OP Malhotra Class 9 ICSE Maths Solutions Chapter 18

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OP Malhotra Class 9 ICSE Maths Solutions Chapter 18 Surface Area and Volume of 3D Solids

Welcome to NCTB Solutions. Here with this post we are going to help 9th class students for the Solutions of OP Malhotra Class 9 ICSE Math Book, Chapter 18, Surface Area and Volume of 3D Solids. Here students can easily find step by step solutions of all the problems for Surface Area and Volume of 3D Solids, Exercise 18a and 18b Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 18 solutions. Here in this post all the solutions are based on ICSE latest Syllabus.

Surface Area and Volume of 3D Solids Exercise 18(a) Solution :

Question no – (1) 

Solution : 

(a) Cuboid whose dimensions are = 3 m, 4 m 5m

∴ Surface area of cuboids

= 2 (3 × 4 + 4 × 5 + 5 × 3) m²

= 2 (12 + 20 + 15)m²

= 2 × 47 m²

= 94 m²

Hence, the surface area of the cuboid will be 94 m²

(b) Cuboid whose dimensions are = 4 cm, 1.7 cm, 2.3 cm.

∴ Surface area of cuboid,

= 2(4 × 1.7 + 1.7 × 2.3 + 2.3 × 4) cm²

= 2 (6.8 + 3.9 + 9.2)cm²

= 2 × 19.91

= 39.82 cm²

Hence, the surface area of the cuboid will be 39.82 cm²

Question no – (2) 

Solution : 

(i) (a) Cube whose dimensions are =  7 cm.

Surface area of cube = 6a²

= 6 × (7)²

= 6 × 49 cm²

= 294 cm²

Hence, the surface area of the cube will be 294 cm²

(b) Cube whose dimensions are = 10 m

∴ Surface area of cube,

= 6 × (10)² m²

= 600 m²

Hence, the surface area of the cube will be 600 m²

(ii) As per the given question, 

To make a model of a cube = 96 cm² of board was used

Length of its edge,

= √Area/6

= √96/6

= √16

= 4 cm

Therefore, the length of one edge of the cube will be 4 cm.

(iii) Given in the question, 

The total surface area of a cube is = 726 cm cm²

∴ Length = √726/6

= √121 cm

= 11 cm

Thus, volume of the cube will be 11 cm.

Question no – (3) 

Solution : 

(i) Volume of cuboid,

= (4 × 3 ×5) cm³

= 60 cm³

(ii) Volume of cuboids,

= (2 × 6 × 8) cm³

= 96 cm³

(iii) Volume of cuboids,

= (7 × 3 × 4) cm³

= 84 cm³

(iv) Volume of cuboids,

= (12 × 9 × 12) cm³

= 1296 cm³

(v) Volume of cuboids,

= 916 × 14 × 18) cm³

= 4032 cm³

(vi) Volume of cuboids,

= x × 28 × 24 = 25984

= x = 25984/28 × 24

= 7 cm

(vii) Volume of cuboids,

= 40 × 24 × x

= 2400 = 40 × 24 × x

= x = 2400/40 × 40

= 2.5 cm

(viii) Width Breadth,

= 5400/60 × 4

= 1.8 cm

Question no – (4) 

Solution : 

(i) Cuboid whose dimensions are = 2, 3 and 4 cm

∴ Diagonal of cuboid,

= √(2)² + 3² + (4)²

= √4 + 9 + 16

= √29

= 5.38 cm

Thus, the length of the diagonal of the cuboid will be 5.38 cm.

(ii) Cuboid whose dimensions are 3, 4 and 5 cm.

∴ Diagonal of cuboid,

= √l² + b² + h²

= √3² + 4² + 5²

= √9 + 16 + 25

= √50

= 707 cm

Hence, the length of the diagonal of the cuboid will be 707 cm.

Question no – (5) 

Solution : 

(i) – (a) Cube whose each edge is 2 m.

∴ Length of diagonal,

= √3 × side

= √3 × 2

= 2 × 1.732

= 3.464 m

∴ Volume of the cube,

= (2)³

= 8 cm³

(i) – (b) Cube whose each edge is 5 m

First, Length of diagonal

= √3 × side

= √3 × 5

= 5 × 91.732)

= 8.66 m

Now, Volume of the cube,

= (5)³

= 125 m³

(i) – (c) Cube whose each edge is 8 cm

First, Length of diagonal,

= √3 × 8

= 8 × (1,732)

= 13.86 cm

Now, Volume of the cube,

= (8)³ cm³

= (64 × 8) cm³

= 512 cm³

(ii) Cube edge is = 12 cm.

