Solids [Surface Area & volume]

Given:

Area of the square = 48 cm²

Let a be the Side of square.

Area of square = side²

⇒ a² = 48 cm²

⇒ a = $\sqrt{48}$ cm

⇒ a = $\sqrt{16 \times 3}$ cm

⇒ a = 4 $\sqrt{3}$ cm

Let s be the side of cube.

Diagonal of cube = side $\sqrt{3}$

⇒ Side of square = Diagonal of cube

⇒ s $\sqrt{3}$ = 4 $\sqrt{3}$ cm

⇒ s $\cancel{\sqrt{3}}$ = 4 $\cancel{\sqrt{3}}$ cm

⇒ side = 4 cm

Each edge of the cube is 4 cm.

Hence, option 2 is the correct option.

Given:

Length of cuboid = 4 cm

Breadth of cuboid = 3 cm

Diagonal of cuboid = 13 cm

Let h be the height of cuboid.

Length of diagonal of cuboid = $\sqrt{l^2 + b^2 + h^2}$

⇒ $\sqrt{4^2 + 3^2 + h^2}$ = 13

⇒ $\sqrt{16 + 9 + h^2}$ = 13

Squaring both sides, we get

⇒ $({\sqrt{16 + 9 + h^2}})^2$ = (13)²

⇒ 16 + 9 + h² = 169

⇒ 25 + h² = 169

⇒ h² = 169 - 25

⇒ h² = 144

⇒ h = $\sqrt{144}$

⇒ h = 12 cm

Volume of cuboid = l x b x h

= 4 x 3 x 12 cm²

= 144 cm²

Volume of cuboid is 144 cm².

Hence, option 3 is the correct option.

Given:

Dimensions of cuboid = 12 cm x 9 cm x 2 cm

Let s be the side of cube.

Volume of cuboid = Volume of cube

⇒ side³ = l x b x h

⇒ s³ = 12 x 9 x 2 cm³

⇒ s³ = 216 cm³

⇒ s = $\sqrt[3]{216}$ cm

⇒ s = 6 cm

The edge of the cube is 6 cm.

Hence, option 1 is the correct option.

Given:

Volume of cuboid = 1296 cm³

Dimensions of cuboid = 3 : 2 : 1

Let the sides of cuboid be 3a, 2a and a.

Volume of cuboid = l x b x h

⇒ 3a x 2a x a = 1296 cm³

⇒ 6a³ = 1296 cm³

⇒ a³ = $\frac{1296}{6}$ cm³

⇒ a³ = 216 cm³

⇒ a = $\sqrt[3]{216}$ cm

⇒ a = 6 cm

Sides are 3a, 2a and a = 3 x 6 cm, 2 x 6 cm and 6 cm

= 18 cm, 12 cm and 6 cm

Hence, option 4 is the correct option.

Given:

Dimensions of tank = 12 m, 10 m and 8 m

Dimensions of field = 40 m, 32 m

Let h be the height of field.

Volume of tank = Volume of soil spread on the field

⇒ l_tank x b_tank x h_tank = l_field x b_field x h_field

⇒ 12 x 10 x 8 = 40 x 32 x h

⇒ 960 = 1,280 x h

⇒ h = $\frac{960}{1,280}$ m

⇒ h = 0.75 m = 75 cm

The rise in level of the field is 75 cm.

Hence, option 3 is the correct option.

Given:

Area of cross section = 12 cm²

Speed of water = 20 cm/s

Volume of water flowing out per second = Area of cross section x speed of water

= 12 x 20 cm³

= 240 cm³/s

Volume of water flowing out per minute = 240 x 60 cm³

= 14,400 cm³

= 0.0144 m³

The total volume of water flowing out in one minute is 0.0144 m³.

Hence, option 1 is the correct option.

Volume of solid = Area of base x height

Area of base = Area of trapezium = $\frac{1}{2}$ x (sum of parallel side) x height

= $\frac{1}{2}$ x (7 + 13) x 3

Volume of solid = $\frac{1}{2}$ x (7 + 13) x 3 x 20

Hence, option 4 is the correct option.

Given:

Total surface area = 1216 cm²

Ratio of dimensions = 5 : 4 : 2

Let the sides of rectangular solid be 5a, 4a and 2a.

Total surface area = 2(lb + bh + hl)

⇒ 2(5a x 4a + 4a x 2a + 2a x 5a) = 1216 cm²

⇒ 2(20a² + 8a² + 10a²) = 1216 cm²

⇒ 2 x 38a² = 1216 cm²

⇒ 76a² = 1216 cm²

⇒ a² = $\frac{1216}{76}$ cm²

⇒ a² = 16 cm²

⇒ a = $\sqrt{16}$ cm

⇒ a = 4 cm

Dimensions of rectangular solid are:

Length = 5a = 5 x 4 cm = 20 cm

Breadth = 4a = 4 x 4 cm = 16 cm

Height = 2a = 2 x 4 cm = 8 cm

Hence, length, breadth and height of rectangular solid are 20 cm, 16 cm and 8 cm, respectively.

