Class 8 SELINA Solutions Maths Chapter 22: Surface Area, Volume and Capacity (Cuboid, Cube and Cylinder)

Correct option: (b) 10 cm

Volume of a cuboid = length × breadth × height

⇒ 4800 cm³ = 24 cm × 20 cm × height

⇒ height = 10 cm

Correct option: (b) 9 cm

Let the length, breadth and height of a cuboid be 3x, 2x and x, respectively.

Volume of a cuboid = 162 cm³

⇒ length × breadth × height = 162 cm³

⇒ 3x × 2x × x = 162 cm³

⇒ 6x³ = 162 cm³

⇒ x³ = 27 cm³

⇒ x = 3 cm

Then, the longest side of a cuboid = 3x = 3(3) = 9 cm

Correct option: (c) three times

Let the length, breadth and height of a cuboid be l, b and h, respectively.

Then, volume of a cuboid = l × b × h

Now, new dimensions are as follows:

length = 2l, breadth  and height = 3h

Then, new volume of a cuboid = 2l ×  × 3h = 3 × l × b × h

= 3 times the original volume

Correct option: (c) nine times

Let the side of a cube = a

Then, the surface area of a cube = 6a²

When each side of a cube is tripled, new side = 3a

Therefore, the new surface area of a cube = 6(3a)² = 6(9a²) = 9(6a²)

Correct option: (c) 15 m²

For a cuboid,

Total surface area = 80 m² = 2(l × b + b × h + h × l)

Lateral surface area = 50 m² = 2(l + b)h

Total surface area – Lateral surface area = 2(l × b + b × h + h × l) – 2(l + b)h

⇒ (80 – 50) m² = 2(l × b)

⇒ 30 m² = 2(l × b)

⇒ 15 m² = l × b

⇒ Area of its base = 15 m²

Let the length, breadth and height of a cuboid be 5x, 3x and 2x, respectively.

Volume of a cuboid = 240 cm³

⇒ length × breadth × height = 240 cm³

⇒ 5x × 3x × 2x = 240 cm³

⇒ 30x³ = 240 cm³

⇒ x³ = 8 cm³

⇒ x = 2 cm

⇒ 5x = 5(2) = 10 cm, 3x = 3(2) = 6 cm and 2x = 2(2) = 4 cm

Thus, the dimensions of a cuboid are 10 cm, 6 cm and 4 cm, respectively.

Now, the total surface area of the cuboid

= 2(l × b + b × h + h × l)

= 2(10 × 6 + 6 × 4 + 4 × 10)

= 2(60 + 24 + 40)

= 2(124)

= 248 cm²

Let the length, breadth and height of a cuboid be 6x, 5x and 3x, respectively.

Total surface area of the cuboid = 504 cm²

⇒ 2(l × b + b × h + h × l) = 504 cm²

⇒ 2(6x × 5x + 5x × 3x + 3x × 6x) = 504 cm²

⇒ 30x² + 15x² + 18x² = 252 cm²

⇒ 63x² = 252 cm²

⇒ x² = 4 cm²

⇒ x = 2 cm

⇒ 6x = 6(2) = 12 cm, 5x = 5(2) = 10 cm and 3x = 3(2) = 6 cm

Thus, the dimensions of a cuboid are 12 cm, 10 cm and 6 cm, respectively.

Now, volume of a cuboid = length × breadth × height

= 12 × 10 × 6 cm³

= 720 cm³

Let the length of edge of a cube = a

Given, volume = 216 cm³

⇒ a³ = 216 cm³

⇒ a = 6 cm

Thus, the length of edge of a cube is 6 cm.

Let the length of edge of a cube = a

Given, volume = 1.728 m³

⇒ a³ = 1.728 m³

⇒ a = 1.2 m

Thus, the length of edge of a cube is 1.2 m.

