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.
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