Given:

Volume of a cube = 729 cm³

Let a be the side of cube.

Volume of a cube = side³

⇒ a³ = 729 cm³

⇒ a = $\sqrt{729}$ cm

⇒ a = 9 cm

Total surface area of a cube = 6a²

= 6 x 9² cm²

= 6 x 81 cm²

= 486 cm²

Hence, the total surface area of a cube is 486 cm².

Given:

Dimensions of hall = 100 m, 60 m and 15 m

Volume of hall = l x b x h

= 100 x 60 x 15 m³

= 90,000 m³

Let n be the number of persons.

Volume of air = Number of person x Air required by each person

⇒ 90,000 m³ = n x 150 m³

⇒ n = $\frac{90,000}{150}$

⇒ n = 600

Hence, the number of people that can sit in the Cinema Hall is 600.

Given:

Number of persons that can sleep in the room = 75

Air required by each person = 16 m³

Let h be the height of the room.

Dimensions of room = 25 m x 9.6 m x h m

Volume of air in the room = Number of persons x Air required by each person

⇒ 25 x 9.6 x h = 75 x 16

⇒ 240 x h = 75 x 16

⇒ 240 x h = 1,200

⇒ h = $\frac{1,200}{240}$

⇒ h = 5 m

Hence, the height of the room is 5 m.

Given:

Edges of three cubes:

Cube 1: s₁ = 3 cm
Cube 2: s₂ = 4 cm
Cube 3: s₃ = 5 cm

Volume of new cube formed = Total volume of the 3 melted cubes

Let S be the edge of the new cube formed from the melted material.

Total Volume of the Three Cubes = V₁ + V₂ + V₃

⇒ s₁³ + s₂³ + s₃³ = S³

⇒ 3³ + 4³ + 5³ = S³

⇒ 27 + 64 + 125 = S³

⇒ S³ = 216

⇒ S = $\sqrt[3]{216}$

⇒ S = 6 cm

Hence, the edge of the new cube is 6 cm.

Given:

Edges of three cubes:

Cube 1: s₁ = x cm
Cube 2: s₂ = 8 cm
Cube 3: s₃ = 10 cm

Edge of the new cube formed from the melted material = 12 cm

Total Volume of the Three Cubes = V₁ + V₂ + V₃

⇒ s₁³ + s₂³ + s₃³ = S³

⇒ x³ + 8³ + 10³ = 12³

⇒ x³ + 512 + 1,000 = 1,728

⇒ x³ + 1,512 = 1,728

⇒ x³ = 1,728 - 1,512

⇒ x³ = 216

⇒ x = $\sqrt[3]{216}$

⇒ x = 6 cm

Hence, x = 6 cm.

Given:

Let a be the side of each cube.

When the cubes are placed adjacently, they form a cuboid with the following dimensions:

Length of cuboid = a + a + a = 3a

Breadth of cuboid = a

Height of cuboid = a

The ratio = $\frac{\text{Tot. surface area of cuboid}}{\text{Sum of tot. surface areas of 3 cubes}}$

$$\begin{aligned}= \frac{2(lb + bh + hl)}{6 \times side^2 + 6 \times side^2 + 6 \times side^2}\\[1em] = \frac{2(3a \times a + a \times a + a \times 3a)}{3 \times 6a^2}\\[1em] = \frac{2(3a^2 + a^2 + 3a^2)}{18a^2}\\[1em] = \frac{2 \times 7a^2}{18a^2}\\[1em] = \frac{14a^2}{18a^2}\\[1em] = \frac{14 \cancel{a^2}}{18 \cancel{a^2}}\\[1em] = \frac{14}{18}\\[1em] = \frac{7}{9}\\[1em]\end{aligned}$$

Hence, the ratio of the total surface area of the resulting cuboid to that of the sum of the total surface areas of the three cubes is 7:9.

Given:

Total cost = ₹ 240

Cost of papering = 75 paise per sq. m. = ₹ 0.75 per sq. m.

Height of the room = 5 m

Ratio of length to breadth = 5 : 3

Let the length and breadth of room be 5a and 3a.

Area of the four walls = 2(l + b)h

= 2(5a + 3a)5

= 2 x 8a x 5 sq. m.

= 80a sq. m.

Total cost = Area of the four walls x Cost of papering

⇒ 240 = 80a x 0.75

⇒ 240 = 60a

⇒ a = $\frac{240}{60}$

⇒ a = 4 m

The length and breadth of room are 5a and 3a = 5 x 4 m and 3 x 4 m

= 20 m and 12 m

Hence, the length and breadth of the room are 20 m and 12 m, respectively.

Given:

Area of playground = 3650 m²

Depth of gravel = 1.2 cm = 0.012 m

Cost of gravel = ₹ 6.40 per cubic metre

Volume of Gravel Needed = Area of playground x Depth

= 3,650 x 0.012 m³

= 43.8 m³

Total cost of the gravel = Volume of Gravel Needed x Cost per cubic metre

= 43.8 x 6.40

= ₹ 280.32

Hence, the cost of covering the playground with gravel is ₹ 280.32.

