Three Dimensional Solids Class 10 OP Malhotra Exe-15A ICSE Maths Solutions

Three Dimensional Solids Class 10 OP Malhotra Exe-15A ICSE Maths Solutions Ch-15 questions as latest prescribe guideline for upcoming exam. In this article you would learn how to solve problems on Surface Area of Cylinder. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.

Three Dimensional Solids Class 10 OP Malhotra Exe-15A ICSE Maths Solutions

Three Dimensional Solids Class 10 OP Malhotra Exe-15A ICSE Maths Solutions Ch-15

Board ICSE
Publications  S Chand
Subject Maths
Class 10th
Chapter-15 Three Dimensional Solids
Writer OP Malhotra
Exe-15A Surface Area of Cylinder
Edition 2024-2025

Surface Area Formula of Cylinder

The Surface Area of Cylinder = Curved Surface + Area of Circular bases
S.A. (in terms of π) = 2πr (h + r) sq.unit
Where, π (Pi) = 3.142 or = 22/7, r  = Radius of the cylinder, h  = Height of the cylinder

Exercise- 15A  (Surface Area of Cylinder)

Three Dimensional Solids Class 10 OP Malhotra Exe-15A ICSE Maths Solutions Ch-15 questions

Que-1: Find the area of curved surface and total surface area of the cylinders whose height and radii are given below :  (i) h = 12 cm, r = 7 cm  (ii) h = 10 cm, r = 7 cm  (iii) h = 5 cm, r = 21 cm  (iv) h = 20 cm, r = 14 cm  (v) h = 16 m, r = 10.5 m

Sol:  (i) We know that,
Curved surface area = 2πrh
Total surface area = 2πr(h + r)
Here, r = 7 cm, h = 12 cm
The curved surface area of a cylinder = 2πrh
= 2×(22/7)×7×12
= 44×12
= 528 cm2
Total surface area of a cylinder = 2πr(h + r)
= 2×(22/7)×7(12+7)
= 44×19
= 836 cm2

(ii) We know that,
Curved surface area = 2πrh
Total surface area = 2πr(h + r)
Here, r = 7 cm, h = 10 cm
The curved surface area of a cylinder = 2πrh
= 2×(22/7)×7×10
= 44×10
= 440 cm2
Total surface area of a cylinder = 2πr(h + r)
= 2×(22/7)×7(10+7)
= 44×17
= 748 cm2

(iii) We know that,
Curved surface area = 2πrh
Total surface area = 2πr(h + r)
Here, r = 21 cm, h = 5 cm
The curved surface area of a cylinder = 2πrh
= 2×(22/7)×21×5
= 44×15
= 660 cm2
Total surface area of a cylinder = 2πr(h + r)
= 2×(22/7)×21(5+21)
= 44×17×3
= 3432 cm2

(iv) We know that,
Curved surface area = 2πrh
Total surface area = 2πr(h + r)
Here, r = 14 cm, h = 20 cm
The curved surface area of a cylinder = 2πrh
= 2×(22/7)×14×20
= 44×40
= 1760 cm2
Total surface area of a cylinder = 2πr(h + r)
= 2×(22/7)×14(20+14)
= 44×68
= 2992 cm2

(v) We know that,
Curved surface area = 2πrh
Total surface area = 2πr(h + r)
Here, r = 10.5 m, h = 16 m
The curved surface area of a cylinder = 2πrh
= 2×(22/7)×10.5×16
= 44×16×1.5
= 1056 m2
Total surface area of a cylinder = 2πr(h + r)
= 2×(22/7)×10.5(16+10.5)
= 44×1.5×26.5
= 1749 m2

Que-2: A cylindrical tank 7 m in diameter, contains water to a depth of 4 m. Find the total area of the wet surface.

Sol:  total surface area of cylindrical tank = CSA of cylindrical tank + area of base of the tank
total surface area of cylindrical tank = 2πrh +πr²
= πr(2h+r)
= 3.14×3.5(2×4+3.5)
= 10.99 × 11.5
= 126.358 m².

Que-3: Find the whole surface of a hollow cylinder open at both ends, whose external diameter is 14 cm, thickness 2 cm, and height 20 cm.

Sol:  External radius R = 14/2 = 7 cm
thickness = 2cm
inner radius r = 7-2 = 5cm
height = 20 cm
Total surface area = 2πRh + 2πrh + 2πR² – 2πr²
= 2π (Rh+rh+R²-r²)
= 2π (7×20+5×20+7²-5²)
= 2π (140+100+49-25)
= 2π (289-25)
= 2π × 264
= 528π cm²

Que-4: Find the radius of the cylinder if area of its curved surface is 110 cm², and height 5 cm.

