Haryana State Board HBSE 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3 Textbook Exercise Questions and Answers.

## Haryana Board 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Exercise 13.3

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

Question 1.

Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.

Solution:

We have,

Slant height of the cone (l) = 10 cm

and diameter of the base of a cone = 10.5 cm

∴ Radius of the base of a cone (r) = \(\frac{22}{7}\)

= 5.25 cm

∴ Curved surface area of the cone = πrl

= \(\frac{22}{7}\) × 5.25 × 10

= 165 cm^{2}

Hence, curved surface area of the cone = 165 cm^{2}

Question 2.

Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.

Solution:

We have,

Slant height of a cone (l) = 21 m

and diameter of base of a cone = 24 m

∴ Radius of base of a cone (r) = \(\frac{24}{2}\) = 12 m

∴ Total surface area of a cone (r) = πr(l + r)

= \(\frac{22}{7}\) × 12(21 + 12)

= \(\frac{22}{7}\) × 12 × 33

= 1244.57 m^{2} (approx.).

Hence, total surface area of a cone = 1244.57 m^{2} (approx.).

Question 3.

Curved surface area of a cone is 308 cm^{2} and its slant height is 14 cm. Find : (i) radius of the base and (ii) total surface area of the cone.

Solution:

We have,

Slant height of a cone (l) = 14 cm

Let radius of the base of a cone be r сm.

(i) curved surface area of a cone = 308 cm^{2}

⇒ πrl = 308

⇒ \(\frac{22}{7}\) × r × 14 = 308

⇒ 44r = 308

⇒ r = \(\frac{308}{44}\)

⇒ r = 7 cm.

(ii) Total surface area of the cone = πr(l + r)

= \(\frac{22}{7}\) × 7(14 + 7)

= 22 × 21

= 462 cm^{2}

Hence, (i) radius of the base of a cone = 7 cm

(ii) total surface area of a cone = 462 cm^{2}.

Question 4.

A conical tent is 10 m high and the radius of its base is 24 m. Find :

(i) slant height of the tent.

(ii) cost of the canvas required to make the tent, if the cost of 1 m^{2} canvas is Rs. 70.

Solution:

We have,

Height of the conical tent (h) = 10 m

and radius of the base of conical tent (r) = 24 m

(i) Let slant height of the cone be l m.

∴ l = \(\sqrt{h^2+r^2}\)

⇒ l = \(\sqrt{10^2+24^2}\)

⇒ l = \(\sqrt{100+576}\)

⇒ l = \(\sqrt{676}\)

⇒ l = 26 m

(ii) Required area of canvas to make the conical tent = Curved surface area of the cone

= πrl = \(\frac{22}{7}\) × 24 × 26

= \(\frac{13728}{7}\) m^{2}

∵ 1 m^{2} canvas cost = Rs. 70

∴ \(\frac{13728}{7}\) m^{2} canvas cost = Rs. \(\frac{70 \times 13728}{7}\)

= Rs. 137280

Hence, (i) Slant height of the tent = 26 m

(ii) Cost of canvas = Rs. 137280.

Question 5.

What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (Use π = 3.14)

Solution:

We have, Height of the conical tent (h) = 8 m

Radius of base of conical tent (r) = 6 m

∴ l = \(\sqrt{h^2+r^2}\)

⇒ l = \(\sqrt{8^2+6^2}\)

⇒ l = \(\sqrt{64+36}\)

⇒ l = \(\sqrt{100}\)

⇒ l = 10 m

Area of tarpaulin required to make the conical tent = Curved surface area of the cone

= πrl = 3.14 × 6 × 10

= 188.4 m^{2}

∵ Width of tarpaulin = 3 m.

∴ Length of tarpaulin = \(\frac{\text { Area of tarpaulin }}{\text { Width of tarpaulin }}\)

\(\frac{188\cdot4}{3}\) = 62.8 m

Wastage in cutting = 20 cm = 0.20 m

Total length of tarpaulin = 62.8 + 0.20

= 63 m

Hence,length of tarpualin = 63 m.

Question 6.

The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of whitewashing its curved surface at the rate of Rs. 210 per 100 m^{2}

Solution:

We have,

Slant height of a conical tomb (l) = 25 m

and diameter of a conical tomb = 14 m

∴ Radius of a conical tomb (r) = \(\frac{14}{2}\) = 7m

∴ Curved surface area of the conical tomb = πrl

= \(\frac{22}{7}\) × 7 × 25

= 550 m^{2}

∵ Cost of white washing of 100 m^{2} of tomb

= Rs. 210

∴ Cost of white washing of 550 m^{2} of tomb

= Rs \(\frac{210 \times 550}{100}\)

= Rs. 1155

Hence,cost of white washing of tomb = Rs. 1155.

Question 7.

A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

Solution:

We have,

Radius of conical cap (r) = 7 cm

and height of the conical cap (h) = 24 cm

∴ l = \(\sqrt{h^2+r^2}\)

⇒ l = \(\sqrt{24^2+7^2}\)

⇒ l = \(\sqrt{576+49}\)

⇒ l = \(\sqrt{625}\)

⇒ l = ±25

⇒ l = 25

∴ Curved surface area of a cap = πrl

= \(\frac{22}{7}\) × 7 × 25 = 550 cm^{2}

Curved surface area of 10 such caps = 550 × 10 = 5500 cm^{2}

Hence, required area of sheet to make 10 such caps = 5500 cm^{2}.

Question 8.

A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard, Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cone is to be painted and the cost of painting is Rs. 12 per m^{2}, what will be the cost of painting all these cones? (Use π = 3.14 and take \(\sqrt{1.04}\) = 1.02)

Solution:

We have,

Height of a cone (h) = 1 m = 100 cm

and diameter of a cone (d) = 40 cm

∴ Radius of a cone (r) = \(\frac{40}{2}\) = 20 cm

∴ l = \(\sqrt{h^2+r^2}\)

⇒ l = \(\sqrt{100^2+20^2}\)

⇒ l = \(\sqrt{10000+400}\)

⇒ l = \(\sqrt{10400}\)

⇒ l = 101.98

⇒ l = 102 (approx.)

Outer surface area of 1 cone = πrl = 3.14 × 20 × 102 = 6405.6 cm^{2}

Outer surface area of 50 cones= 6405.6 × 50 = 320280 cm^{2}

\(\frac{320280}{10000}\) m^{2}

= 32.028 m^{2}

Cost of painting of 1 m^{2} = Rs. 12

Cost of painting of 32.028 m^{2} = Rs. 12 × 32.028

= Rs. 384.336

= Rs. 384.34

Hence, cost of painting of 50 cones = Rs. 384.34.