Given:

Volume of the square plate = 2880 cm³

Thickness of the square plate = 8 mm = 0.8 cm

Side of the square plate = x cm

Volume of the square plate = side x side x thickness

⇒ 2,880 = x $\times$ x $\times$ 0.8

⇒ 2,880 = x² $\times$ 0.8

⇒ x² = $\frac{2,880}{0.8}$

⇒ x² = 3600

⇒ x = $\sqrt{3600}$

⇒ x = 60 cm

Hence, the value of x is 60 cm.

(i) Given:

External dimensions of wooden box = 27 cm, 19 cm and 11 cm.

Thickness of the wood = 1.5 cm

External volume of box = l x b x h

= 27 x 19 x 11

= 5,643 cm ³

Internal volume of box = l x b x h

= (27 - 2 x 1.5) x (19 - 2 x 1.5) x (11 - 2 x 1.5)

= (27 - 3) x (19 - 3) x (11 - 3)

= 24 x 16 x 8

= 3,072 cm ³

Volume of wood = External volume of box - Internal volume of box

= 5,643 cm ³ - 3,072 cm ³

= 2,571 cm ³

Hence, volume of wood in the box is 2,571 cm ³.

(ii) Cost of wood = ₹ 1.20 per cm³

Total cost = Volume of wood x Cost of wood

= ₹ 2,571 x 1.20

= ₹ 3,085.20

Hence, total cost of the box is ₹ 3,085.20.

(iii) Side of cube = 4 cm

Let n be the number of cubes.

Internal volume of box = Number of cubes x Volume of cube

⇒ 3,072 = n x side³

⇒ 3,072 = n x 4³

⇒ 3,072 = n x 64

⇒ n = $\frac{3,072}{64}$

⇒ n = 48

Hence, the number of 4 cm cubes that could be placed into the box is 48.

Given:

Dimensions of the tank = 20 m x 12 m x 8 m

Width of the iron sheet = 2.5 m

Rate of iron sheet = ₹ 12.50 per metre

Area of sheet = Surface area of the tank

⇒ Length_sheet x Width_sheet = Area of 4 walls of the tank + Area of base

⇒ Length_sheet x 2.5 m = 2(l + b)h + l x b

⇒ Length_sheet x 2.5 m = 2(20 + 12)8 + 20 x 12 m²

⇒ Length_sheet x 2.5 m = 2 x 32 x 8 + 240 m²

⇒ Length_sheet x 2.5 m = 512 + 240 m²

⇒ Length_sheet x 2.5 m = 752 m²

⇒ Length_sheet = $\frac{752}{2.5}$ m

⇒ Length_sheet = 300.8 m

Cost of the sheet = Length of the sheet x Rate of iron sheet

= 300.8 x 12.50

= ₹ 3,760

Hence, the total cost of the iron sheet is ₹ 3,760.

Given:

External length of the box = 78 cm

External breadth of the box = 19 cm

Thickness of the wood = 1.5 cm

Volume of the box = 15 dm³ = 15,000 cm³

Let h be the external height of the box.

Volume of the box = l x b x h

⇒ (78 - 2 x 1.5) x (19 - 2 x 1.5) x (h - 2 x 1.5) cm³ = 15,000 cm³

⇒ (78 - 3) x (19 - 3) x (h - 3) cm³ = 15,000 cm³

⇒ 75 x 16 x (h - 3) cm³ = 15,000 cm³

⇒ 1,200 x (h - 3) cm³ = 15,000 cm³

⇒ (h - 3) = $\frac{15,000}{1,200}$ cm

⇒ h - 3 = 12.5 cm

⇒ h = 12.5 + 3 cm

⇒ h = 15.5 cm

Hence, the exterior height of the box is 15.5 cm.

(i) Given:

Area of the square = 1875 cm²

Let a be the side of the square.

Let s be the side of cube.

Area of the square = side²

⇒ a² = 1875 cm²

⇒ a = $\sqrt{1875}$ cm

⇒ a = 25 $\sqrt{3}$ cm

Side of the square = Diagonal of the cube

Diagonal of the cube = side $\sqrt{3}$

⇒ s $\sqrt{3}$ = 25 $\sqrt{3}$ cm

⇒ side $\cancel{\sqrt{3}}$ = 25 $\cancel{\sqrt{3}}$ cm

⇒ side = 25 cm

Hence, the side of the cube is 25cm.

(ii) Total surface area of the cube = 6 x side²

= 6 x (25)² cm²

= 6 x 625 cm²

= 3,750 cm²

Hence, the total surface area of the cube is 3,750 cm².

The solid can be divided into two cuboids.

