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

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If we reduce the diameter by 25%, the radius will also be reduced by 25%.

let initial radius of sphere = r

initial CSA = 4πr²

after reducing it 25%,

25% of r = (25/100)×r = 0.25r

final radius = r - 0.25r = 0.75r

final CSA = 4π(0.75r²) = 4πr²×(0.5625)

decrease in CSA = initial CSA - final CSA = 4πr² - 4πr² × (0.5625)

⇒decrease in CSA = 4πr² × (1-0.5625) = 4πr² × 0.4375

% decrease in CSA =

Thus CSA decreases by 43.75%.

let initial radius of sphere = r

initial CSA = 4πr²

after reducing it 25%,

25% of r = (25/100)×r = 0.25r

final radius = r - 0.25r = 0.75r

final CSA = 4π(0.75r²) = 4πr²×(0.5625)

decrease in CSA = initial CSA - final CSA = 4πr² - 4πr² × (0.5625)

⇒decrease in CSA = 4πr² × (1-0.5625) = 4πr² × 0.4375

% decrease in CSA =

Thus CSA decreases by 43.75%.

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Let the initial radius be=r

On decreasing by 25%,new radius=r-((25/100)xr)=3r/4

Initial surface area=4πr²

Final surface area=4π(3r/4)²=4π9r/16

% of new surface area=56.25%

% of decrease in area=(100-56.25)%=43.75%

On decreasing by 25%,new radius=r-((25/100)xr)=3r/4

Initial surface area=4πr²

Final surface area=4π(3r/4)²=4π9r/16

% of new surface area=56.25%

% of decrease in area=(100-56.25)%=43.75%