Question

The radius of a sphere increases by 25%. Find the percentage increase in its surface area​

Answer 1

Step-by-step explanation:

Let radius be r

Therefore,surface area=4pi×r^2

when radius is increased by 25%.

therefore new Radius=r+r×25/100

R=5/4 r (new radius=R)

New surface area=4 pi×R^2=4×pi×5/4 r×5/4 r

Area=25pi×r^2/4

Increase in area=25pir^2/4- 4pir^2

=25pir^2-16pir^2 /4

9pir^2/4

Therefore,increase %=9pir^2/4÷4pir^2 ×100

=9pir^2/4 ×1/4pir^2 ×100

=225/4=56.25%

Answer 2

Radius increases by 25% then,

r'=r+r(25/100)

=r+r/4

=(5r/4)

.

New surface area of sphere is

=4π(5r/4)^2

=25πr^2/4

increase in surface area =(new surface area - old surface area)/old surface area

==(25πr^2/4 -4πr^2)/4πr^2

={πr^2( 25/4-4)}/4πr^2

=( 9/4 πr^2) / 4πr^2

={(9/4)/4 }

=9/16

for percentage

=(9/16)*100

=2.25*25%

=56.25%

$=4\pi { \frac{5r}{4} } \binom{5r}{4} ^{2} </p><p> =25πr^2/4</p><p>increase in surface area =(new surface area - old surface area)/old surface area</p><p> ==(25πr^2/4 -4πr^2)/4πr^2</p><p> ={πr^2( 25/4-4)}/4πr^2</p><p> =( 9/4 πr^2) / 4πr^2</p><p> ={(9/4)/4 }</p><p> =9/16</p><p> for percentage </p><p> =(9/16)*100</p><p> =2.25*25%</p><p>=56.25%</p><p>$