Dimensions of cuboid 1:

Length(l₁) = 9 cm Breadth(b₁) = 4 cm Height(h₁) = 3 cm

Dimension of cuboid 2:

Length(l₂) = 6 cm Breadth(b₂) = 4 cm Height(h₂) = 3 cm

Volume of solid = V₁ + V₂

= l₁b₁h₁ + l₂b₂h₂

= 9 x 4 x 3 + 6 x 4 x 3 cm³

= 108 + 72 cm³

= 180 cm³

Hence, the volume of the solid is 180 cm³.

Length of the pool (L) = 40 m

Width of the pool (W) = 15 m

Depth at the shallow end = 1.5 m

Depth at the deep end = 3 m

The cross-section of the pool is a trapezium with:

Parallel sides (depths) as 1.5 m and 3 m.

Height of the trapezium equal to the width of the pool (15 m).

Area of the trapezium = $\frac{1}{2}$ x (sum of parallel lines) x height

= $\frac{1}{2}$ x (1.5 + 3) x 15 m²

= $\frac{1}{2}$ x 4.5 x 15 m²

= $\frac{1}{2}$ x 67.5 m²

= 33.75 m²

Volume of the pool = Area of the trapezium x length

= 33.75 x 40 m³

= 1350 m³ = 13,50,000 litres

Hence, the amount of water required to fill the swimming pool is 13,50,000 litres.

(i) Length of the tunnel = 40 m

Height of the tunnel = 2.4 m

Cross-section dimensions: AB = 7 m, CD = 5 m

AM = BN = $\frac{AB - CD}{2}$ = $\frac{7 - 5}{2}$ = $\frac{2}{2}$ = 1 m

In Δ ADM,

Using the pythagorean theorem in triangle ADM,

Base² + Height² = Hypotenuse²

⇒ AM² + DM² = AD²

⇒ 1² + (2.4)² = AD²

⇒ 1 + 5.76 = AD²

⇒ 6.76 = AD²

⇒ AD = $\sqrt{6.76}$

⇒ AD = 2.6 m

Area of the internal surface of the tunnel (excluding the floor) = Area of vertical walls - Area of floor

= 2.6 x 40 + 5 x 40 + 2.6 x 40 m²

= 104 + 200 + 104 m²

= 408 m²

Total cost of painting = Area of vertical walls x Cost of painting

= ₹ 408 x 5

= ₹ 2,040

Hence, total cost of painting the internal surface = ₹ 2,040.

(ii) Area of floor = 7 x 40 m²

= 280 m²

Total cost of paving = Area of floor x rate of paving

= ₹ (280 x 18)

= ₹ 5,040

Hence, total cost of paving = ₹ 5,040.

(i) Cross-sectional area = 3.2 cm²

Speed of water = 5 m/s = 500 cm/s

Volume of water discharged per second = Cross-sectional area x Speed

= 3.2 x 500

= 1,600

Hence, the volume of water discharged per second is 1600 cm^3.

(ii) Volume of water discharged per min = 1600 x 60

= 96,000 cm³

= $\frac{96,000}{1,000}$ = 96 litres

Hence, the volume of water discharged per minute is 96 litres.

Given:

Cross-sectional area of the hose-pipe = 2 cm²

Volume of water = 1,500 liters = 1,500,000 cm³

Time = 5 minutes = 5 x 60 seconds = 300 seconds

Volume of water flowing per second = $\frac{1,500,000}{300}$

= 5,000 cm³/ sec

Volume of water flowing per second = Area of cross-section x Speed of water

⇒ 5,000 = 2 x Speed of water

⇒ Speed of water = $\frac{5,000}{2}$

⇒ Speed of water = 2,500 cm/s = 25 m/s

Hence, the speed of water through the pipe is 25 m/s.

(i) Area of cross-section = Area of rectangle ABCE + Area of Δ DEF

= length x breadth + $\frac{1}{2}$ x base x height

= 12 x 10 + $\frac{1}{2}$ x (16 - 10) x (12 - 7.5) m²

= 120 + $\frac{1}{2}$ x 6 x 4.5 m²

= 120 + 3 x 4.5 m²

= 120 + 13.5 m²

= 133.5 m²

Hence, the area of cross-section is 133.5 m².

(ii) Length of the piece of metal = 4 m = 400 cm

Volume of the piece of metal = Area of cross-section x length

= 133.5 x 400 cm³

= 53,400 cm³

Weight of 1 cubic cm of metal = 6.6 g

Weight of 53,400 cubic cm metal = 6.6 x 53,400 g

= $\frac{352,440}{1000}$ kg = 352.44 ≈ 352 kg.

Hence, the volume of the piece of metal is 53,400 cm³ and weight of the piece of metal is 352 kg.

