### NCERT Solutions For Class 9 Math Chapter – 13 Exercise – 13.2

__NCERT Solutions For Class 9 Math Chapter – 13 Exercise – 13.2__

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**Q1. 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:-**

** **

Curved surface area of cylinder = 2πrh

= 2πrh = 88

= 2 x 22/7 x r x 14 = 88

R = 88 x 7/2 x 22 x 14

R = 1 cm

Diameter = 2 x r

= (2 x 1 ) cm

= 2 cm

**Q2. It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a meta sheet How man square meters of the sheet are required for the same?**

**Solution:-**

** **

Total

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

= 2 x 22 x 0.10 x 1.70 m^{2}

= 7.48 m^{2}.

Hence the required sheet = 7.48 m^{2}.

**Q3. A metal pipe is 77 cm long The inner ft diameter of a cross section is 4 cm, the**

**outer diameter being 4.4cm. (see fig. 13.11). Find its**

**(i) inner curved surface area,**

**(ii) outer curved surface area**

**(iii) total surface area**

**(Assume π=22/7)**

**Solution:**

Let r_{1} and r_{2} Inner and outer radii of cylindrical pipe

r_{1 }= 4/2 cm = 2 cm

r_{2 }= 4.4/2 cm = 2.2 cm

Height of cylindrical pipe, h = length of cylindrical pipe = 77 cm

(i) curved surface area of outer surface of pipe = 2πr_{1}h

= 2×(22/7)×2×77 cm^{2}

= 968 cm^{2}

(ii) curved surface area of outer surface of pipe = 2πr_{2}h

= 2×(22/7)×2.2×77 cm^{2}

= (22×22×2.2) cm^{2}

= 1064.8 cm^{2}

(iii) Total surface area of pipe = inner curved surface area+ outer curved surface area+ Area of both circular ends of pipe.

= 2r_{1}h+2r_{2}h+2π(r_{1}^{2}-r_{2}^{2})

= 9668+1064.8+2×(22/7)×(2.2^{2}-2^{2})

= 2031.8+5.28

= 2038.08 cm^{2}

Therefore, the total surface area of the cylindrical pipe is 2038.08 cm^{2}.

**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}? (Assume π = 22/7)**

**Solution:**

A roller is shaped like a cylinder.

Let h be the height of the roller and r be the radius.

h = Length of roller = 120 cm

Radius of the circular end of roller = r = (84/2) cm = 42 cm

Now, CSA of roller = 2πrh

= 2×(22/7)×42×120

= 31680 cm^{2}

Area of field = 500×CSA of roller

= (500×31680) cm^{2}

= 15840000 cm^{2}

= 1584 m^{2}.

Therefore, area of playground is 1584 m^{2}. Answer!

**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 Rs. 12.50 per m ^{2}.**

**(Assume π = 22/7)**

**Solution:**

Let h be the height of a cylindrical pillar and r be the radius.

Given:

Height cylindrical pillar = h = 3.5 m

Radius of the circular end of pillar = r = diameter/2 = 50/2 = 25cm = 0.25m

CSA of pillar = 2πrh

= 2×(22/7)×0.25×3.5

= 5.5 m^{2}

Cost of painting 1 m^{2} area = Rs. 12.50

Cost of painting 5.5 m^{2} area = Rs (5.5×12.50)

= Rs.68.75

Therefore, the cost of painting the curved surface of the pillar at the rate of Rs. 12.50 per m^{2} is Rs 68.75.

**6. Curved surface area of a right circular cylinder is 4.4 m ^{2}. If the radius of the base of the base of the cylinder is 0.7 m, find its height. (Assume π = 22/7)**

**Solution:**

Let h be the height of the circular cylinder and r be the radius.

Radius of the base of cylinder, r = 0.7m

CSA of cylinder = 2πrh

CSA of cylinder = 4.4m^{2}

Equating both the equations, we have

2×(22/7)×0.7×h = 4.4

Or h = 1

Therefore, the height of the cylinder is 1 m.

**7. The inner diameter of a circular well is 3.5m. It is 10m deep. Find**

**(i) its inner curved surface area,**

**(ii) the cost of plastering this curved surface at the rate of Rs. 40 per m ^{2}.**

**(Assume π = 22/7)**

**Solution:**

Inner radius of circular well, r = 3.5/2m = 1.75m

Depth of circular well, say h = 10m

(i) Inner curved surface area = 2πrh

= (2×(22/7 )×1.75×10)

= 110

Therefore, the inner curved surface area of the circular well is 110 m^{2}.

