Question

A solid is in form of a right circular with a a hemisphere at one end and a at the other end the radius of the common base is 3.5 and the height of the cylindrical and conicsl portion are 10cm.and 6cm,respectively. Find the total surface area of the solid

Answer 1

radius(r) is 8 cm, height of cylinder(H) is 10 cm and height of cone(h) is 6cm.
Also, l = √ r2 + h2
⇒ l = √ 82 + 62
⇒ l = √ 64 + 36
⇒ l = √ 100
⇒ l = 10
Now, total surface area of solid = surface area of cone + surface area of cylinder + surface area of sphere
⇒ total surface area of solid = πrl + 2πrH + 2πr2 = πr(l + 2H + 2r)
= 3.14 × 8(10 + 2 × 10 + 2 × 8)
= 3.14 × 8 × 46
= 1155.55 cm2

Answer 2

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 

➠l² = 12.25 + 36

➠l² = 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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