The curved surface area of frustum cone is 360 $i n^{2}$ . The diameter of a top cone is 12 and the bottom cone is 20 in. Find the slant height. (Use $π$ = 3.14). Round off your answer to the nearest whole number.

7 in

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3 in

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4 in

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5 in

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Solution

The correct option is A 7 in
Curved surface area = $π$ (R + r)s
Diameter = radius $\times 2$
Radius, R = 10 in, r = 6 in
360 = $3.14 \times (10 + 6) \times s$
360 = $3.14 \times (16) \times s$
$360 / 3.14 \times 16$ = s
Slant Height = 7 in

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The curved surface area of frustum cone is 360 in2. The diameter of a top cone is 12 and the bottom cone is 20 in. Find the slant height. (Use π = 3.14). Round off your answer to the nearest whole number.

The correct option is A 7 inCurved surface area = π(R + r)sDiameter = radius ×2Radius, R = 10 in, r = 6 in360 = 3.14×(10+6)×s360 = 3.14×(16)×s360/3.14×16 = sSlant Height = 7 in

The curved surface area of frustum cone is 360 $i n^{2}$ . The diameter of a top cone is 12 and the bottom cone is 20 in. Find the slant height. (Use $π$ = 3.14). Round off your answer to the nearest whole number.

The correct option is A 7 in
Curved surface area = $π$ (R + r)s
Diameter = radius $\times 2$
Radius, R = 10 in, r = 6 in
360 = $3.14 \times (10 + 6) \times s$
360 = $3.14 \times (16) \times s$
$360 / 3.14 \times 16$ = s
Slant Height = 7 in