Given:

Dimensions of the rectangular tank = 80 cm x 60 cm x 60 cm

Cross-sectional area of the pipe = 1.5 cm²

Speed of water = 3.2 m/s = 3.2 × 100 cm/s = 320 cm/s

Volume of the rectangular tank = 80 x 60 x 60 cm³

= 2,88,000 cm³

Volume of water flowing per second = Cross-sectional area x Speed of water

= 1.5 x 320 cm³/s

= 480 cm³/s

Volume of water flowing per minute = 480 x 60 cm³

= 28,800 cm³/min

Total time required to fill the tank = $\frac{\text{Volume of tank}}{\text{Volume per minute}}$

= $\frac{288,000}{28,800}$ min

= 10 minutes

Hence, the time taken by the pipe to fill the tank is 10 minutes.

Given:

Radius of a sphere = 2r

By formula,

Volume of sphere = $\frac{4}{3}$ π(radius)³

= $\frac{4}{3}$ x π x (2r)³

= $\frac{4}{3}$ x π x 8r³

= $\frac{4 \times 8}{3}$ πr³

= $\frac{32}{3}$ πr³

Hence, option 4 is the correct option.

Given,

Total surface area of a cube = 96 cm²

By formula,

Total surface area of a cube = 6 x side²

⇒ 6 x side² = 96

⇒ side² = $\frac{96}{6}$

⇒ side² = 16

⇒ side = $\sqrt{16}$

⇒ side = 4 cm.

Volume of cube = side³

= 4³

= 64 cm³.

Hence, option 3 is the correct option.

Given,

Side of the original cube = 16 cm

Let s be the side of each smaller cube.

As we know,

Volume of a cube = side³

Volume of the original cube = Volume of 8 identical cubes

⇒ (16)³ = 8 x s³

⇒ 4096 = 8 x s³

⇒ s³ = $\frac{4096}{8}$

⇒ s³ = 512

⇒ s = $\sqrt[3]{512}$

⇒ s = 8 cm.

Hence, option 4 is the correct option.

Given,

The diameters of two solid spheres are in the ratio 5 : 7.

Let the diameters of two spheres be 5a and 7a.

So, their radius will be r₁ = $\frac{5a}{2}$ and r₂ = $\frac{7a}{2}$

By formula,

Curved surface area of sphere = 4πr²

$$\begin{aligned}\text{Ratio of their curved surfaces areas }= \frac{4πr_1^2}{4πr_2^2}\\[1em] = \frac{4 \times π \times \Big(\frac{5a}{2}\Big)^2}{4 \times π \times \Big(\frac{7a}{2}\Big)^2}\\[1em] = \frac{\Big(\frac{5a}{2}\Big)^2}{\Big(\frac{7a}{2}\Big)^2}\\[1em] = \frac{\Big(\frac{25a^2}{4}\Big)}{\Big(\frac{49a^2}{4}\Big)}\\[1em] = \frac{25}{49}\\[1em] = 25 : 49.\end{aligned}$$

Hence, option 4 is the correct option.

Let the original radius and height of a cylinder be r and h units, respectively.

Given, radius of the cylinder is doubled.

∴ New radius = 2r

Let H be the new height of the cylinder.

By formula,

Curved surface area = 2π x radius x height

Original curved surface area = 2πrh

New curved surface area = 2π x (2r) x H = 4πrH

Given,

Curved surface area remains same.

⇒ 2πrh = 4πrH

⇒ H = $\frac{2πr}{4πr}h = \frac{h}{2}$.

Hence, option 3 is the correct option.

Let the original cuboid have length l, breadth b, and height h.

So, sides of new cuboid = 2l, 2b and 2h.

As we know that surface area of the cuboid = 2(lb + bh + hl)

The surface area of new cuboid = 2[(2l) x (2b) + (2b) x (2h) + (2h) x (2l)]

= 2(4lb + 4bh + 4hl)

= 2 x 2 x [2(lb + bh + hl)]

= 4 x surface area of the cuboid

= 2 x 2 x surface area of the cuboid

∴ Both the statements are false.

Hence, option 2 is the correct option.

Let A₁ be the surface area of the original hemisphere.

A₁ = πr² + 2πr² = 3πr².

Let A₂ be the surface area of the resulting hemisphere.

A₂ = π(2r)² + 2π(2r)²

= 3π(2r)²

= 3π x 4r²

= 12πr².

Ratio between the surface area of the original hemisphere and the resulting hemisphere is :

$$\begin{aligned}\text{Ratio } = \frac{A_1}{A_2}\\[1em] = \frac{3πr^2}{12πr^2}\\[1em] = \frac{3}{12}\\[1em] = \frac{1}{4}.\end{aligned}$$

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is the correct option.

Let r be the radius of the sphere. 

From figure,

The radius of the cylinder is also r and the height of the cylinder is 2r.

By formula,

Volume of a cylinder (V_cylinder) = πr²h

= πr²(2r)

= 2πr³.

By formula,

Volume of a sphere (V_sphere) = $\frac{4}{3}$ πr³

Ratio of the volume of the cylinder to the volume of the sphere :

$$\begin{aligned}\text{Ratio} = \frac{V_{cylinder}}{V_{sphere}}\\[1em] = \frac{2πr^3}{\frac{4}{3} πr^3}\\[1em] = \frac{2}{\frac{4}{3}}\\[1em] = \frac{2 \times 3}{4}\\[1em] = \frac{6}{4}\\[1em] = \frac{3}{2} \\[1em] = 3 : 2.\end{aligned}$$

∴ A is false, but R is true.

