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(A) Let the initial radius of the cone be $r$ and the initial slant height be $l$.The initial curved surface area of the cone is $A_1 = \pi rl$.Accord

Vedclass · Accepted Answer

$140$

Vedclass · Answer

(A) Let the initial radius of the cone be $r$ and the initial slant height be $l$.The initial curved surface area of the cone is $A_1 = \pi rl$.According to the problem,the new radius $r' = r + 0.20r = 1.2r$.The new slant height $l' = 2l$.The new curved surface area of the cone is $A_2 = \pi r' l' = \pi (1.2r)(2l) = 2.4 \pi rl$.The increase in the curved surface area is $\Delta A = A_2 - A_1 = 2.4 \pi rl - \pi rl = 1.4 \pi rl$.The percentage increase is $\frac{\Delta A}{A_1} 	imes 100 = \frac{1.4 \pi rl}{\pi rl} 	imes 100 = 140 \%$.Therefore,the curved surface area of the cone will increase by $140 \%$.

If the base radius of a cone is increased by $20 \%$ and its slant height is doubled,then by what percentage will the area of its curved surface increase?

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