Question

if the radius of Sphere is increased by 9.09% then surface area of Sphere is increased by how much percentage​

Answer 1

Answer:

19%

Step-by-step explanation:

we know that 9.09% in fraction is 1/11

now that radius is is increased by 1/11 meaning radius if previous radius is 11 then the final radius will be 12.

initial surface area = 4π(11)^2

final surface area = 4π(12)^2

therefore, the % change in surface area

=  (4π(12)^2- 4π(11)^2)/ 4π(11)^2×100

=23/121×100= 19%

Answer 2

The surface area of Sphere is increased by 19 %

Solution:

The surface area of sphere is:

$SA = 4 \pi r^2$

Where, "r" is the radius of sphere

Radius of Sphere is increased by 9.09%

Then,

$R = \frac{9.09}{100}r + r \\\\R = 1.0909r$

Now,

$SA = 4 \times \pi \times (1.0909r)^2$

Surface area of Sphere is increased by how much percentage​

$Percentage = \frac{ 4 \times \pi \times (1.0909r)^2 - 4 \pi r^2 }{ 4 \pi r^2 } \times 100 \% \\\\\\Percentage = (1.0909^2 - 1 ) \times 100 \% \\\\\\Percentage = 19.006 \approx 19 \%$

Thus, surface area of Sphere is increased by 19 %

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