# CLASS 9 MATH NCERT SOLUTION FOR CHAPTER – 13 SURFACE AREAS AND VOLUMES EX – 13.4

## Surface Areas and Volumes

1. Find the surface area of a sphere of radius:

• (i) 10.5cm
• (ii) 5.6cm
• (iii) 14cm

(Assume π=22/7)

Solution:

Formula: Surface area of sphere (SA) = 4πr2

(i) Radius of sphere, r = 10.5 cm

SA = 4×(22/7)×10.5= 1386

Surface area of sphere is 1386 cm2

(ii) Radius of sphere, r = 5.6cm

Using formula, SA = 4×(22/ 7)×5.6= 394.24

Surface area of sphere is 394.24 cm2

(iii) Radius of sphere, r = 14cm

SA = 4πr2

= 4×(22/7)×(14)2

= 2464

Surface area of sphere is 2464 cm2

2. Find the surface area of a sphere of diameter:

(i) 14cm (ii) 21cm (iii) 3.5cm

(Assume π = 22/7)

Solution:

(i) Radius of sphere, r = diameter/2 = 14/2 cm = 7 cm

Formula for Surface area of sphere = 4πr2

= 4×(22/7)×72 = 616

Surface area of a sphere is 616 cm2

(ii) Radius (r) of sphere = 21/2 = 10.5 cm

Surface area of sphere = 4πr2

= 4×(22/7)×10.5= 1386

Surface area of a sphere is 1386 cm2

Therefore, the surface area of a sphere having diameter 21cm is 1386 cm2

(iii) Radius(r) of sphere = 3.5/2 = 1.75 cm

Surface area of sphere = 4πr2

= 4×(22/7)×1.752 = 38.5

Surface area of a sphere is 38.5 cm2

3. Find the total surface area of a hemisphere of radius 10 cm. [Use π=3.14]

Solution:

Radius of hemisphere, r = 10cm

Formula: Total surface area of hemisphere = 3πr2

= 3×3.14×102 = 942

The total surface area of given hemisphere is 942 cm2.

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:

Let rand r2 be the radii of spherical balloon and spherical balloon when air is pumped into it respectively. So

r= 7cm

r= 14 cm

Now, Required ratio = (initial surface area)/(Surface area after pumping air into balloon)

= 4r12/4r22

= (r1/r2)2

= (7/14)= (1/2)2 = ¼

Therefore, the ratio between the surface areas is 1:4.

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 Rs 16 per 100 cm2. (Assume π = 22/7)

Solution:

Inner radius of hemispherical bowl, say r = diameter/2 = (10.5)/2 cm = 5.25 cm

Formula for Surface area of hemispherical bowl = 2πr2

= 2×(22/7)×(5.25)2 = 173.25

Surface area of hemispherical bowl is 173.25 cm2

Cost of tin-plating 100 cm2 area = Rs 16

Cost of tin-plating 1 cm2 area = Rs 16 /100

Cost of tin-plating 173.25 cmarea = Rs. (16×173.25)/100 = Rs 27.72

Therefore, the cost of tin-plating the inner side of the hemispherical bowl at the rate of Rs 16 per 100 cm2 is Rs 27.72.

6. Find the radius of a sphere whose surface area is 154 cm2. (Assume π = 22/7)

Solution:

Let the radius of the sphere be r.

Surface area of sphere = 154 (given)

Now,

4πr= 154

r= (154×7)/(4×22) = (49/4)

r = (7/2) = 3.5

The radius of the sphere is 3.5 cm.

7. The diameter of the moon is approximately one fourth of the diameter of the earth.

Find the ratio of their surface areas.

Solution:

If diameter of earth is said d, then the diameter of moon will be d/4 (as per given statement)

Radius of moon = ½×d/4 = d/8

Surface area of moon = 4π(d/8)2

Surface area of earth = 4π(d/2)2

The ratio between their surface areas is 1:16.

8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface of the bowl. (Assume π =22/7)

Solution:

Given:

Inner radius of hemispherical bowl = 5cm

Thickness of the bowl = 0.25 cm

Outer radius of hemispherical bowl = (5+0.25) cm = 5.25 cm

Formula for outer CSA of hemispherical bowl = 2πr2, where r is radius of hemisphere

= 2×(22/7)×(5.25)2 = 173.25

Therefore, the outer curved surface area of the bowl is 173.25 cm2.

9. A right circular cylinder just encloses a sphere of radius r (see fig. 13.22). Find

(i) surface area of the sphere,

(ii) curved surface area of the cylinder,

(iii) ratio of the areas obtained in(i) and (ii).

Solution:

(i) Surface area of sphere = 4πr2, where r is the radius of sphere

(ii) Height of cylinder, h = r+r =2r

CSA of cylinder formula = 2πrh = 2πr(2r) (using value of h)

= 4πr2

(iii) Ratio between areas = (Surface area of sphere)/CSA of Cylinder)

= 4r2/4r= 1/1

Ratio of the areas obtained in (i) and (ii) is 1:1.

#### CLASS 9 MATH NCERT SOLUTION FOR CHAPTER – 13 SURFACE AREAS AND VOLUMES EX – 13.5

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