Class 9 NCERT Solutions- Chapter 13 Surface Areas And Volumes - Exercise 13.4

In Class 9 Mathematics, Chapter 13 of the NCERT textbook focuses on Surface Areas and Volumes, a crucial topic for understanding the properties of three-dimensional shapes. Exercise 13.4 delves into the specific problems involving the calculation of the surface areas and volumes of various solids. This chapter is essential as it lays the foundation for more advanced geometry and helps in solving real-world problems involving the measurements of solids.

Surface Areas and Volumes

Surface areas and volumes are fundamental concepts in geometry that involve calculating the extent of a surface and the amount of space a three-dimensional object occupies respectively. Understanding these concepts is critical for practical applications such as construction, manufacturing, and various scientific fields. Exercise 13.4 includes problems that enhance students' skills in applying formulas for the surface areas and volumes of cubes, cuboids, cylinders, cones, and spheres.

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

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

(Assume π=22/7)

Solution:

We know that, Surface area of sphere (SA) = 4πr²

(i) Radius of sphere, r = 10.5 cm

SA = 4×(22/7)×10.5² = 1386

Therefore, Surface area of sphere is 1386 cm²

(ii) Radius of sphere, r = 5.6cm

SA = 4×(22/ 7)×5.6² = 394.24

Therefore, Surface area of sphere is 394.24 cm²

(iii) Radius of sphere, r = 14cm

SA = 4×(22/7)×(14)²

= 2464

Therefore, Surface area of sphere is 2464 cm²

Question 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

We know that, Surface area of sphere = 4πr²= 4×(22/7)×7² = 616

Surface area of a sphere is 616 cm²

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

We know that, Surface area of sphere = 4πr²

= 4×(22/7)×10.5² = 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

We know that, Surface area of sphere = 4πr²

= 4×(22/7)×1.75² = 38.5

Surface area of a sphere is 38.5 cm²

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

Solution:

Given, Radius of hemisphere, r = 10cm

Formula: Total surface area of hemisphere = 3πr²

= 3×3.14×10² = 942

Therefore, total surface area of given hemisphere is 942 cm².

Question 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:

We assume that r1 and r2 be the radii of spherical balloon and spherical balloon when air is pumped into it respectively. So,

r1 = 7cm

r2 = 14 cm

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

= 4(r1)²/4(r2)²

= (r1/r2)²

= (7/14)² = (1/2)² = ¼

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

Question 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:

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

We know, Formula for Surface area of hemispherical bowl = 2πr²

= 2×(22/7)×(5.25)² = 173.25cm²

Cost of tin-plating 100 cm² area = Rs 16

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

Cost of tin-plating 173.25 cm² area = 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.

Question 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)

So,

4πr² = 154

r² = (154×7)/(4×22) = (49/4)

r = (7/2) = 3.5cm

Therefore, the radius of the sphere is 3.5 cm.

Question 7. The diameter of the moon is approximately one-fourth of the diameter of the earth. Find the ratio of their surface areas.

Solution:

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

Radius of earth = d/2

So, 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.

Question 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

We know, 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.

Question 9. A right circular cylinder just encloses a sphere of radius r (see fig.). 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) As Height of cylinder, h = r+r =2r

And Radius of cylinder = r

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

= 4πr²

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

= 4r²/4r² = 1/1

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

Read More:

• Surface Areas and Volumes
• Practice Problems on Surface Area and Volume Formulas