Let the length of edge of a cube = a

Given, total surface area of a cube = 216 cm²

⇒ 6a² = 216 cm²

⇒ a² = 36 cm²

⇒ a = 6 cm

Then, volume of a cube = 6³ = 216 cm³

Dimensions of a wall:

length = 9 m = 900 cm

height = 6 m = 600 cm

breadth = 20 cm

Then, volume of a cuboid = (900 × 600 × 20) cm³

Dimensions of a brick:

length = 30 cm

breadth = 15 cm

height = 10 cm

Then, volume of a brick = (30 × 15 × 10) cm³

Edge of a cube = 14 cm

Then, volume of a cube = (14 cm)³ = 2744 cm³

Edge of each small melted cube = 2 cm

Then, volume of each small melted cube = (2 cm)³ = 8 cm³

Dimensions of a cuboidal box:

length = 40 cm, breadth = 30 cm and height = 50 cm

Then, total surface area of a cuboidal box

= 2(l × b + b × h + h × l)

= 2(40 × 30 + 30 × 50 + 50 × 40) cm²

= 2(1200 + 1500 + 2000) cm²

= 9400 cm²

= 0.94 m²

Now, total surface area of 20 such boxes = (20 × 0.94) m² = 18.8 m²

Cost of 1 m² of metal sheet = Rs. 45

Therefore, cost of 18.8 m² metal sheet = Rs. (45 × 18.8) = Rs. 846

For a resulting cuboid,

Length = 4 × 9 cm = 36 cm

Breadth = 9 cm

Height = 9 cm

Thus, the dimensions of the cuboid obtained are 36 cm, 9 cm and 9 cm, respectively.

Then, its total surface area = 2(l × b + b × h + h × l)

= 2(36 × 9 + 9 × 9 + 9 × 36)

= 2(324 + 81 + 324)

= 1458 cm²

Volume of a cuboidal box = 36 cm × 9 cm × 9 cm = 2916 cm³

For a rectangular box,

Length, l = 32 cm

Breadth, b = 24 cm

Height, h = 8 cm

The maximum length of a rod that can be kept in a rectangular box is its diagonal.

And, length of the diagonal of a cuboid

Diagonal of a cube

m

⇒ Edge of a cube = 25 m

Therefore, surface area of a cube = 6(edge)²

= 6(25)²

= 6 × 625

= 3750 m²

For a rectangular room,

Length = 4.5 m

Breadth = 4 m

Height = 3 m

Area to be white-washed

= Lateral surface area of the room + Area of the roof

= [2(l + b)×h] + (l × b)

= [2(4.5 + 4)3] + (4.5 × 4)

= 51 + 18

= 69 m²

Cost of white-washing per square metre = Rs. 15

Therefore, cost of white-washing 69 m² = Rs. 15 × 69 = Rs. 1035

Correct option: (a) 1000

Capacity of a hall = length × breadth × height

= 40 m × 25 m × 5 m

= 5000 m³

Amount of air required by each person = 5 m³

Therefore, number of persons accommodated in the hall

Correct option: (d) 80 cm × 45 cm × 58 cm

Internal dimensions of a rectangular box,

Length = (82 – 2 × 1) cm = 80 cm

Breadth = (47 – 2 × 1) cm = 45 cm

Height = (60 – 2 × 1) cm = 58 cm

Correct option: (d) 800 cm³

External dimensions of a closed small box,

Length = 12 cm

Breadth = 12 cm

Height = 10 cm

Internal dimensions of a closed small box,

Length = (12 – 2 × 1) cm = 10 cm

Breadth = (12 – 2 × 1) cm = 10 cm

Height = (10 – 2 × 1) cm = 8 cm

⇒ Internal volume = (10 × 10 × 8) cm³ = 800 cm³

Correct option: (b) 17 m²

Remaining area of the walls

= Lateral surface area of the room – Area of one door – Area of two windows

= [2(3 + 2)2 – (2 × 1) – 2(1 × 0.5)] m²

= (20 – 2 – 1) m²

= 17 m²

Dimensions of a room,

Length, l = 5 m

Breadth, b = 4.5 m

Height, h = 3.6 m

Then, lateral surface area of the room = 2(l + b) h m²

= 2(5 + 4.5) × 3.6 m²

= 68.4 m²

Area of one door = 1.5 m × 2.4 m = 3.6 m²

Area of two windows = 2(1 m × 0.75 m) = 1.5 m²

(i) Area of its walls, excluding doors and windows

= Lateral surface area of the room – Area of one door – Area of two windows

= (68.4 – 3.6 – 1.5) m²

= 63.3 m²

(ii) Cost of distempering its walls per m² = Rs. 4.50

Then, the cost of distempering 63.3 m² = Rs. 4.50 × 63.3 = Rs. 284.85

(iii) Area of the roof = 5 m × 4.5 m = 22.5 m²

Cost of painting its roof per m² = Rs. 9

Then, the cost of painting 22.5 m² = Rs. 9 × 22.5 = Rs. 202.50

Dimensions of a dining hall,

Length, l = 75 m

Breadth, b = 60 m

Height, h = 16 m

Then, lateral surface area of the room = 2(l + b) h m²

= 2(75 + 60) × 16 m²

= 4320 m²

Area of one door = 4 m × 3 m = 12 m²

Area of one window = 3 m × 1.6 m = 4.8 m²

(i) Area of the walls to be papered = Area of the dining hall, excluding doors and windows

