#### Class nine Maths NCERT Solu Surface Areas Volumes

## Chapter 13: Surface Areas and Volumes

**TEXTBOOK’S
EXERCISE – 13.4 **

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

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

**Sol.
**(i) r = 10.5 cm

Surface area of the sphere

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

= 4 × 22/7
× 10.5 × 10.5 cm^{2} = 1386 cm^{2}

(ii) r = 5.6 cm

Surface area of the sphere

= 4πr^{2} = 4 × 22/7 × (5.6)^{2} cm^{2}

= 4 × 22 7
× 5.6 × 5.6 cm^{2}

= 394.24 cm^{2}

(iii)
*r *= 14 cm

Surface area of the
sphere = 4π*r*^{2}

= 4 × 22/7 × 14 × 14 cm^{2} = 88 × 28 cm^{2}

= 2464 cm2

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

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

**Sol.
**(i) r = 14/2
cm = 7 cm

Surface area of the sphere

= 4πr2 = 4 × 22/7 ×
72 cm^{2}

= 4 × 22/7
× 7 × 7 cm^{2}

=
88 × 7 cm^{2} = 616 cm^{2 }

(ii) r = 21/2 cm = 10.5 cm

Surface area of the sphere

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

= 4 × 22 7
× 10.5 × 10.5 cm^{2} = 1386 cm^{2 }

(iii) r = 3.5/2 m = 1.75 m

Surface area of the sphere

= 4πr^{2} =
4 × 22/7 × (1.75)^{2} m^{2}

= 4 × 22/7 × 1.75 × 1.75 m^{2} = 38.5 m^{2}

**3. Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)**

**Sol.
**r = 10 cm

Total surface area of the hemisphere = 3πr2

= 3 × 3.14 × (10)^{2}
cm^{2}

= 3 × 3.14 × 100 cm^{2}
= 942 cm^{2}

**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. **

**Sol.
When r = 7 cm **

The surface area of the balloon

= 4πr^{2} =
4 × π × 7 × 7 cm^{2}

**When
r = 14 cm **

The surface area of the balloon

= 4πr^{2} =
4 × π × 14 × 14 cm^{2 }

The required ratio of the surface areas of the balloon