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Question -

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).



Answer -


Given,

Radius of the commonbase (r) = 3.5 cm

Height of thecylindrical part (h) = 10 cm

Height of the conicalpart (H) = 6 cm

Let, ‘l’ be the slantheight 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 surfacearea of the cone (S1) = πrl

S1 = π(3.5)(6.95)

S1 =76.38 cm

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

S2 = 2π(3.5)2

S2 =77 cm2

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

S2 = 2π(3.5)(10)

S2 =220 cm2

Thus, the totalsurface area (S) = S+ S2 + S3

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

Therefore, the totalsurface area of the solid is 373.38 cm2

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