= Lateral surface area of the room – Area of five doors – Area of four windows

= 4320 – 5(12) – 4(4.8) m²

= 4320 – 60 – 19.2 m²

= 4240.80 m²

Cost of papering its walls per m² = Rs. 12

Then, the cost of papering 4240.80 m² = Rs. 12 × 4240.80 = Rs. 50889.60

(ii) Area of the floor = 75 m × 60 m = 4500 m²

Cost of carpeting the floor per m² = Rs. 25

Then, the cost of carpeting 4500 m² = Rs. 25 × 4500 = Rs. 1,12,500

External dimensions of a closed box,

Length = 80 cm

Breadth = 75 cm

Height = 60 cm

⇒ External volume of the box = (80 × 75 × 60) cm³ = 3,60,000 cm³

Internal dimensions of a closed box,

Length = (80 – 2 × 2) cm = 76 cm

Breadth = (75 – 2 × 2) cm = 71 cm

Height = (60 – 2 × 2) cm = 56 cm

⇒ Internal volume of the box = (76 × 71 × 56) cm³ = 3,02,176 cm³

Volume of the wood required to make a closed box

= External Volume – Internal volume

= (3,60,000 – 3,02,176) cm³

= 57,824 cm³

External dimensions of a closed box,

Length = 66 cm

Breadth = 36 cm

Height = 21 cm

⇒ External volume of the box = (66 × 36 × 21) cm³ = 49,896 cm³

Internal dimensions of a closed box,

Length = (66 – 2 × 0.5) cm = 65 cm

Breadth = (36 – 2 × 0.5) cm = 35 cm

Height = (21 – 2 × 0.5) cm = 20 cm

⇒ Internal volume of the box = (65 × 35 × 20) cm³ = 45,500 cm³

(i) The capacity of the box = Internal volume of the box = 45,500 cm³

(ii) Volume of metal sheet used = External Volume – Internal volume

= (49,896 – 45,500) cm³

= 4396 cm³

(iii) Weight of 1 cm³ of metal = 3.6 g

⇒ Weight of 4396 cm³ of metal = 4396 × 3.6 g = 15825.6 g

Internal dimensions of a closed box,

Length = 1 m = 100 cm

Breadth = 80 cm

Height = 25 cm

⇒ Internal volume of the box = (100 × 80 × 25) cm³ = 2,00,000 cm³

The capacity of the box = Internal volume of the box

= 2,00,000 cm³

= 0.2 m³

Internal dimensions of a closed box,

Length = 1 m = 100 cm

Breadth = 80 cm

Height = 25 cm

⇒ Internal volume of the box = (100 × 80 × 25) cm³ = 2,00,000 cm³

External dimensions of a closed box,

Length = (100 + 2 × 2.5) cm = 105 cm

Breadth = (80 + 2 × 2.5) cm = 85 cm

Height = (25 + 2 × 2.5) cm = 30 cm

⇒ External volume of the box = (105 × 85 × 30) cm³ = 2,67,750 cm³

Volume of wood used to make the box

= External Volume – Internal volume

= (2,67,750 – 2,00,000) cm³

= 67,750 cm³

= 0.06775 m³

Dimensions of an open tank,

Length = 10 m

Breadth = 7.5 m

Height = 3.8 m

Area of the metal sheet required to make an open tank

= Lateral surface area of a tank + Area of the base

= 2(10 + 7.5) × 3.8 + (10 × 7.5) m²

= (133 + 75) m²

= 208 m²

Dimensions of an open tank,

Length = 30 m

Breadth = 24 m

Height = 4.5 m

Area of the iron sheet required to make an open tank

= Lateral surface area of a tank + Area of the base

= 2(30 + 24) × 4.5 + (30 × 24) m²

= (486 + 720) m²

= 1206 m²

Cost of iron sheet per m² = Rs. 65

⇒ Cost of 1206 m² of iron sheet = Rs. 65 × 1206 = Rs. 78,390

Volume of the resulting cube

= Volume of cube of edge 6 cm + Volume of cube of edge 8 cm + Volume of cube of edge 10 cm

= (6³ + 8³ + 10³) cm³

= (216 + 512 + 1000) cm³

= 1,728 cm³

Then, edge of the resulting cube =

Let the edges of two cubes be 3x and 2x respectively.