(ii)Cost of plastering 1 m^{2} area = Rs.40

Cost of plastering 110 m^{2} area = Rs (110×40)

= Rs.4400

Therefore, the cost of plastering the curved surface of the well is Rs. 4400.

**8. In a hot water heating system, there is cylindrical pipe of length 28 m and diameter 5 cm. Find**

**the total radiating surface in the system. (Assume π = 22/7)**

**Solution:**

Height of cylindrical pipe = Length of cylindrical pipe = 28m

Radius of circular end of pipe = diameter/ 2 = 5/2 cm = 2.5cm = 0.025m

Now, CSA of cylindrical pipe = 2πrh, where r = radius and h = height of the cylinder

= 2×(22/7)×0.025×28 m^{2}

= 4.4m^{2}

The area of the radiating surface of the system is 4.4m^{2}.

**9. Find**

**(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in**

**diameter and 4.5m high.**

**(ii) How much steel was actually used, if 1/12 of the steel actually used was wasted in making the tank. (Assume π = 22/7)**

**Solution:**

Height of cylindrical tank, h = 4.5m

Radius of the circular end , r = (4.2/2)m = 2.1m

(i) the lateral or curved surface area of cylindrical tank is 2πrh

= 2×(22/7)×2.1×4.5 m^{2}

= (44×0.3×4.5) m^{2}

= 59.4 m^{2}

Therefore, CSA of tank is 59.4 m^{2}.

(ii) Total surface area of tank = 2πr(r+h)

= 2×(22/7)×2.1×(2.1+4.5)

= 44×0.3×6.6

= 87.12 m^{2}

Now, Let S m^{2} steel sheet be actually used in making the tank.

S(1 -1/12) = 87.12 m^{2}

This implies, S = 95.04 m^{2}

Therefore, 95.04m^{2 }steel was used in actual while making such a tank.

**10. In fig. 13.12, you see the frame of a lampshade. It is to be covered with a decorative cloth.**

**The frame has a 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. (Assume π = 22/7)**

**Solution:**

Say h = height of the frame of lampshade, looks like cylindrical shape

r = radius

Total height is h = (2.5+30+2.5) cm = 35cm and

r = (20/2) cm = 10cm

Use curved surface area formula to find the cloth required for covering the lampshade which is 2πrh

= (2×(22/7)×10×35) cm^{2}

= 2200 cm^{2}

Hence, 2200 cm^{2} cloth is required for covering the lampshade.

**11. The students of 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? (Assume π =22/7)**

**Solution:**

Radius of the circular end of cylindrical penholder, r = 3cm

Height of penholder, h = 10.5cm

Surface area of a penholder = CSA of pen holder + Area of base of penholder

= 2πrh+πr^{2}

= 2×(22/7)×3×10.5+(22/7)×3^{2}= 1584/7

Therefore, Area of cardboard sheet used by one competitor is 1584/7 cm^{2}

So, Area of cardboard sheet used by 35 competitors = 35×1584/7 = 7920 cm^{2}

Therefore, 7920 cm^{2} cardboard sheet will be needed for the competition.

## Exercise 13.3 Page No: 221

**1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area (Assume π=22/7)**

**Solution:**

Radius of the base of cone = diameter/ 2 = (10.5/2)cm = 5.25cm

Slant height of cone, say l = 10 cm

CSA of cone is = πrl

= (22/7)×5.25×10 = 165

Therefore, the curved surface area of the cone is 165 cm^{2}.

**2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m. (Assume π = 22/7)**

**Solution:**

Radius of cone, r = 24/2 m = 12m

Slant height, l = 21 m

Formula: Total Surface area of the cone = πr(l+r)

Total Surface area of the cone = (22/7)×12×(21+12) m^{2}

= 1244.57m^{2}

**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.**

**(Assume π = 22/7)**

**Solution:**

Slant height of cone, l = 14 cm

Let the radius of the cone be r.

(i) We know, CSA of cone = πrl

Given: Curved surface area of a cone is 308 cm^{2}

(308 ) = (22/7)×r×14

308 = 44 r

r = 308/44 = 7

Radius of a cone base is 7 cm.

(ii) Total surface area of cone = CSA of cone + Area of base (πr^{2})

Total surface area of cone = 308+(22/7)×7^{2} = 308+154

Therefore, the total surface area of the cone is 462 cm^{2}.

