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A tent is in the form of a right circular cylinder, surmounted by a cone. The diameter of the cylinder is $24$ m. The height of the cylindrical portion is $11$ m, while the vertex of the cone is $16$ m above the ground. Find the curved surface area of the cylindrical portion.

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(246 π) m^{2}

(264 π) m^{2}

(426 π) m^{2}

(462 π) m^{2}

answer is B.

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Detailed Solution

Given that, the diameter of the cylinder d

= 24

m.
The height of the cylinder = 11m.
The radius

r = \frac{d}{2} = \frac{24}{2}

r = 12

m
The curved surface area of the cylinder is

2 (π) rh .

Substitute values in the formula,

= 2 \times (π) \times 12 \times 11

= 264 (π) m^{2}

Hence, the curved surface area of the cylinder is

(264 π) m^{2}

.
Therefore, option 2 is correct.

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