The circumference of the base of a conical tent of height 8 m is 32 πm. Find its curved surface area.

Correct — D. Circumference gives the radius, the radius with the height gives the slant height, and those give the curved surface.

2πr = 32π → r = 16 m

l = √(r² + h²) = √(16² + 8²)
= √(256 + 64) = √320 = 8√5 m

Curved surface area = πrl
= π × 16 × 8√5 = 128√5 πm² → option (d)

Why the others are wrong

• (a)
  Option (a) reads 164√5 πm². The radical is right but the coefficient is not — πrl needs 16 × 8 = 128, and 164 is not a product of r = 16, h = 8 or l = 8√5.
• (b)
  Option (b) reads 162√3 πm². A √3 would need r² + h² to be three times a perfect square; here it is 320 = 64 × 5, so the surface area carries √5.
• (c)
  Option (c) reads 132√3 πm², so both parts are wrong: the radical should be √5, and the coefficient should be 16 × 8 = 128.

A cone's curved surface area is πrl, where l is the slant height and l² = r² + h². The question is a chain: circumference gives r, r with h gives l, r with l gives the area.

The circumference is printed as 32π m, with π kept as a symbol. Cancelling π on both sides of 2πr = 32π leaves r = 16 at once, so no numerical value of π is needed anywhere — and the options keep π as a symbol too.

Total surface area, πr(l + r), is a different quantity. For this tent it would add the base, another 256π m².

Key facts

• The curved surface area of a cone is πrl, with l the slant height.
• Slant height satisfies l² = r² + h², giving l = √(16² + 8²) = 8√5 m here.
• √320 simplifies to 8√5 because 320 = 64 × 5.

Common traps

• Dropping the π from 32π m and solving 2πr = 32, which gives r = 16⁄π.
• Using the height 8 in place of the slant height inside πrl.
• Computing the total surface area when only the curved surface is asked.

No directly related past PYQ was found.