The frustum cone curved surface area is 2,489 $m^{2}$ . The upper and lower circle area of a cone is 24 and 45 m. Find the total surface area.

2,558

m^{3}

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2,558 m

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2,398

m^{3}

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2,408

m^{3}

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Solution

The correct option is B 2,558 $m^{3}$
Total surface area = $π (R + r) s + A_{1} + A_{2}$
Curved surface area = $π (R + r) s \Rightarrow 2, 489 m^{2}$
$A_{1} = 45 m, A_{2} = 24 m$
Therefore, Total surface area $= 2, 489 + 45 + 24$
TSA = 2,558 $m^{3}$

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