The curved surface area of frustum cone is 400 $m^{2}$ The diameter of a cone is 0.5 and 2 m. Find the slant height. (Use $π$ = 3.14).

123.47 m

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124.47 m

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101.91 m

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126.47 m

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Solution

The correct option is B 101.91 m
Curved surface area = $π$ (R + r)s
Diameter = radius $\times 2$
Radius, R = 0.25 m, r = 1 m
400 = $3.14 \times (0.25 + 1) \times s$
400 = $3.14 \times (1.25) \times s$
400 = 3.925 $\times s$
S = 400/3.925
Slant Height = 101.91 m

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The curved surface area of frustum cone is 400 m2 The diameter of a cone is 0.5 and 2 m. Find the slant height. (Use π = 3.14).

The correct option is B 101.91 mCurved surface area = π(R + r)sDiameter = radius ×2Radius, R = 0.25 m, r = 1 m400 = 3.14×(0.25+1)×s400 = 3.14×(1.25)×s400 = 3.925 ×sS = 400/3.925Slant Height = 101.91 m

The curved surface area of frustum cone is 400 $m^{2}$ The diameter of a cone is 0.5 and 2 m. Find the slant height. (Use $π$ = 3.14).

The correct option is B 101.91 m
Curved surface area = $π$ (R + r)s
Diameter = radius $\times 2$
Radius, R = 0.25 m, r = 1 m
400 = $3.14 \times (0.25 + 1) \times s$
400 = $3.14 \times (1.25) \times s$
400 = 3.925 $\times s$
S = 400/3.925
Slant Height = 101.91 m