Then,

Let the edges of two cubes be 3x and 2x respectively.

Then,

Let the length, breadth and height of a rectangular solid be 4x, 3x and 2x, respectively.

Surface area of a rectangular solid = 2548 cm²

⇒ 2(4x × 3x + 3x × 2x + 4x × 2x) = 2548 cm²

⇒ 2(12x² + 6x² + 8x²) = 2548 cm²

⇒ 26x² = 1274 cm²

⇒ x² = 49 cm²

⇒ x = 7 cm

⇒ length = 4(7) = 28 cm, breadth = 3(7) = 21 cm and height = 2(7) = 14 cm

Then, the volume of the rectangular solid = (28 × 21 × 14) cm³ = 8232 cm³

Let the length, breadth and height of a rectangular solid be 4x, 3x and 2x, respectively.

Volume of a rectangular solid = 3000 m³

⇒ (4x × 3x × 2x) = 3000 m³

⇒ 24x³ = 3000 m³

⇒ x³ = 125 m³

⇒ x = 5 m

⇒ length = 4(5) = 20 m, breadth = 3(5) = 15 m and height = 2(5) = 10 m

Then, the surface area of the rectangular solid

= 2(20 × 15 + 15 × 10 + 20 × 10) m²

= 2(300 + 150 + 200) m²

= 2(650) m²

= 1300 m²

Correct option: (d) 440 cm²

Curved surface area of a cylinder = 2π × radius × height

Correct option: (b) 22 cm³

Volume of a cylinder = π × (radius)² × height

Correct option: (d) none of these

Lateral surface area of a cylinder = 2π × radius × height

Correct option: (d) π(R² – r²) h

R = outer radius

r = inner radius

h = height of the pipe

Volume of the cylindrical pipe = πR²h – πr²h = π(R² – r²) h

Correct option: (c) 80 cm²

When a rectangular piece of paper is rolled along its width,

Circumference of the base of cylinder = 8 cm = 2πr

Then, curved surface area of the cylinder formed

= Circumference of the base of cylinder × height

= 8 × 10

= 80 cm²

*Options (c) and (d) are same.

Let the height of the cylinder = h

Radius of the cylinder, r = 7 cm

Total surface area of the cylinder = 1100 cm²

⇒ 2πr(h + r) = 1100

⇒ h + 7 = 25

⇒ h = 18 cm

Therefore, the height of the cylinder is 18 cm.

Let the radius and height of the cylinder be ‘r’ and ‘h’ respectively.

Then,

⇒ 2h = h + r

⇒ h = r

⇒ h : r = 1 : 1

Let the radius and height of the cylinder be ‘r’ and ‘h’ respectively.

(i) Given,

Circumference of the base of cylinder = 88 cm

⇒ 2πr = 88 cm

⇒ r = 14 cm

(ii)

Total surface area of the cylinder = 6512 cm²

⇒ 2πr(h + r) = 6512

⇒ h + 14 = 74

⇒ h = 60

Then, the volume of the cylinder = πr²h

= 36,960 cm³

Let the radius and height of the cylinder be ‘r’ and ‘h’ respectively.

Then, r + h = 37 cm

Given,

Total surface area of the cylinder = 1628 cm²

⇒ 2πr(h + r) = 6512

⇒ r = 7 cm

⇒ height, h = 37 – 7 = 30 cm

Then, the volume of the cylinder = πr²h

= 4620 cm³

Radius of a cylindrical pillar, r = 21 cm = 0.21 m

Height of a cylindrical pillar, h = 4 m

(i) Curved surface area of the cylindrical pillar = 2πrh

= 5.28 m²

(ii) Curved surface area of 36 such cylindrical pillars

= 36 × 5.28 m²

= 190.08 m²

Cost of polishing per m² = Rs. 12

Then, the cost of polishing 190.08 m² = Rs. 12 × 190.08 = Rs. 2280.96

For first cylinder, radius = r₁ and height = h₁

For second cylinder, radius = r₂ and height = h₂

Then,

When a thin rectangular card board is rolled along its length.