Sol:  The curved surface area of the cylinder = 110 cm²
Height = 5 cm
The curved surface area of the cylinder = 2πrh
Curved surface area of cylinder = 2πr × h = 110 cm2
⇒ 2πr × 5 = 110
⇒ 22/7 × r = 11
⇒ r = 7/2 = 3.5 cm

Que-5: Find the height of the cylinder if area f its curved surface is 13.2 cm², and radius 6 cm.

Sol: The curved surface area of the cylinder = 13.2 cm²
Radius = 6 cm
The curved surface area of the cylinder = 2πrh
Curved surface area of cylinder = 2πr × h = 13.2 cm2
⇒ 2π(6) × h = 13.2
⇒ 22/7 × h = 13.2/12
⇒ h = 1.1 × 7/22
h = 7/20 = 0.035
h = 3.5 mm.

Que-6: A gardener roller is 75 cm in diameter and 105 cm in width. What area does it cover in 14 revolutions ?

Sol:  d = 75 cm
r = 75/2 cm
width or h = 105 cm
csa = 2πrh
= 2 × 22/7 × 75/2 × 105
= 2 × 11/7 × 75 × 15
= 24750
we got the area covered in 1 revolution is 24750 cm²
area covered in 14 revolutions = 24750 × 14 = 346500 cm²
= 34.65 m²

Que-7: 10 cylindrical pillars of the building have to be painted. If the diameter of each pillar is 50 cm and the height 4m, what will be the cost of painting these at the rate of 50 paise per m² ?

Sol:  Radius of cylindrical pillars = 50/2 = 25cm = 25/100 = 0.25m
Height of cylindrical pillars = 4 m
CSA = 2πrh
2 × 3.14 × 0.25 × 4
→ 6.28 × 1
→ 6.28 m²
Total cost = 6.28 × 0.50 × 10
= 62.8 × 0.5
= Rs31.4

Que-8: The diameter of a roller 120 cm long is 84 cm. If it takes 500 complete revolutions to level a playground, determine the cost of levelling at the rate of 30 paise per m² ?

Sol:  Length of the roller, h = 120 cm
Radius of the roller, r = 84/2 cm = 42 cm
Curved Surface Area of the roller = 2πrh
= 2 × 22/7 × 42 cm × 120 cm
= 31680 cm²
Area of the playground = Area leveled by the cylinder in 500 revolutions
= 500 × 31680 cm²
= 15840000 cm²
= 15840000/10000 m²  [Since 1cm² = 1/10000 m²]
= 1584 m²
30 paisa per m² = 1584 × 0.30
= Rs475.20.

Que-9: The curved surface of a cylinder is 1000 cm². A wire of diameter of 5 mm is wound round is so as to cover it completely. Find the length of the wire.

Sol: Let the no. of turns the wire had made on the surface be n.
For n turns, the wire should make upto 5n mm or 0.005n m height
Since the wire should completely cover the cylinder,
∴ Height of cylinder = 0.005n m
Let the length of wire needed be l m
l = 2π × radius × n
l/n = 2π × radius
1000cm² = 0.1m² = 2π × radius × height
0.1 = (l/n) × 0.005n
l = 0.1/0.005 = 20m.

Que-10: An iron pipe 20 cm long has exterior diameter equal to 25 cm. If the thickness of the pipe is 1 cm, find the whole surface of the pipe.

Sol: The pipe is 20cm long (l)
extension diameter is 25cm, thickness is 10m.
∴ exterior radius (R) = 25/2 cm
interior radius (r) = (25/2) − 1 = 23/2 cm
So, the total surface area is
= exterior surface of the pipe + interior surface of the pipe + two ends of the pipe (thickness area)
= 2π(R).l + 2π(r).l + 2(πr²−πr²)
= 2π [l(R+r)+(R+r)(R−r)]
=2π (R+r) (l+R−r)
= 2×(22/7)×(25/2+23/2)(20+25/2−23/2)
= (44/7)×24×21
= 44×72
= 3168
So, the total surface face area is 3168 sq.cm.

Que-11: 50 circular plates, each of radius 7 cm and thickness 0.5 cm, are placed one above the other to form a right circular cylinder. Find its total surface area.

Sol:  Given that 50 circular plates each with diameter = 14cm
Radius of circular plates (r) = 7cm
Thickness of plates= 0.5
Since these plates are placed one above other so total thickness of plates = 0. 5 x 50
= 25cm.
Total surface area of a cylinder = 2πrh + 2πr2
= 2πrh + 2πr2
= 2πr (h+r)
= 2 x 22/7 x 7(25+7)
T.S.A = 1408cm2
∴Total surface area of circular plates is  1408cm2

–: End of Three Dimensional Solids Class 10 OP Malhotra Exe-15A ICSE Maths Solutions Ch-15 :–

Return to :  OP Malhotra S Chand Solutions for ICSE Class-10 Maths

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