∴ Volume of cube,

= (12)³ cm³

= 1728 cm³

∴ Total surface area,

= 6 × (12)²

= 864 cm²

Hence, the volume of the cube is 1728 cm³ and total surface area of the cube is 864 cm²

Question no – (6) 

Solution : 

(i) – (a) Cubes whose volumes are 216 m³

∴ Length of each edge,

= 3 √216

= 3√6 × 6 × 6

= (6 × 6 × 6 ) × 1/3

= (6)³ × 1/2

= 6 m

Thus, each edge of the cube will be 6 m.

(i) – (b) Cube whose volume is 2197 m³

∴ Length of each edge,

= 3√2179

= (13)³ × 1/3

= 13 m

∴ Each edge of the cube will be 13 m.

(ii) Answer true or false

Cuboid Length,

= 1 m = 100 cm

= 0.5 m

Now, breadth = 50 cm

∴ Volume

= (100 × 50 × 50)

= 250000 cm³

Thus, the statement is True.

(iii) Let, height be ‘h’

Diagonal of cuboid = √l² + b² + h²

= 5√2 = √5² + 4² + h²

= 50 = 25 + 16 + h²

= h² = 50 – 41

= 9

= h = 3 dm

Hence, the height of solid will be 3 dm.

Question no – (7)

Solution : 

As per the question we get,

The volume of a rectangular solid is 3600cm³

length =  20 cm

Height = 9 cm

∴ Breadth, 

= 3600/20 × 9

= 20 cm

Therefore, width of the rectangular solid will be 20 cm.

Question no – (8)

Solution : 

According to the question,

The length of rectangular solid = 25 cm

The breadth of rectangular solid =  20 cm.

The volume is = 7000 cm³

∴ Height of rectangular solid,

= 70000/25 × 20

= 14 cm

Therefore, Height of rectangular solid will be 14 cm.

Question no – (9)

Solution : 

First, Side of cube,

= 20/4

= 5 cm

∴ Total surface area of cube,

= 6 × 5²

= 150 cm²

Hence, the total area of the 6 faces will be 150 cm²

Question no – (10)

Solution : 

First, Depth of gravel,

= 1 cm

= 1/100

Now, Volume gravel,

= 4800 × 1/100

= 48m³

∴ Total cost,

= (48 × 4.80) Rs

= 230. 40 Rs

Hence, the cost of covering the playground will be 230. 40 Rs.

Question no – (11)

Solution : 

First, Area of base,

= (7 × 6)m²

= 42m²

Now, Volume of the water in the tank, 

= (42 × 5)

= 210 m²

Therefore,  210 m² cubic metres of water are there in the tank.

Question no – (12)

Solution : 

First, Volume of cuboid,

= 20 × 16 × 24

= 7680 cm³

Now, Volume of cube,

= (4)³ cm³

= 64 cm³

∴ Number of cubes kept in the box,

= 7680/64

= 120

Therefore, 120, 4 cm cubes could be put into the box.

Question no – (13)

Solution : 

Let, length = 5x, breadth = 4x, height = 2x

First, Surface area of cuboid,

= 2(5x + 4x + 2x + 2x + 5x)

= 2 × 38x²

= 76x²

= 76x² = 1216

= x² = 1216/76

= 16

= x = 4

∴ Length = 5x = 5 × 4 = 20 cm

∴ Breadth = 4x = 4 × 4 = 16 cm

∴ Height = 2x = 2 × 4 = 8 cm

Question no – (14)

Solution : 

Here, level of water decreased,

= (5 – 3.80)m

= 1.20 m

∴ Volume of water ruins out,

= l × b × h

= 40 × 7 × 1.20

= 336 m³

Therefore, 336 m³ cubic metre of water runs out.

Question no – (15)

Solution : 

(i) Volume of board,

= (90 × 78 × 42) cm³ ….(according to the question)

And, Rectangular block,

height = 2 1/2 cm = 5/2 cm

breadth = 2 cm

height,  = 1 1/2 cm = 3/2

Now, Volume of 1 block,

= (5/2 × 2 × 3/2)

= 15/2 cm²

∴ Number of block packed  in the box,

= 9 × 78 × 42 × 2/15

= 39312

Thus, 39312 rectangular blocks can be packed.

