The document Ex 13.4 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev is a part of the Class 9 Course Class 9 Mathematics by Full Circle.

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**Question 1. Find the surface area of a sphere of radius: (i) 10.5 cm (ii) 5.6 cm (iii) 14 cm Solution: **(i) Here r = 10.5 cm

âˆ´ Surface area of the sphere = 4Ï€r

= 4 x (22/7) x (10.5)^{2} cm^{2}

= 4 x (22/7) x (105/10) x (105/10) cm^{2}

= 22 x 3 x 21 cm^{2} = 1386 cm^{2}

(ii) Here r = 5.6 cm

(iii) Here, r = 14 cm

âˆ´ Surface area = 4Ï€r^{2} = 4 x (22/7) (14)^{2} cm^{2}

= 4 x (22/7) x 14 x 14 cm^{2}

= 4 x 22 x 2 x 14 cm^{2} = 2464 cm^{2}

**Question 2. Find the surface area of a sphere of diameter: (i) 14 cm (ii) 21 cm (iii) 3.5 m Solution:** (i) Here, Diameter = 14 cm

â‡’ Radius = (14/2)= 7 cm

âˆ´ Surface area = 4Ï€r

= 4 x (22/7) x 7 x 7 cm

(ii) Here, Diameter = 21 cm â‡’ r = (21/2) cm

âˆ´ Surface area = 4Ï€r^{2} = 4 x (22/7) (21/2)^{2} cm^{2}

= 4 x (22/7) x (21/2) x (21/2) cm^{2}

= 22 x 3 x 21 cm^{2} = 1386 cm^{2}

(iii) Here, Diameter = 3.5 m

**Question 3. Find the total surface area of a hemisphere of radius 10 cm. (Use Ï€ = 3.14) Solution:** Here, radius (r) = 10 cm

âˆ´

Plane surface area of the hemisphere = Ï€r^{2 }= 3.14 x (10)^{2} cm^{2}

= (314/100) x 10 x 10 cm^{2}

= 314 cm^{2}

âˆ´ Total surface area = 628 cm^{2} + 314 cm^{2} = 942 cm^{2}

**REMEMBER**

(i) Curved surface area of a hemisphere = 2Ï€r^{2}

(ii) Total surface area of a hemisphere = 3Ï€r^{2 }

**Question 4. The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases. Solution:** Case I: When radius (r

= 4 x 22 x 14 x 2 cm^{2} = 2464 cm^{2}

The required ratio

â‡’ The required ratio = 1 : 4

Alternate method

**Question 5. A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tinplating it on the inside at the rate of **â‚¹ **16 per 100 cm ^{2}. Solution**: Inner diameter of the hemisphere = 10.5 m

âˆ´ Radius (r) = 10.5/2 cm = 105/20 cm

âˆµ Curved surface area of a hemisphere = 2Ï€r^{2}

âˆ´ Inner curved surface area of hemispherical bowl

Cost of tinplating:

**Question 6. Find the radius of a sphere whose surface area is 154 cm ^{2}. Solution:** Let the radius of the sphere be â€˜râ€™ cm.

âˆ´ Surface area = 4Ï€r^{2}

â‡’ 4Ï€r^{2 }= 154

â‡’ 4 x 22/7 x r^{2} = 154

Thus, the required radius of the sphere is 3.5 cm

** Question 7. The diameter of the moon is apÏ€roximately one-fourth of the diameter of the earth. Find the ratio of their surface areas. Solution:** Let the radius of the earth = r

âˆ´ Radius of the moon = (r/4)

âˆµ Surface area of a sphere = 4Ï€r

Since, the earth as well as the moon are considered to be spheres.

âˆ´ Surface area of the earth = 4Ï€r

or [Surface area of the moon] : [Surface area of the earth] = 1 : 16

Thus, the required ratio = 1 : 16

**Question 8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl. Solution: **Inner radius (r) = 5cm

Thickness = 0.25 cm

âˆ´ Outer radius (R) = [5.00 + 0.25] cm = 5.25 cm

âˆ´ Outer surface area of the bowl = 2Ï€r

**Question 9. A right circular cylinder just encloses a sphere of radius r (see the figure) 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) For the sphere: Radius = r

âˆ´ Surface area of the sphere = 4 Ï€r^{2}

(ii) For the right circular cylinder:

âˆµ Radius of the cylinder = Radius of the sphere

âˆ´ Radius of the cylinder = r

Height of the cylinder = Diameter of the sphere

â‡’ Height of the cylinder (h) = 2r

Since, curved surface area of a cylinder = 2Ï€rh = 2Ï€r(2r) = 4Ï€r^{2}

**VOLUME OF A CUBOID**

All solid bodies occupy space. The amount of space occupied by a solid is called its volume. The units of volume are cubic centimetres (written as cm^{3}) or cubic metres (written as m^{3}).

Note: If an object is hollow then its interior is empty, and can be filled with fluid (air or some liquid) which will take the shape of the container. The volume of the substance that can fill the interior is called the capacity of the container.

**REMEMBER**

I . Volume of cuboid = [Base area] x [height]

II. Volume of the cube = Edge x Edge x Edge = (side)^{3}

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