**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.**

**(Assume π=22/7)**

**Solution:**

Let ABC be a conical tent

Height of conical tent, h = 10 m

Radius of conical tent, r = 24m

Let the slant height of the tent be l.

(i) In right triangle ABO, we have

AB^{2 }= AO^{2}+BO^{2}(using Pythagoras theorem)

l^{2} = h^{2}+r^{2}

= (10)^{2}+(24)^{2}

= 676

l = 26

Therefore, the slant height of the tent is 26 m.

(ii) CSA of tent = πrl

= (22/7)×24×26 m^{2}

Cost of 1 m^{2} canvas = Rs 70

Cost of (13728/7)m^{2} canvas is equal to Rs (13728/7)×70 = Rs 137280

Therefore, the cost of the canvas required to make such a tent is Rs 137280.

**5. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m? 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:**

Height of conical tent, h = 8m

Radius of base of tent, r = 6m

Slant height of tent, l^{2} = (r^{2}+h^{2})

l^{2 }= (6^{2}+8^{2}) = (36+64) = (100)

or l = 10

Again, CSA of conical tent = πrl

= (3.14×6×10) m^{2}

= 188.4m^{2}

Let the length of tarpaulin sheet required be L

As 20 cm will be wasted, therefore,

Effective length will be (L-0.2m).

Breadth of tarpaulin = 3m (given)

Area of sheet = CSA of tent

**[(L–0.2)×3] = 188.4**

L-0.2 = 62.8

L = 63

Therefore, the length of the required tarpaulin sheet will be 63 m.

**6. The slant height and base diameter of conical tomb are 25m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 m ^{2}. (Assume π = 22/7)**

**Solution:**

Slant height of conical tomb, l = 25m

Base radius, r = diameter/2 = 14/2 m = 7m

CSA of conical tomb = πrl

= (22/7)×7×25 = 550

CSA of conical tomb= 550m^{2}

Cost of white-washing 550 m^{2} area, which is Rs (210×550)/100

= Rs. 1155

Therefore, cost will be Rs. 1155 while white-washing tomb.

**7. A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. (Assume π =22/7)**

**Solution:**

Radius of conical cap, r = 7 cm

Height of conical cap, h = 24cm

Slant height, l^{2} = (r^{2}+h^{2})

= (7^{2}+24^{2})

= (49+576)

= (625)

Or l = 25 cm

CSA of 1 conical cap = πrl

= (22/7)×7×25

= 550

CSA of 10 caps = (10×550) cm^{2} = 5500 cm^{2}

Therefore, the area of the sheet required to make 10 such caps is 5500 cm^{2}.

**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 cones 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 √(1.04) =1.02)**

**Solution**:

Given:

Radius of cone, r = diameter/2 = 40/2 cm = 20cm = 0.2 m

Height of cone, h = 1m

Slant height of cone is l, and l^{2 }= (r^{2}+h^{2})

Using given values, l^{2} = (0.2^{2}+1^{2})

= (1.04)

Or l = 1.02

Slant height of the cone is 1.02 m

Now,

CSA of each cone = πrl

= (3.14×0.2×1.02)

= 0.64056

CSA of 50 such cones = (50×0.64056) = 32.028

CSA of 50 such cones = 32.028 m^{2}

Again,

Cost of painting 1 m^{2} area = Rs 12 (given)

Cost of painting 32.028 m^{2} area = Rs (32.028×12)

= Rs.384.336

= Rs.384.34 (approximately)

Therefore, the cost of painting all these cones is Rs. 384.34.

