Math, asked by ranjnasharma1977, 1 year ago

The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease

Answers

Answered by oyo
3
let the diameter be d
radius=d/2
surface area=4π(d/2)^2
new radius =d/2-25/100×d2=175d/200
new surface area=4π×(175d/200)^2
decresed area=4π(d/2)^2-4π×(175 d/200)^2=16/49
percentage decreases=16/49×100=32.65%

ranjnasharma1977: Thanks..but the answer is 43.75%
Answered by nickkaushiknick
2

Answer:

43.75%

Step-by-step explanation:

Let the Original diameter be 4r

∴ Radius = 2r                                        [∵ radius = diameter/2]

∴ Surface Area = 4 × π × ( 2r )²           [S.A. of Sphere = 4π(radius)² ]

                         = 16πr²

Now, Diameter decreased by 25%

25% of 4r = r

New diameter = 4r - r = 3r

∴ New Radius = 3r/2                             [∵ radius = diameter/2]

New Surface Area = 4 × π × (3r/2)²     [S.A. of Sphere = 4π(radius)² ]

                               = 9πr²

Surface area decreased = Old Surface Area - New Surface Area

                                        = 16πr² - 9πr²

                                        = 7πr²

Percentage Decreased = (Decreased Area × 100)/ Old Area

                                       = ( 7πr² × 100)/16πr²

                                       = 43.75%



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