Height of the cylinder = 22 cm

And, circumference of the base of cylinder = length of the cylinder

⇒ 2πr = 44 cm

⇒ r = 7 cm

Then, the volume of the cylinder formed = πr²h

Let the radius and height of the cylinder be ‘r’ and ‘h’ respectively.

Then,

⇒ 5h = 3h + 3r

⇒ 2h = 3r

⇒ h : r = 3 : 2

**Back answer is 2 : 3, which is incorrect.**

For first cylinder, radius = r₁ and height = h₁

For second cylinder, radius = r₂ and height = h₂

Then,

Correct option: (b) 125 cm²

Lateral surface area of a cube = 100 cm²

⇒ 4(edge)² = 100

⇒ (edge)² = 25

⇒ edge = 5 cm

Then, the volume of the cube = (edge)³ = (5)³ = 125 cm³

Correct option: (c) 8 : 1

For an original cuboid,

Length = l, breadth = b and height = h

For a new cuboid,

Length = 2l, breadth = 2b and height = 2h

Correct option: (b) 2 times

For an original cylinder, radius = r and height = h

Then, for a new cylinder, radius = 2r and height = h/2

Then,

⇒ Volume of new cylinder = 2 × Volume of original cylinder

Correct option: (a) 6 cm

Dimensions of a cuboid,

Length = 9 cm

Breadth = 8 cm

Height = 3 cm

Now, Volume of a cube formed = Volume of a cuboid

⇒ (edge)³ = length × breadth × height

⇒ (edge)³ = 9 cm × 8 cm × 3 cm = 216 cm³

⇒ edge = 6 cm

**Cylinder has only two dimensions – radius and height. Hence, cuboid is incorrectly written as cylinder in the question.**

Correct option: (d) 8 cm

For a cuboid, height = 7 cm

Since, the base of a cuboid is a square.

⇒ length = breadth = s

Given, Volume = 448 cm³

⇒ length × breadth × height = 448

⇒ s × s × 7 = 448

⇒ s² = 64

⇒ s = 8 cm

Therefore, each side of the base measures 8 cm.

Dimensions of a cuboid,

Length, l = 8 m

Breadth, b = 12 m

Height, h = 3.5 m

Then, the total surface area of a cuboid

= 2(l × b + b × h + h × l)

= 2(8 × 12 + 12 × 3.5 + 3.5 × 8)

= 2(96 + 42 + 28)

= 2(166)

= 332 m²

Dimensions of a cuboid,

Length, l = 8 m

Breadth, b = 12 m

Height, h = 3.5 m

Then, the lateral surface area of a cuboid

= 2(l + b)h

= 2(8 + 12) × 3.5

= 140 m²

Dimensions of a wall,

Length = 16 m = 1600 cm

Breadth = 22.5 cm

Height = 3 m = 300 cm

⇒ Volume of a wall = (1600 × 22.5 × 300) cm³

Dimensions of a brick,

Length = 25 cm

Breadth = 11.25 cm

Height = 6 cm

⇒ Volume of a brick = (25 × 11.25 × 6) cm³

Let the length, breadth and height of a cuboid be 6x, 5x and 3x, respectively.

Total surface area of the cuboid = 504 cm²

⇒ 2(l × b + b × h + h × l) = 504

⇒ 2(6x × 5x + 5x × 3x + 3x × 6x) = 504

⇒ 30x² + 15x² + 18x² = 252

⇒ 63x² = 252

⇒ x² = 4

⇒ x = 2 cm

⇒ 6x = 6(2) = 12 cm, 5x = 5(2) = 10 cm and 3x = 3(2) = 6 cm

Thus, the dimensions of a cuboid are 12 cm, 10 cm and 6 cm, respectively.

