# Surface Areas and Volumes – Exercise 13.4 – (MATHEMATICS) – 9th Class

**Exercise 31.4**

Assume unless stated otherwise.

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

(i) 10.5 cm

(ii) 5.6 cm

(iii) 14 cm.

### Show Answer

**Sol.**

(i) Radius of a sphere (r) = 10.5 cm

∴ Surface area of the sphere = 4πr²

= 1386 cm²

(ii) Radius of a sphere (r) = 5.6 cm

∴ Surface area of the sphere = 4πr2

= 394.24 cm²

(iii) Radius of a sphere (r) = 14 cm

Surface area of the sphere = 4πr2

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

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

### Show Answer

**Sol.** (i) Diameter (d) = 14 cm

Radius (r)

Now, surface area of the sphere = 4πr^{2}

(ii) Diameter (d) = 21 cm

∴ Radius (r) =

Now, surface area of the sphere = 4nr2

(iii) Diameter (d) = 3.5 m

∴ Radius (r) =

Now, surface area of the sphere = 4nr2

**3. Find the total surface area of a hemisphere of radius 10 cm.**

(Use π = 3.14).

### Show Answer

**Sol.** Total surface area of hemisphere = 3πr²

= 3 × 3.14 × 10 × 10 cm²

= 942 cm².

**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.**

### Show Answer

**Sol.** S_{1} = Surface area with radius 7 cm = 4π(7)² cm².

S_{2} = Surface area with radius 14 cm = 4π(14)^{2} cm^{2}.

∴

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

### Show Answer

**Sol.** Inner diameter = 10.5 cm ⇒ Inner radius = 5.25 cm.

Inner surface area =

Cost of tin-plating = Rs.

**6. Find the radius of a sphere whose surface area is 154 cm².**

### Show Answer

**Sol.** Let r be the radius of the sphere.

Surface area of the sphere = 154 cm²

⇒

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

### Show Answer

**Sol.** Let diameter of the earth = *d* units

Radius of the earth = units

∴ Diameter of the moon = units

⇒ Radius of the moon = units.

∴

∴ Surface area of the moon : surface area of the earth

= 1 : 16.

**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.**

### Show Answer

**Sol.** Inner radius = 5 cm.

∴ Outer radius = (5 + 0.25) cm = 5.25 cm.

∴ Outer curved surface area =

= 173.25 cm^{2}

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

### Show Answer

**Sol.**(i) Surface area of the sphere = 4πr²

(ii) For cylinder: radius of base = r, height = 2r.

∴ Curved surface area of the cylinder = 2π(r)(2r)

= 4πr^{2}

(iii) Ratio of the required areas is 1 : 1.