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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 heights 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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Solution
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,
I2 = r2 + H2
I2 = 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 = `22/7 xx 3.5 xx 6.95`
S1 = 76.45 cm2
Curved surface area of cylinder (S2) = 2πrh
S2 = 2π(3.5)(10)
S2 = `2 xx 22/7 xx 3.5 xx 10`
S2 = 220 cm2
Curved surface area of hemisphere(S3) = 2πr2
S3 = `2 xx 22/7 xx 3.5^2`
S3 = 77 cm2
Total surface area of the solid is given by
S = S1 + S2 +S3
= 76.45 + 220 + 77
= 373.45 cm2
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