Question

The radius of hemisphere is decreased by 10%. Find the percentage change in its surface area?

Answer 1

Answer:

Step-by-step explanation:

Surface area of hemisphere  = 3$\pi. R^{2}$;

R is the radius of the hemisphere:

Now radius is decreased by 10%:

New radius is 0.9 R;

New surface area of hemisphere is = 3.$\pi. (0.9R)^{2} ;$;

Percentage change is the surface area = $\frac{Changes in the surface.area}{ Old.surface.area}$ * 100;

= $\frac{3.\pi.(1-<strong>0.81)</strong> R^{2}  }{3.\pi. R^{2}  }$ * 100;

= 0.19 * 100;

Percentage change in the area  = 19%

Answer 2

Answer:

The radius of hemisphere is decreased by 10%. The percentage change in its surface area = 19% reduction

Step-by-step explanation:

There can be two cases Hemisphere can be Opened or closed

Surface Area of open hemisphere = 2πR²

Surface Area of open hemisphere = 3πR²

Case 1  : open hemisphere

Let say initial radius = R

Initial Surface Area of open hemisphere = 2πR²

Radius after 10 % decrease = R - (10/100)R = 0.9R

Surface Area of open hemisphere after radius reduction = 2π(0.9)R²

= 0.81(2πR²)

Decease in surface area = 2πR² - 0.81(2πR²) = (0.19)(2πR²)

% reduction in surface Area = ((0.19)(2πR²)/2πR² ) * 100 = 19 %

19 % reduction

Case 2 : closed Hemisphere

Let say initial radius = R

Initial Surface Area of Closed hemisphere = 3πR²

Radius after 10 % decrease = R - (10/100)R = 0.9R

Surface Area of closed hemisphere after radius reduction = 3π(0.9)R²

= 0.81(3πR²)

Decease in surface area = 3πR² - 0.81(3πR²) = (0.19)(3πR²)

% reduction in surface Area = ((0.19)(3πR²)/3πR² ) * 100 = 19 %

19 % reduction