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