## Exercise 13.4 Page No: 225

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

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

**(Assume π=22/7)**

**Solution**:

Formula: Surface area of sphere (SA) = 4πr^{2}

(i) Radius of sphere, r = 10.5 cm

SA = 4×(22/7)×10.5^{2 }= 1386

Surface area of sphere is 1386 cm^{2}

(ii) Radius of sphere, r = 5.6cm

Using formula, SA = 4×(22/ 7)×5.6^{2 }= 394.24

Surface area of sphere is 394.24 cm^{2}

(iii) Radius of sphere, r = 14cm

SA = 4πr^{2}

= 4×(22/7)×(14)^{2}

= 2464

Surface area of sphere is 2464 cm^{2}

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

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

**(Assume π = 22/7)**

**Solution:**

(i) Radius of sphere, r = diameter/2 = 14/2 cm = 7 cm

Formula for Surface area of sphere = 4πr^{2}

= 4×(22/7)×7^{2} = 616

Surface area of a sphere is 616 cm^{2}

(ii) Radius (r) of sphere = 21/2 = 10.5 cm

Surface area of sphere = 4πr^{2}

= 4×(22/7)×10.5^{2 }= 1386

Surface area of a sphere is 1386 cm^{2}

Therefore, the surface area of a sphere having diameter 21cm is 1386 cm^{2}

(iii) Radius(r) of sphere = 3.5/2 = 1.75 cm

Surface area of sphere = 4πr^{2}

= 4×(22/7)×1.75^{2} = 38.5

Surface area of a sphere is 38.5 cm^{2}

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

**Solution:**

Radius of hemisphere, r = 10cm

Formula: Total surface area of hemisphere = 3πr^{2}

= 3×3.14×10^{2} = 942

The total surface area of given hemisphere is 942 cm^{2}.

**4. The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.**

**Solution:**

Let r_{1 }and r_{2} be the radii of spherical balloon and spherical balloon when air is pumped into it respectively. So

r_{1 }= 7cm

r_{2 }= 14 cm

Now, Required ratio = (initial surface area)/(Surface area after pumping air into balloon)

= 4r_{1}^{2}/4r_{2}^{2}

= (r_{1}/r_{2})^{2}

= (7/14)^{2 }= (1/2)^{2} = ¼

Therefore, the ratio between the surface areas is 1:4.

**5. A hemispherical bowl made of brass has inner diameter 10.5cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm ^{2}. (Assume π = 22/7)**

**Solution:**

Inner radius of hemispherical bowl, say r = diameter/2 = (10.5)/2 cm = 5.25 cm

Formula for Surface area of hemispherical bowl = 2πr^{2}

= 2×(22/7)×(5.25)^{2} = 173.25

Surface area of hemispherical bowl is 173.25 cm^{2}

Cost of tin-plating 100 cm^{2} area = Rs 16

Cost of tin-plating 1 cm^{2} area = Rs 16 /100

Cost of tin-plating 173.25 cm^{2 }area = Rs. (16×173.25)/100 = Rs 27.72

Therefore, the cost of tin-plating the inner side of the hemispherical bowl at the rate of Rs 16 per 100 cm^{2} is Rs **27.72.**

**6. Find the radius of a sphere whose surface area is 154 cm ^{2}. (Assume π = 22/7)**

**Solution:**

Let the radius of the sphere be r.

Surface area of sphere = 154 (given)

Now,

4πr^{2 }= 154

r^{2 }= (154×7)/(4×22) = (49/4)

r = (7/2) = 3.5

The radius of the sphere is 3.5 cm.

**7. The diameter of the moon is approximately one fourth of the diameter of the earth.**

**Find the ratio of their surface areas.**

**Solution:**

If diameter of earth is said d, then the diameter of moon will be d/4 (as per given statement)

Radius of earth = d/2

Radius of moon = ½×d/4 = d/8

Surface area of moon = 4π(d/8)^{2}

Surface area of earth = 4π(d/2)^{2}

The ratio between their surface areas is 1:16.

**8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface of the bowl. (Assume π =22/7)**

**Solution:**

Given:

Inner radius of hemispherical bowl = 5cm

Thickness of the bowl = 0.25 cm

Outer radius of hemispherical bowl = (5+0.25) cm = 5.25 cm

Formula for outer CSA of hemispherical bowl = 2πr^{2}, where r is radius of hemisphere

= 2×(22/7)×(5.25)^{2} = 173.25

Therefore, the outer curved surface area of the bowl is 173.25 cm^{2}.

**9. A right circular cylinder just encloses a sphere of radius r (see fig. 13.22). Find**

**(i) surface area of the sphere,**

**(ii) curved surface area of the cylinder,**

**(iii) ratio of the areas obtained in(i) and (ii).**

**Solution:**

(i) Surface area of sphere = 4πr^{2}, where r is the radius of sphere

(ii) Height of cylinder, h = r+r =2r

Radius of cylinder = r

CSA of cylinder formula = 2πrh = 2πr(2r) (using value of h)

= 4πr^{2}

(iii) Ratio between areas = (Surface area of sphere)/CSA of Cylinder)

= 4r^{2}/4r^{2 }= 1/1

Ratio of the areas obtained in (i) and (ii) is 1:1.

** **