Hence, option 2 is the correct option.

Given:

Volume of iron = 192 cm³

Length of the tube = 8 cm

Internal side of the square = 5 cm

Let x be the thickness of the tube.

External side of the square = (5 + 2x) cm

Internal volume of the tube = Internal Side x Internal Side x length

= 5 x 5 x 8 cm³

= 200 cm³

External volume of the tube = External side x External side x length

= (5 + 2x) x (5 + 2x) x 8 cm³

= (4x² + 20x + 25) x 8 cm³

= 32x² + 160x + 200 cm³

Volume of iron = External volume - Internal volume

⇒ (32x² + 160x + 200) - 200 cm³ = 192 cm³

⇒ 32x² + 160x + 200 - 200 - 192 = 0

⇒ 32x² + 160x - 192 = 0

Divide through by 32:

⇒ x² + 5x - 6 = 0

⇒ x² + 6x - 1x - 6 = 0

⇒ x(x + 6) - 1(x + 6) = 0

⇒ (x + 6)(x - 1) = 0

⇒ x = - 6 or 1

Since thickness cannot be negative, x = 1.

Hence, the thickness of the pipe is 1 cm.

Given:

Total surface area of the cuboid = 648 cm²

Let a be the side of each cube.

When four identical cubes are placed adjacently, the cuboid's dimensions are:

Length = a + a + a + a = 4a

Breadth = a

Height = a

Total surface area of the cuboid = 2(lb + bh + hl)

⇒ 2(4a x a + a x a + a x 4a) = 648

⇒ 2(4a² + a² + 4a²) = 648

⇒ 2 x 9a² = 648

⇒ 18a² = 648

⇒ a² = $\frac{648}{18}$

⇒ a² = 36

⇒ a = $\sqrt{36}$

⇒ a = 6

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

$\text{The ratio} = \frac{\text{Total surface area of the cuboid}}{\text{Surface Area of one cube}}$

$$\begin{aligned}= \frac{648}{6 \times side^2}\\[1em] = \frac{648}{6 \times 6^2}\\[1em] = \frac{648}{6 \times 36}\\[1em] = \frac{648}{216}\\[1em] = \frac{3}{1}\\[1em]\end{aligned}$$

Hence, the edge of each cube is 6 cm, and the ratio of the surface area of the resulting cuboid to that of one cube is 3:1.

Given:

Length of the sheet = 32 cm

Breadth of the sheet = 26 cm

Side of square = 3 cm

Inner length of the sheet = 32 - 3 - 3 cm = 26 cm

Inner breadth of the sheet = 26 - 3 - 3 cm = 20 cm

Height of the container (equal to the side of the square) = 3 cm

Volume of container = l x b x h

= 26 x 20 x 3 cm³

= 1,560 cm³

Hence, the volume of the container is 1,560 cm³.

Given:

Length of the pool = 18 m

Width of the pool = 8 m

Depth at the deep end = 2 m

Depth at the shallow end = 1.2 m

Volume of the pool = Area of base x width

Area of base(ABCD) = Area of trapezium = $\frac{1}{2}$ x (sum of parallel side) x height

∵ Height (Length of pool) = 18 m

= $\frac{1}{2}$ x (2 + 1.2) x 18 m²

= $\frac{1}{2}$ x 3.2 x 18 m²

= 1.6 x 18 m²

= 28.8 m²

Volume of the pool = 28.8 x 8 m³

= 230.4 m³

Hence, the capacity of the pool is 230.4 m³.

Given:

The dimensions of step 1 = 30 cm x 40 cm x (30 + 30) cm

The dimensions of step 2 = 40 cm x 30 cm x 30 cm

The dimensions of step 3 = 40 cm x 30 cm x 20 cm

Total volume of victory stand = Volume of step 1 + Volume of step 2 + Volume of step 3

(∵ Volume of cuboid = l x b x h)

Total volume of victory stand = 40 x 30 x 60 + 40 x 30 x 30 + 40 x 30 x 20 cm³

= 72,000 + 36,000 + 24,000 cm³

= 1,32,000 cm³

We know that

Total surface area = Area of front and back + Area of vertical faces + Area of top faces

Putting the values, we get

= 2(30 x 20 + 30 x 60 + 30 x 30) + (20 x 40 + 40 x 40 + 30 x 40 + 30 x 40) + 3(30 x 40) cm²

= 2(600 + 1,800 + 900) + (800 + 1,600 + 1,200 + 1,200) + 3 x 1,200 cm²

= 2 x 3,300 + 4,800 + 3,600 cm²

= 6,600 + 4,800 + 3,600 cm²

= 15,000 cm²

Hence, the volume of the victory stand is 1,32,000 cm³ and the surface area is 15,000 cm².

