**Rd Sharma Solutions Class 8 Chapter 22 Mensuration – III Surface Area and Volume of Right Circular Cylinder**

Welcome to NCTB Solution. Here with this post we are going to help 8th Class students for the Solutions of RD Sharma Class 8 Mathematics, Chapter 22, Mensuration – III (Surface Area and Volume of Right Circular Cylinder). Here students can easily find Exercise wise solution for chapter 22, Mensuration – III (Surface Area and Volume of Right Circular Cylinder). Students will find proper solutions for Exercise 22.1 and 22.2 Our teacher’s solved every problem with easily understandable methods so that every students can understand easily.

**Mensuration – III Surface Area and Volume of Right Circular Cylinder Exercise 22.1 Solution :**

**Question no – (1) **

**Solution : **

Given, r = 7/2 cm

Curved surface are,

= 2πrh = 2 × 22/7 × 7/2 × 60

= 1320 cm^{2}

TSA = 2 π r (h + r)

= 2 × 22/7 × 7/2 (60 + 7/2)

= 22 (127/2)

= 1397 cm^{2}

Hence, the Curved surface area is 1320 cm^{2} and total surface area is 1397 cm^{2}

**Question no – (2) **

**Solution : **

Length = h

According to question,

**∴** 2πrh = 132

or, 2 × 22/7 × 0.35 × h = 132

or, h = 132/ 2 × 22 × 0.5

= h = 60 cm.

Therefore, its length will be 60 cm.

**Question no – (3) **

**Solution : **

Let, radii = 14

According to Question,

= πr^{2} = 616

or, r^{2} = 616/22/7 = 616 × 7 / 22 = 196

or, r = √196 = 14

**∴** CSA = 2πrh

= 2 × 22/7 × 14 × 2.5

= 220 cm^{2}

Thus, the curved surface area will be 220 cm^{2}

**Question no – (4) **

**Solution :**

Let, radius = r

2πr = 88

or, r = 88/ 2 × 22/7

= 88 × 7 / 2 × 22

= 14 cm

Volume = πr^{2}h

= 22/7 × (14)^{2} × 60

= 36960 m

Curved Surface Area,

= 2πrh = 2 × 22/7 × 14 × 60

= 5280 m

**Question no – (5) **

**Solution : **

As per the question,

A rectangular strip 25 cm × 7 cm is rotated about the longer side.

**∴** Total Surface Area,

= 2πr (h + r)

= 2 × 22/7 × 7 (25 + 7)

= 2 × 22 × 32

= 1408 cm^{2}

Therefore, the total surface area of the solid will be 1408 cm^{2}

**Question no – (6) **

**Solution : **

Let, radius = r

According to question,

2πr = 44

or, r = 44/2 × π

= 44 × 7/2 × 22

= 7 cm.

V = πr^{2}h = 22/7 × (7)^{2} × 20

= 22 × 7 × 20

= 3080 cm^{3}

Thus, total surface area of the cylinder will be 3080 cm^{3}

**Question no – (7) **

**Solution : **

Let, common factor = r

ratio = 2 : 3

radius of 1st cylinder = 2r

radius of 2nd cylinder = 3r

common factor = h

ratio = 5 : 3

height = h

height of 1st cylinder = 5h

height of 2nd cylinder = 3h

**∴** Volume of 1st cylinder/ Volume of 2nd cylinder

= π ×(2r)^{2} × 5h / π × (3r)^{2} × 3h

= 20r^{2}/ 27r^{2}

= 20/27

= 20 : 27

Therefore, the ratio of their curved surface areas will be 20 : 27

**Question no – (8) **

**Solution : **

CSA : TSA = 1 : 2

CSA = 616/2 = 308

Now, according to question,

2πrh/2πr(h + r) = 1/2

or, h / h + r = 1/2

or, 2h = h + r

or, 2h – h = r

or, h = r

**∴** CSA = 2 π r h = 2 π r . r = 2πr^{2} = 308

According to question,

2πr^{2} = 308

or, r^{2} = 308 × 7 / 2 × 22 = 49

or, r = √49 = 7cm

h = r = 7cm

r = πr^{2}h = 22/7 × 7 × 7 ×

**∴** r = 1078 cm^{2}

**Question no – (9) **

**Solution : **

As per the question,

Curved surface area of a cylinder is = 1320 cm^{2}

Its base has diameter = 21 cm

Let, height = h cm

Now, according to the question,

2 π r h = 1320

or, h = 1320/2 × 22/7 × 21/2

= 1320/ 22 × 3

= 1320/60

= 20 cm

Therefore, the height of the cylinder will be 20 cm.

**Question no – (10) **

**Solution : **

Let, radius = r

Area of 2 bases of cylinder = 2πr^{2}

CSA of cylinder = 2πrh

Now, according to the question,

3(2πr^{2}) = 2(2πrh)

or, 6πr^{2} = 4πrh

or, 6r = 4h

**Question no – (11) **

**Solution : **

CSA = 2πrh = 2 × 22/7 × 3 × 21

= 396 m^{2}

**∴** Cost of plastering,

= 396 × 9.50

= 3762 Rs.