(ii) Volume of cube = (4)³

= 64cm³

Number of cubes to be packed = 30 × 78 × 42/4

Now, length wise no of 4 complete cubes,

= 90/4

= 22

Breadth wise, no of complete cubes

= 78/4

= 19

Length wise,

= 42/4

= 10.5

= 10

∴ Total No of cubes.

= (22 × 19 × 10)

= 4180

Therefore, 4180 cubes can be packed.

Question no – (16)

Solution : 

First, Volume of water in the tank,

= (72 × 60 × 18)cm³

= 77760 cm³

Now, volume of rises,

= (48 × 36 × 15) cm³

= 25920 cm³

∴ Water level rises,

= 25920/(72 × 60)

= 6 cm

Thus, the amount the water level rises is 6 cm.

Question no – (17) 

Solution : 

Here, Inner volume,

= (20 × 12.5 × 9.5) cm³

= 2375 cm³

And, outer length

= 20 + (2 × 1.25)

= 22.5 cm

Outer breadth,

= 12.5 (2 × 1.5)

= 15 cm

Outer height,

= 9.5 + (2 × 1.25)

= 12 cm

∴ Volume,

= (22.5 × 15 × 12) cm³

= 4050 cm³

∴ Volume of wood used,

= (4050 – 2375)

= 16575 cm³

Therefore, the volume of wood required will be16575 cm³

Question no – (19) 

Solution : 

First, Area of filed,

= (30 × 15)

= 540 m²

Now, Area of pit,

= (6 × 4)

= 24 cm²

∴ Volume of dug filed,

= (6 × 4 × 3)

= 72 cm³

∴ Area of the remaining field Living pit,

= (540 – 24)

= 516 m²

∴ Rise level,

= 72/516 × 100 cm²

= 13.9 cm

Therefore, the rise in the level of the remaining area of the field will be 13.9 cm.

Question no – (20) 

Solution : 

First, Area of field,

= (20 × 14)

= 280 m²

Now, Area of face pit, 

= (6 × 3)

= 18m²

∴ Volume of dug out field,

= (6 × 3 × 2.5)

= 45 m³

∴ Area of remaining excluding pit,

= (280 – 18)

= 262 m²

∴ Height of field level,

= 45/262 m

= 45 × 100/262

= 17.18 m

Therefore, the field level will rise up to 17.18 m.

Question no – (22) 

Solution : 

After placing cube in container, volume of water,

= (6 × 6 × 1 + 2)

= (36 + 2) cm³

= 238 cm³

Therefore, the volume of the cube will be 238 cm³

Question no – (23) 

Solution :

After joining, length,

= (8 + 8)

= 16 cm

breadth = 8 cm

height = 8 cm

Therefore, Surface area,

= 2 (16 × 8 + 8 × 8 + 8 × 16) cm²

= 2 (128 + 64 + 128)

= (2 × 320) cm²

= 640 cm²

Hence, the surface area of the resulting cuboid will be 640 cm²

Question no – (24) 

Solution : 

Let, cuboid length = l

Cuboid breadth = b

Cuboid height = h

Now, from question,

= x = lh

= y = bh

= z = lb

∴ Volume = lbh

∴ xyx = lb + bh + lh

= l² b2h² = (lbh)²

= v² = xy²

Question no – (26) 

Solution : 

First, Total area of floor,

= (50 × 9)

= 450 m²

Now, Total volume of air of the room,

= (108 × 50)m³

= 5400 m³

And, length of room = 25 m

∴ Breadth = 450/25

= 18 m

∴ Height,

= 5400/25 × 18

= 12 m

Therefore, its breadth is 18 m and height is 12 m.

Question no – (27) 

Solution : 

Rectangular sheet,

Length = 42 cm

Breath = 36 cm

∴ After cutting, length

= 42 – (2 × 6)

= 42 – 12

= 30 cm

∴ Breadth = 36 – 2 × 6

= 24 cm

height = 6 cm

∴ Volume,

= (30 × 24 × 6) cm³

= 4320 cm³

Therefore, the volume of the box will be 4320 cm³

Question no – (28) 

Solution : 

First, Edge of each cube,

= 3 √243

= (7)3 l/3

= 7 cm

After joining, length of cuboid = (7 + 7) cm

Breadth = 7 cm

Height = 7 cm

∴ Surface area of cuboid = 2 (lb + bh + hl)

= 2 (14 × 7 + 7 × 7 + 7 × 14) cm²

= 2 × (98 + 49 + 98)

= 2 × 245

= 490 cm²

Therefore, the surface area of the resulting cuboid will be 490 cm²

Next Chapter Solution : 

👉 Chapter 19 👈