Given:

Perimeter of each face = 32 cm

Perimeter of square = 4 x side

⇒ 4 x side = 32 cm

⇒ side = $\frac{32}{4}$ cm

⇒ side = 8 cm

Surface area of cube = 6 x side²

= 6 x (8)² cm²

= 6 x 64 cm²

= 384 cm²

Volume of cube = side³

= 8³ cm³

= 512 cm³

Hence, the surface area is 384 cm² and the volume is 512 cm³.

Given:

Dimension of the auditorium = 40 m x 30 m x 12 m

Area of the floor = l x b

= 40 x 30 m²

= 1,200 m²

Each student requires 1.2 m² of floor area.

Thus, maximum number of students = $\frac{1,200}{1.2}$

= 1,000

Volume of auditorium = l x b x h

= 40 x 30 x 12 m³

= 14,400 m³

Air available per student = $\frac{14,400}{1000}$ = 14.4 m³

Hence, the maximum number of students that can be accommodated in the auditorium = 1,000 and the volume of air available per student = 14.4 m³.

Given:

Dimensions of the box = 12 cm $\times$ x cm $\times$ 9 cm

Length of longest diagonal = $\sqrt{l^2 + b^2 + h^2}$

$$\begin{aligned}⇒ \sqrt{12^2 + x^2 + 9^2} = 17\\[1em] ⇒ \Big[\sqrt{12^2 + x^2 + 9^2}\Big]^2 = (17)^2\\[1em] ⇒ 12^2 + x^2 + 9^2 = 289\\[1em] ⇒ 144 + x^2 + 81 = 289\\[1em] ⇒ 225 + x^2 = 289\\[1em] ⇒ x^2 = 289 - 225\\[1em] ⇒ x^2 = 64\\[1em] ⇒ x = \sqrt{64}\\[1em] ⇒ x = 8\end{aligned}$$

Hence, the value of x is 8 cm.

(i) Given:

Dimensions of box = 30 cm x 24 cm x 15 cm

Edge of cube = 3 cm

Number of cubes which can be placed along length = $\frac{30}{3}$ = 10

Number of cubes which can be placed along breadth = $\frac{24}{3}$ = 8

Number of cubes which can be placed along height = $\frac{15}{3}$ = 5

The total number of cubes placed = 10 x 8 x 5 = 400

Hence, the number of cubes = 400.

(ii) Given:

Dimensions of box = 30 cm x 24 cm x 15 cm

Edge of cube = 4 cm

Number of cubes which can be placed along length = $\frac{30}{4}$ = 7.5 = 7 (taking only integer value)

Number of cubes which can be placed along breadth = $\frac{24}{4}$ = 6

Number of cubes which can be placed along height = $\frac{15}{4}$ = 3.75 = 3 (taking only integer value)

The total number of cubes placed = 7 x 6 x 3 = 126

Hence, the number of cubes = 126.

(iii) Given:

Dimensions of box = 30 cm x 24 cm x 15 cm

Edge of cube = 5 cm

Number of cubes which can be placed along length = $\frac{30}{5}$ = 6

Number of cubes which can be placed along breadth = $\frac{24}{5}$ = 4.8 = 4 (taking only integer value)

Number of cubes which can be placed along height = $\frac{15}{5}$ = 3

The total number of cubes placed = 6 x 4 x 3 = 72

Hence, the number of cubes = 72.

Given:

Dimensions of the rectangular field = 112 m x 62 m

Edge of each cubical tank = 6 m

Let h be the rise in level.

Volume of 4 tanks = Volume of the removed earth

4 x Volume of 1 tank = (Area of field - 4 x Area of tank) x Rise in level

⇒ 4 x (6)³ = (112 x 62 - 4 x 6²) x h

⇒ 4 x 216 = (6,944 - 4 x 36) x h

⇒ 864 = (6,944 - 144) x h

⇒ 864 = 6,800 x h

⇒ h = $\frac{864}{6,800}$

⇒ h = 0.127 m

⇒ h = 12.7 cm

Hence, the rise in level of the field is 12.7 cm.

Let a be the side of the original cube.

Side of the new cube = a + 3

Volume of the new cube = a³ + 2457

⇒ a³ + 2457 = (a + 3)³

⇒ a³ + 2457 = a³ + 3³ + 3 x a² x 3 + 3 x a x 3²

⇒ $\cancel{a^3}$+ 2457 = $\cancel{a^3}$+ 27 + 9a² + 27a

⇒ 2457 = 27 + 9a² + 27a

⇒ 9a² + 27a - 2457 + 27 = 0

⇒ 9a² + 27a - 2430 = 0

⇒ a² + 3a - 270 = 0

⇒ a² + 18a - 15a - 270 = 0

⇒ a(a + 18) - 15(a + 18) = 0

⇒ (a + 18)(a - 15) = 0

⇒ a = -18 or 15

Since the side cannot be negative, the side of the original cube, a = 15 cm.

