The document Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.

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**Question 1. The curved surface area of a right circular cylinder of height 14 cm is 88 cm ^{2}. Find the diameter of the base of the cylinder. Solution:** Let â€˜r â€™ be the radius of the cylinder.

Here, height (h) = 14 cm and curved surface area = 88 cm

âˆµ Curved surface area of a cylinder = 2Ï€rh

âˆ´ 2 Ï€rh = 88

â‡’ 2 x (22/7)r x 14 = 88

â‡’

â‡’ Diameter = 2 x r = 2 x 1 cm = 2 cm

** Question 2. It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same? Solution: **Here, height (h) = 1 m

âˆµ Diameter of the base = 140 cm = 1.40 m

âˆ´ Radius (r) = (1.40/2) = 0.70 m

Total surface area of the cylinder = 2Ï€r (h + r)

= 2 x(22/7)x 0.70 (1 + 0.70) m

= 2 x 22 x 0.10 (1.70) m

Hence, the required sheet is 7.48 m^{2 }

**Question 3. A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see figure) Find its. (i) inner curved surface area, (ii) outer curved surface area, (iii) total surface area. Solution: **Length of the metal pipe = 77 cm

âˆµ It is in the form of a cylinder.

âˆ´ Height (h) of the cylinder = 77 cm Inner diameter = 4 cm

â‡’ Inner radius (r) =(4/2)= 2 cm

Outer diameter = 4.4 cm

â‡’ Outer radius (R) = (4.4/2)= 2.2 cm

(i) âˆµ Curved surface area = 2Ï€rh = 2 x (22/7) x 2 x 77 cm^{2}

^{ }= 2 x 22 x 2 11 cm^{2} = 968 cm^{2}

(ii) Outer curved surface area = 2Ï€rh ^{ }

(iii) Total surface area

= [Inner curved surface] + [outer curved surface area] + [Two base circular lamina]

**Question 4. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m ^{2}. Solution:** The roller is in the form of a cylinder diameter of the roller = 84 cm

â‡’ Radius of the cylinder =(84/2)

= 42 cm

Length of the roller = 120 cm

â‡’ Height of the cylinder (h) = 120 cm

âˆ´ Curved surface area of the roller (cylinder) = 2 Ï€rh

= 2 x (22/7)x 42 x 120 cm

= 2 x 22 x 6 x 120 cm

âˆ´ Area of the playground levelled in one revolution of the roller = 31680 cm

= (31680 / 10000) m^{2}

â‡’ Area levelled in 500 revolutions = 500 x (31680 / 10000) m^{2}

**Question 5. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of **â‚¹ **12.50 per m ^{2}. Solution:** Diameter of the pillar = 50 cm

â‡’ Radius (r) = (50/2) = 25 cm = (1/4)m

Height (h) = 3.5 m = 3.50 m

âˆµ Curved surface area of a cylinder = 2Ï€rh

âˆ´ Curved surface area of a cylinder = 2 x (22/7) x(1/4)x 3.50 m

â‡’ Curved surface area to be painted = (11/2)m^{2}

Cost of painting âˆµ Rate of painting = â‚¹ 1250 per m^{2}

âˆ´ Cost of Painting 11/2 m^{2}

**Question 6. Curved surface area of a right circular cylinder is 4.4 m ^{2}. If the radius of the base of the cylinder is 0.7 m, find its height. Solution:** Radius (r) = 0.7 m Let height of the cylinder be â€˜hâ€™ metres.

âˆ´ Curved surface area = 2Ï€rh

But the curved surface area is 4.4 m^{2}.

Thus, the required height is 1 metre.

**Question 7. The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find: (i) its inner curved surface area. (ii) the cost of plastering this curved surface at the rate of ****40 per m ^{2}. Solution: **Inner diameter of the well = 3.5 m

â‡’ Radius of the well (cylinder) (r) =(3.5/2)

.

Depth of the well (= height of the cylinder) h = 10 m

âˆ´ (i) Inner curved surface area = 2Ï€rh

(ii) Cost of plastering

âˆµ Rate of plastering = â‚¹40 per m^{2}

âˆ´ Total cost of plastering 110 m^{2} = â‚¹110 x 40 = â‚¹4400

**Question 8. In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system. Solution: **Length of the cylindrical pipe = 28 m

â‡’ h = 28 m

Diameter of the pipe = 5 cm

âˆ´ Radius (r) = (5/2) cm = (5/200) m

âˆ´ Curved surface area = 2Ï€rh

= (440/100)m^{2} = 4.40 m^{2}

Thus, the total radiating surface is 4.40 m^{2}

**Question 9. Find: (i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high. (ii) how much steel was actually used if (1/12) of the steel actually used was wasted in making the tank. Solution:** The storage tank is in the form of a cylinder, and diameter of the tank = 4.2 m

âˆ´ The radius (r) = (42/2) = 2.1 m

Height (h) = 4.5 m

Now, (i) Lateral (or curved) surface area = 2Ï€rh

= 2 x (22/7) x 2.1 x 4.5 m^{2}

= 2 x 22 x 0.3 x 4.5 m^{2}

= 59.4 m^{2}

(ii) Total surface area of the tank = 2Ï€r(r + h)

= 2 x (22/7) x 2.1 (2.1 + 4.5) m^{2}

= 44 x 0.3 x 6.6 m^{2 }= 87.12 m^{2 }

Let actual area of the steel used = x m^{2}

steel was wasted.

âˆ´ Area of steel that wasted = (1/12) x x

= (x/12)m^{2}

â‡’ Area of steel used

â‡’

Thus, the required area of the steel that was actually used is 95.04 m^{2}

**Question 10. In the figure you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade. Solution:** The lampshade is in the form of a cylinder, where

r= (20/2)= 10 cm Height = 30 cm.

âˆµ A margin of 2.5 cm is to be added to top and bottom.

âˆ´ Total height of the cylinder h = 30 cm + 2.5 cm + 2.5 cm = 35 cm

Now, curved surface area = 2 Ï€rh = 2 x (22/7)x 10 x (35) cm^{2 }

= 2 x 22 x 10 x 5 cm^{2 }

= 2200 cm^{2}

Thus, the required area of the cloth = 2200 cm^{2}

**Question 11. The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition? Solution: **Here, the penholders are in the form of cylinders Radius of a cylinder (r) = 3 cm

Height of a cylinder (h) = 10.5 cm

Since, a penholder must be open from the top.

âˆ´ Surface area of a penholder (cylinder) = [Lateral surface area] + [Base area]

â‡’ Surface area of 35 penholders =

[âˆµ There are 35 competitors]

Thus, 7920 cm^{2} of cardboard was required to be bought.

**SURFACE AREA OF A RIGHT CIRCULAR CONE**

If we rotate a right angled triangular lamina AOB about OA, a cone is generated, as shown in the adjoining figure. OA is its height (h) and OB is a radius of its base (r) such that OA âŠ¥ OB. Its base is a circle with centre O. The length AB = l is called its slant height.

We have:

(i) l (slant height) =

(ii) Curved (= Lateral surface) area = Ï€rl

(iii) Total surface area = Lateral surface area + Base area = Ï€rl + Ï€r^{2} = Ï€r(l + r)

**Note:** In the following Ï€roblems assume Ï€ = (22/7)unless stated otherwise.

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