Thus, the cost of plastering will be Rs. 3762

**Question no – (12) **

**Solution : **

TSA = 2πrh + πr^{2}

= [2 × 22/7 × 10 × 14] + [22/7 × (10)^{2}]

= 6160/7 + 2200/7

= 8360/7

= 1193.2 cm^{2}

**Question no – (14) **

**Solution : **

Radius = 84/2 = 42 cm

Area = 2πrh

= 2 × 22/7 × 42 × 120

= 31680 cm^{2}

**∴** Area of the playground,

= 31680 × 500

= 15840000 cm^{2}

Therefore, area of the playground will be 15840000 cm^{2}

**Question no – (15) **

**Solution : **

CSA = 2πrh= (2 × 22/7 × 0.5/2 × 4)

**∴** Cost = 21 × 2.50 × 22/7 × 0.5 × 4

= 3 × 2.50 × 22 × 0.5 × 4

= 330 Rs.

Therefore, the cost of cleaning will be Rs. 330

**Question no – (18) **

**Solution : **

According to the question,

height and radius are 7.5 cm and 3.5 cm

**∴** Required ratio = 2πr(h + r)/ 2πrh

= h + r/h = 75 + 3.5/ 75

= 22 : 15

**Question no – (19) **

**Solution : **

Given, h = 1.5m = 140cm

r = 70 cm

**∴ **Area to be tin-coated = 2(2πrh + πr^{2})

= 2πr (2h + r)

= 2 × 22/7 × 70 (280 + 70)

= 154000 cm^{2}

**∴** Cost of coating,

= 3.50 × 154000/1000

= 539 Rs.

Thus, the cost of tin-coating will be 539 Rs.

**Mensuration – III Surface Area and Volume of Right Circular Cylinder Exercise 22.2 Solution :**

**Question no – (1)**

**Solution : **

**(i) Given, r = 3.5 cm,**

h = 40 cm

**∴** Volume of cylinder = πr^{2}h

= 22/7 × 3.5 × 3.5 × 10

= 1540 cm^{3}

Therefore, the volume of the cylinder will be 1540 cm^{3}

**(ii) r = 2.8 m,**

h = 15 m

**∴** Volume of cylinder,

= πr^{2}h = 22/7 × 2.8 × 2.8 × 15

= 369.5 cm^{3}

Hence, the volume of the cylinder will be 369.5 cm^{3}

**Question no – (2) **

**Solution : **

**(i)** Given, d = 21,

r = 21/2

**∴** Volume is,

= πr^{2}h = 22/7 × 21/2 × 21/2 × 10

= 3465cm^{3}

Therefore, the volume of cylinder will be 3465 cm^{3}

**(ii)** Given, d = 7 m,

r = 7/2

h = 24 m

**∴** Volume is,

= πr^{2}h = 22/7 × 7/2 × 7/2 × 24

= 924 m^{3}

Hence, the volume of cylinder will be 924 m^{3}

**Question no – (3) **

**Solution : **

Let, radii = 14

According to Question,

= πr^{2} = 616

or, r^{2} = 616/22/7 = 616 × 7 / 22 = 196

or, r = √196 = 14

**∴** CSA = 2πrh

= 2 × 22/7 × 14 × 2.5

= 220 cm^{2}

Therefore, the volume of the cylinder will be 220 cm^{2}

**Question no – (4) **

**Solution : **

Let, height = h

According to question,

2πr = 88

or, r = 88 × 7/2 × 22

= 14 cm

Volume = πr^{2}h

= 22/7 × 14 × 14 × 60

= 36960 cm^{3}

CSA = 2πrh

= 2 × 22/7 × 14 × 60

= 580 cm^{3}

Thus, the volume of the cylinder will be 580 cm^{3}

**Question no – (5) **

**Solution : **

Height = h = 21 cm = 210cm

Outer radii (R) = 10/2 = 5 cm

Inner radii = r = 6/2 = 3cm

**∴** Volume of copper used,

= πR^{2}h – πr^{2}h = πh[R^{2} – r^{2}]

= 22/7 × 210 [5^{2} – 3^{2}] = 660 [25 – 9]

= 660 × 16

= 10560 cm^{3}

Therefore, the volume of the copper will be 10560 cm^{3}

**Question no – (6) **

**Solution : **

**(i) CSA = 2 π r h**

= 2 × 22/7 × 7 × 15

= 660 cm^{2}

**(ii) T.S.A = 2 π r h + πr ^{2} 660 + (22/7 × 7 × 7)**

= 660 + 154

= 814 cm^{2}

**(iii) Volume = πr ^{2}h**

= 154 × 15

= 2310 cm^{3}

**Question no – (8) **

**Solution : **

radil = r = 7/2 = 3.5 cm

∴ Volume is = πr^{2}h = 22/7 × 3.5 × 3.5 × 60

= 2310 cm^{3}

= 2.31 lit.

Therefore, the volume of a cylinder will be 2.31 lit.

**Other Chapter Solution : **

👉 Chapter 1 👈