere (S2) = 2πr2

S2 = 2π(3.5)2

S2 = 77 cm2

Next, the curved surface area of the cylinder (S3) = 2πrh

S2 = 2π(3.5) (10)

S2 = 220 cm2

Thus, the total surface area (S) = S1 + S2 + S3

S = 76. 38 + 77 + 220 = 373.38 cm2

Therefore, the total surface area of the solid is 373.38 cm2

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