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 cm2
TSA = 2 π r (h + r)
= 2 × 22/7 × 7/2 (60 + 7/2)
= 22 (127/2)
= 1397 cm2
Hence, the Curved surface area is 1320 cm2 and total surface area is 1397 cm2
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,
= πr2 = 616
or, r2 = 616/22/7 = 616 × 7 / 22 = 196
or, r = √196 = 14
∴ CSA = 2πrh
= 2 × 22/7 × 14 × 2.5
= 220 cm2
Thus, the curved surface area will be 220 cm2
Question no – (4)
Solution :
Let, radius = r
2πr = 88
or, r = 88/ 2 × 22/7
= 88 × 7 / 2 × 22
= 14 cm
Volume = πr2h
= 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 cm2
Therefore, the total surface area of the solid will be 1408 cm2
Question no – (6)
Solution :
Let, radius = r
According to question,
2πr = 44
or, r = 44/2 × π
= 44 × 7/2 × 22
= 7 cm.
V = πr2h = 22/7 × (7)2 × 20
= 22 × 7 × 20
= 3080 cm3
Thus, total surface area of the cylinder will be 3080 cm3
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
= 20r2/ 27r2
= 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 = 2πr2 = 308
According to question,
2πr2 = 308
or, r2 = 308 × 7 / 2 × 22 = 49
or, r = √49 = 7cm
h = r = 7cm
r = πr2h = 22/7 × 7 × 7 ×
∴ r = 1078 cm2
Question no – (9)
Solution :
As per the question,
Curved surface area of a cylinder is = 1320 cm2
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 = 2πr2
CSA of cylinder = 2πrh
Now, according to the question,
3(2πr2) = 2(2πrh)
or, 6πr2 = 4πrh
or, 6r = 4h
Question no – (11)
Solution :
CSA = 2πrh = 2 × 22/7 × 3 × 21
= 396 m2
∴ Cost of plastering,
= 396 × 9.50
= 3762 Rs.
Thus, the cost of plastering will be Rs. 3762
Question no – (12)
Solution :
TSA = 2πrh + πr2
= [2 × 22/7 × 10 × 14] + [22/7 × (10)2]
= 6160/7 + 2200/7
= 8360/7
= 1193.2 cm2
Question no – (14)
Solution :
Radius = 84/2 = 42 cm
Area = 2πrh
= 2 × 22/7 × 42 × 120
= 31680 cm2
∴ Area of the playground,
= 31680 × 500
= 15840000 cm2
Therefore, area of the playground will be 15840000 cm2
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 + πr2)
= 2πr (2h + r)
= 2 × 22/7 × 70 (280 + 70)
= 154000 cm2
∴ 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 = πr2h
= 22/7 × 3.5 × 3.5 × 10
= 1540 cm3
Therefore, the volume of the cylinder will be 1540 cm3
(ii) r = 2.8 m,
h = 15 m
∴ Volume of cylinder,
= πr2h = 22/7 × 2.8 × 2.8 × 15
= 369.5 cm3
Hence, the volume of the cylinder will be 369.5 cm3
Question no – (2)
Solution :
(i) Given, d = 21,
r = 21/2
∴ Volume is,
= πr2h = 22/7 × 21/2 × 21/2 × 10
= 3465cm3
Therefore, the volume of cylinder will be 3465 cm3
(ii) Given, d = 7 m,
r = 7/2
h = 24 m
∴ Volume is,
= πr2h = 22/7 × 7/2 × 7/2 × 24
= 924 m3
Hence, the volume of cylinder will be 924 m3
Question no – (3)
Solution :
Let, radii = 14
According to Question,
= πr2 = 616
or, r2 = 616/22/7 = 616 × 7 / 22 = 196
or, r = √196 = 14
∴ CSA = 2πrh
= 2 × 22/7 × 14 × 2.5
= 220 cm2
Therefore, the volume of the cylinder will be 220 cm2
Question no – (4)
Solution :
Let, height = h
According to question,
2πr = 88
or, r = 88 × 7/2 × 22
= 14 cm
Volume = πr2h
= 22/7 × 14 × 14 × 60
= 36960 cm3
CSA = 2πrh
= 2 × 22/7 × 14 × 60
= 580 cm3
Thus, the volume of the cylinder will be 580 cm3
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,
= πR2h – πr2h = πh[R2 – r2]
= 22/7 × 210 [52 – 32] = 660 [25 – 9]
= 660 × 16
= 10560 cm3
Therefore, the volume of the copper will be 10560 cm3
Question no – (6)
Solution :
(i) CSA = 2 π r h
= 2 × 22/7 × 7 × 15
= 660 cm2
(ii) T.S.A = 2 π r h + πr2 660 + (22/7 × 7 × 7)
= 660 + 154
= 814 cm2
(iii) Volume = πr2h
= 154 × 15
= 2310 cm3
Question no – (8)
Solution :
radil = r = 7/2 = 3.5 cm
∴ Volume is = πr2h = 22/7 × 3.5 × 3.5 × 60
= 2310 cm3
= 2.31 lit.
Therefore, the volume of a cylinder will be 2.31 lit.
Other Chapter Solution :
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