# The diameter of a sphere is decreased by 25%.by what % does it's curved surface area decrease ?

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Initial r = d/2

Initial S.A. = 4*Pi*(d^2)/4

New dia. = d-(25% of d)

= d-(d/4) = 3d/4

New r = 3d/8

Thus, New S.A. = 4*Pi*(9d^2)/64

Now %age increase = [ (New S.A.)/(Initial S.A.) ] * 100

= (9/64) * 4 * 100

= 56.25 %

So, %age decrease = 100 - 56.25 = 43.75 %

diam. decreased by 25%

i.e., (100-25)%=75% of the diam. is left

Therefore, 75/100 of diam.=3diam./4= 3d/4 × 1/2= 3d/8 (r=d/2)

NEW CSA,

4π{(3d/4)/2}²

4π(3d/8)²=4π×9d²/64

=9πd²/16

Now, decrease%= decrease in CSA × 100%/ original CSA

= (Original CSA-new CSA)×100/ Original CSA

={(πd²-9πd²/16) × 100}/πd²

=7πd²/16 × 1/πd² × 100= 43.75 % decrease in CSA