NCERT Class 9 Solutions: Surface Areas and Volumes (Chapter 13) Exercise 13.4 – Part 2

Curved surface area of hemisphere is half of sphere.

Curved Surface Area and Total Surface Area of Sphere

Curved surface area of hemisphere is half of sphere.

Curved surface area of hemisphere is half the area of sphere. The total surface area also includes the base of radius r at the bottom therefore TSA of Sphere = 2πr2+πr2=3πr2

Q-3 Find the total surface area of a hemisphere of radius 10cm . (Use π=3.14 )Solution:

Hemisphere of radius r=10cm

Therefore, total surface area of hemisphere = 3πr2 = (3×3.14×10×10)cm2 = 942cm2

Q-4 The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.Solution:

Now, the initial radius is r and the increased radius of balloons is R . Where, r=7cmandR=14cm

Therefore, ratio of the surface area

  • 4πr24πR2

  • r2R2

  • 7×714×14=14

So, the surface areas are in ratio 1:4 .

Q-5 A hemispherical bowl made of brass has inner diameter 10.5cm . Find the cost of tin-plating it on the inside at the rate of s.16per 100cm2 .


A bowl is a hemisphere with no base. So we are interested only in the curved surface area.

Radius of the bowl (r) =10.52cm=5.25cm ( radius=diameter2 )

Therefore, curved surface area of the hemispherical bowl

  • =2πr2

  • =(2×227×5.25×5.25)cm2

  • =173.25cm2

Tin-plating costs for 100 cm2 is Rs.16

So, cost of tin plating 1cm2=Rs16100

Finally, total cost of tin-plating the hemisphere bowl with area 173.25 cm2 is 173.25×16100=Rs27.72

Q-6 Find the radius of a sphere whose surface area is 154cm2 .Solution:

Consider, r be the radius of the sphere.

We know that surface area = 154cm2

  • Therefore, 4πr2=154

  • 4×227×r2=154

  • r2=1544×227

  • r2=154×74×22

  • r2=494

  • r=72

  • 3.5cm

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