Haryana State Board HBSE 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4 Textbook Exercise Questions and Answers.

## Haryana Board 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Exercise 13.4

Assume π = \(\frac{22}{7}\), unless stated otherwise.

Question 1.

Find the surface area of a sphere of radius:

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

Solution:

(i) We have,

Radius of sphere (r) = 10.5 cm

∴ Surface area of a sphere = 4πr^{2}

= 4 × \(\frac{22}{7}\) × (10.5)^{2}

= \(\frac{9702}{7}\) cm^{2} Ans.

Hence,surface area of a sphere = 1386 cm^{2}.

(ii) We have,

Radius of sphere (r) = 5.6 cm^{2}

∴ Surface area of a sphere = 4πr^{2}

= 4 × \(\frac{22}{7}\) × (5.6)^{2}

= \(\frac{2759.68}{7}\)

= 394.24 cm^{2}

Hence,surface area of a sphere

= 394.24 cm^{2}.

(iii) We have,

Radius of sphere (r) = 14 cm

∴ Surface area of a sphere = 4πr^{2}

= 4 × \(\frac{22}{7}\) × (14)^{2}

\(\frac{17248}{7}\) = 2464 cm^{2}

Hence, surface area of a sphere = 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) We have,

Diameter of a sphere (d) = 14 cm

∴ Radius of a sphere (r) = \(\frac{14}{2}\) = 7 cm

∴ Surface area of a sphere = 4πr^{2}

= 4 × \(\frac{22}{7}\) × (7)^{2}

= \(\frac{4312}{7}\)

= 616 cm^{2}

Hence,surface area of a sphere = 616 cm^{2}.

(ii) We have,

Diameter of a sphere (d) = 21 cm

∴ Radius of a sphere (r) = \(\frac{21}{2}\) = 10.5 cm

∴ Surface area of a sphere = 4πr^{2}

= 4 × \(\frac{22}{7}\) × (10.5)^{2}

= \(\frac{9702}{7}\) cm^{2} = 1386 m^{2}

Hence,surface area of a sphere = 1386 cm^{2}.

(iii) We have,

Diameter of a sphere (d) = 3.5 m

∴ Radius of a sphere (r) = \(\frac{3.5}{2}\) = 1.75 m

∴ Surface area of a sphere = 4πr^{2}

= 4 × \(\frac{22}{7}\) × (1.75)^{2}

= \(\frac{269.5}{7}\)

= 38.5 m^{2}

Hence,surface area of a sphere= 38.5 m^{2}.

Question 3.

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

Solution:

We have, Radius of a hemisphere (r) = 10 cm

∴ Total surface area of a hemisphere= 3πr^{2}

= 3 × 3.14 × (10)^{2}

= 942 cm^{2}

Hence, total surface area of a hemisphere = 942 cm^{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:

We have,

Radius of a spherical balloon (r_{1}) = 7 cm

∴ Surface area of a spherical balloon (S_{1}) = 4π(r_{1})^{2} = 4 × \(\frac{22}{7}\) × (7)^{2}

= 616 cm^{2}

and increased radius of a spherical ballon (r_{2})= 14 cm

∴ Surface area of a spherical ballon in this case (S_{2}) = 4π(r_{1})^{2}

= 4 × \(\frac{22}{7}\) × (14)^{2}

= 2464 cm^{2}

Now,

\(\frac{S_1}{S_2}=\frac{616}{2464}\)

⇒ \(\frac{S_1}{S_2}=\frac{1}{4}\)

⇒ S_{1} : S_{2} = 1 : 4

Hence, required ratio of the balloon in two cases = 1 : 4.

Question 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 cm^{2}.

Solution:

We have,

Diameter of hemispherical bowl

(d) = 10.5 cm

∴ Radius of the hemispherical bowl (r) = \(\frac{10.5}{2}\) = 5.25 cm

∴ Inner curved surface area of the bowl = 2πr^{2}

= 2 × \(\frac{22}{7}\) × (5.25)^{2}

= \(\frac{44 \times 27.5625}{7}\)

= 173.25 cm^{2}

Rate of tin-plating = Rs. 16 per 100 cm^{2}

∴ Cost of tin-plating inside of bowl = \(\frac{173.25 \times 16}{100}\)

= Rs. 27.72

Hence, cost of tin-plating the inside of a bowl = Rs. 27.72.

Question 6.

Find the radius of a sphere whose surface area is 154 cm^{2}.

Solution:

Let the radius of a sphere be r cm.

Surface area of a sphere = 154 cm^{2}, (given)

⇒ 4πr^{2} = 154

⇒ 4 × \(\frac{22}{7}\) × r^{2} = 154

⇒ \(\frac{88}{7}\) × r^{2} = 154

⇒ r^{2} = \(\frac{154 \times 7}{88}\)

⇒ r^{2} = \(\frac{7 \times 7}{4}\)

⇒ r = \(\sqrt{\frac{7 \times 7}{2 \times 2}}\)

⇒ r = \(\frac{7}{2}\) = 3.5 cm

Hence, radius of the sphere = 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 diameter of the Earth be 2x.

∴ Radius of the Earth (R) = x

∴ Surface area of the Earth (S_{1}) = 4πR^{2} = 4πx^{2}

According to question,

Diameter of the Moon = \(\frac{1}{4}\) × diameter of the Earth

= \(\frac{1}{4}\) × 2x = \(\frac{x}{4}\)

∴ Radius of the Moon (r) = \(\frac{\frac{x}{2}}{2}=\frac{x}{4}\)

∴ Surface area of the Moon (S_{2}) = 4πr^{2}

Hence, required ratio of the their surface areas = 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:

We have,

Inner radius of the bowl = 5 cm

and thickness of the bowl = 0.25 cm

Fig. 13.34

∴ Outer radius of the bowl (r) = 5 + 0.25

= 5.25 cm

∴ Outer curved surface area of the bowl = 2πr^{2}

= 2 × \(\frac{22}{7}\) × (5.25)^{2}

= \(\frac{44 \times 27.5625}{7}\)

= 44 × 3.9375

= 173.25 cm^{2}

Hence, outer curved surface area of the bowl = 173.25 cm^{2}.

Question 9.

A right circular cylinder just encloses a sphere of radius r (see Fig. 13.35). Find :

(i) surface area of the sphere, (ii) curved surface area of the cylinder, (iii) ratio of the areas obtained in (i) and (ii).

Fig. 13.35

Solution:

We have,

(i) Radius of sphere = r

∴ Surface area of the sphere (S_{1}) = 4πr^{2}.

(ii) Radius of the cylinder = radius of the sphere = r.

Height of the cylinder (h) = 2 × r = 2r

∴ Curved surface area of the cylinder

(S_{2}) = 2πrh = 2π × r × 2r

= 4πr^{2}

Hence, (i) surface area of sphere = 4πr^{2}

(ii) curved surface area of the cylinder = 4πr^{2}

(iii) required ratio of the areas obtained in (i) and (ii) = 1 : 1.