The diameter of a sphere is decreased by 20%.By what percent does its Curved surface area decrease?
Answers
Answered by
108
Let the diameter of sphere be "d"
surface area of sphere , S = 4πr²
= π (2r)²
= π(d)²
Diameter of sphere decreases by 20%
New diameter = d - (d/5) = [5d- d] / 5 = 4d/5
New surface area , S' = π(4d/5)²
= (16/25) πd²
Change in surface area of sphere = S - S'
= πd² – (16/25) πd²
= (9/25) πd²
Decrease in surface area = [(9/25πd²) / πd²] x 100 = 9 / 25 × 100 = 36 %
Therefore the decrease in S.A of sphere is 36%
surface area of sphere , S = 4πr²
= π (2r)²
= π(d)²
Diameter of sphere decreases by 20%
New diameter = d - (d/5) = [5d- d] / 5 = 4d/5
New surface area , S' = π(4d/5)²
= (16/25) πd²
Change in surface area of sphere = S - S'
= πd² – (16/25) πd²
= (9/25) πd²
Decrease in surface area = [(9/25πd²) / πd²] x 100 = 9 / 25 × 100 = 36 %
Therefore the decrease in S.A of sphere is 36%
Answered by
8
Answer:let diameter =x
Radius =x/2
C.S.A=4πr^2
=4π*x^2/4
=πx^2
New d=x-x*20/100
=4x/5
R=4x/10
C.S.A=4πr^2
=4π*4x/10*4x/10
=64πx^2/100
Decrease C.S.A=64πx^2/100--πx^2
=36πx^2
decrease%=36πx^2/100/πx^2*100
Therefore 100 & 100 will be cancel and πx^2 &πx^2 wii be cancel then,
36% Answer
Similar questions