Now, volume of a cuboid = length × breadth × height

= 12 × 10 × 6 cm³

= 720 cm³

External dimensions of an open wooden box,

Length = 65 cm

Breadth = 34 cm

Height = 25 cm

⇒ External volume of the box = (65 × 34 × 25) cm³ = 55,250 cm³

Internal dimensions of a closed box,

Length = (65 – 2 × 2) cm = 61 cm

Breadth = (34 – 2 × 2) cm = 30 cm

Height = (25 – 2 × 1) cm = 23 cm

⇒ Internal volume of the box = (61 × 30 × 23) cm³= 42,090 cm³

Capacity of the box = Internal volume of the box = 42,090 cm³

Volume of wood used = External Volume – Internal volume

= (55250 – 42090) cm³

= 13160 cm³

Let the radius and the height of the cylindrical toy be r and h respectively.

Curved surface area of the cylindrical toy = 132 cm²

⇒ 2πrh = 132

Volume of the cylindrical toy = 462 cm³

⇒ πr²h = 462

⇒ r = 7 cm

⇒ Diameter = 2r = 14 cm

Now, 2πrh = 132

Therefore, the diameter and the length of the cylindrical toy are 14 cm and 3 cm respectively.

Let the height of the rectangular hall be ‘h’ m.

Given, Perimeter of the rectangular hall = 250 m

⇒ 2(l + b) = 250

Now, the lateral surface area of a rectangular hall = 2(l + b)h

= 250h m²

Given, the cost of painting (250h) cm² at the rate of 10 per m² = 15,000

⇒ 10 × (250h) = 15000

⇒ h = 6 m

Let the breadth of the hall be ‘b’ m.

Then, its length, l = 2b m

Height, h = 3 m

Given, the lateral surface area of a hall = 108 m²

⇒ 2(l + b)h = 108

⇒ 2(2b + b) × 3 = 108

⇒ 3b × 3 = 54

⇒ 9b = 54

⇒ b = 6 m

⇒ l = 2b = 12 m

Therefore, volume of the rectangular hall

= (12 × 6 × 3) m³

= 216 m³

Side of the original cube = 12 cm

Now, volume of the original cube = volume of 8 new cubes

⇒ 12³ = 8(Volume of new cube)

⇒ 1728 = 8(Volume of new cube)

⇒ Volume of new cube = 216

⇒ (side of new cube)³ = 216

⇒ side of new cube = 6 cm

Then,

For a garden roller,

Diameter = 1.4 m

⇒ Radius, r = 0.7 m

And, height, h = 2 m

Now, curved surface area of a garden roller = 2πrh

Therefore, area covered in 50 revolutions = 50 × 8.8 m² = 440 m²

For each pillar,

Radius, r = 28 cm = 0.28 m

And height, h = 4 m

Now, curved surface area of each pillar = 2πrh

Then, curved surface area of 24 such pillars = 24 × 7.04 = 168.96 m²

Given, cost of painting pillars per m² = Rs. 8

Therefore, the cost of painting 168.96 m² = 8 × 168.96 m² = Rs. 1351.68.

Given, the total surface area of a cylinder = 616 cm²

Therefore, volume of the cylinder = πr²h

Area of face I = 120 cm²

⇒ l × b = 120 cm²       …(i)

Area of face II = 72 cm²

⇒ b × h = 72 cm²       …(ii)

Area of face III = 60 cm²

⇒ l × h = 60 cm²       …(iiii)

Multiplying (i), (ii) and (iii),

l² × b² × h² = 120 × 72 × 60

Edge of each cube = 5 cm

Then the dimensions of a resulting cuboid are as follows:

Length, l = 8(5) = 40 cm

Breadth, b = 5 cm

Height, h = 5 cm

Therefore, the total surface area of a cuboid

= 2(l × b + b × h + h × l)

= 2(40 × 5 + 5 × 5 + 5 × 40)

= 2(200 + 25 + 200)

= 2(425)

= 850 cm²

And, volume of a cuboid = l × b × h

= (40 × 5 × 5) cm³

= 1000 cm³

Dimensions of a rectangular swimming pool are as follows:

Length, l = 20 m

Breadth, b = 10 m

Height, h = 2 m

Then, the area of the pool to be tiled

= Lateral surface area of the pool + Area of the base

= 2(l + b)h + (l × b)

= 2(20 + 10) × 2 + (20 × 10)

= 120 + 200

= 320 m²

= 3200000 cm²

Dimensions of a tile:

Length = 40 cm

Breadth = 40 cm

Then, the area of each tile = 40 cm × 40 cm = 1600 cm²

Now,