Volume of original cube = side³

= (15)³ cm³

= 3375 cm³

When the length of side is reduced by 20%.

New side of the cube = side - 20% of side

= 15 - $\frac{20}{100}$ x 15

= 15 - $\frac{1}{5}$ x 15

= 15 - 3

= 12 cm

New volume of the cube = side³

= (12)³ cm³

= 1728 cm³

Decrease in volume = 3375 - 1728 cm³

= 1647 cm³

Hence, the side of the original cube is 15 cm and the decrease in volume is 1647 cm³.

Given:

Dimensions of the tank = 30 cm x 20 cm x 12 cm

Side of the cube = 10 cm

Initial depth of water in the tank = 6 cm

Initial volume of water in tank = 30 x 20 x 6 = 3600 cm³

Volume of cube = side cm³ = 10 x 10 x 10 = 1000 cm³

Volume of water required to fill the tank to a depth of 10 cm = 30 x 20 x 10 = 6000 cm³

Additional volume of water required to just submerge the cube = Volume of water at 10 cm depth - Volume of water at 6 cm depth - Volume of cube

= 6000 - 3600 - 1000

= 1400 cm³ = 1.4 litre

Hence, the volume of water that must be poured into the tank so that the metal cube is just submerged is 1.4 litre.

Given:

Dimensions of the solid cuboid = 72 cm x 30 cm x 75 cm

Side of the cube = 6 cm

Let n be the number of cubes.

Volume of solid cuboid = Number of cubes x Volume of one cube

⇒ l x b x h = n x side³

⇒ 72 x 30 x 75 = n x (6)³

⇒ 162,000 = n x 216

⇒ n = $\frac{162,000}{216}$

⇒ n = 750

Total surface area of one cube = 6 x side²

= 6 x (6)² cm²

= 216 cm²

Total surface area of 750 cubes = 750 x 216 cm²

= 162,000 cm²

= 16.2 m²

Rate of polishing = ₹ 150 per sq.m

Total cost = Total surface area x Rate of polishing

= ₹ 16.2 x 150

= ₹ 2,430

Hence, the number of cubes formed = 750 and the total cost of polishing = ₹ 2,430.

Given:

Dimensions of tank = 50 cm x 32 cm x 24 cm

Volume of tank = 50 x 32 x 24 cm³

= 38,400 cm³

= 38.4 litre (∵ 100 cm³ = 1 litre)

Average consumption of car = 15 km per litre

Maximum distance covered by car = Volume of tank x Average consumption of car

= 38.4 x 15

= 576 km

Hence, the maximum distance that can be covered by the car is 576 km.

Given:

The ratio of dimensions of the rectangular box = 4 : 2 : 3

Let the dimension be 4a, 2a and 3a.

Total surface area of box = 2(lb + bh + hl)

= 2(4a x 2a + 2a x 3a + 3a x 4a)

= 2(8a² + 6a² + 12a²)

= 2 x 26a²

= 52a²

The difference between cost of covering it with paper at 12 per m² and with paper at the rate of 13.50 per m² = ₹ 1,248

⇒ Total surface area x 13.5 per m² - Total surface area x 12 per m² = ₹ 1,248

⇒ 52a² x 13.5 - 52a² x 12 = ₹ 1,248

⇒ 52a² (13.5 - 12) = ₹ 1,248

⇒ 52a² x 1.5 = ₹ 1,248

⇒ 78a² = ₹ 1,248

⇒ a² = $\frac{1,248}{78}$

⇒ a² = 16

⇒ a = $\sqrt{16}$

⇒ a = 4 m

So, the dimensions are 4a, 2a and 3a

= 4 x 4 m, 2 x 4 m and 3 x 4 m

= 16 m, 8 m and 12 m

Hence, the dimensions of the box are 16 m , 8 m and 12 m.

Let the dimensions of the cuboid be l, b and h cm and diagonal be d cm.

Given,

Volume = 780 cm³

TSA = 562 cm²

d = $13\sqrt{2}$

By formula,

d² = l² + b² + h²

$(13\sqrt{2})^2$ = l² + b² + h²

338 = l² + b² + h²..........(1)

We know that,

Volume of cuboid = l × b × h

780 = l × b × h ..........(2)

By formula,

TSA = 2(lb + bh + hl)

562 = 2(lb + bh + hl)..........(3)

By using identity,

(l + b + h)² = l² + b² + h² + 2(lb + bh + hl)

By substituting the values we get,

(l + b + h)² = 338 + 562

(l + b + h)² = 900

l + b + h = $\sqrt{900}$

l + b + h = 30..........(4)

From equation (2) & (4)

Sum of the dimensions = 30

Product of the dimensions = 780

780 = 5 x 12 x 13

and,

5 + 12 + 13 = 30.

∴ The dimensions are 5, 12 and 13 cm.

l = 5 cm, b = 12 cm, h = 13 cm.

Hence, dimensions of the cuboid = 5 cm, 12 cm and 13 cm.

Given,

Dimensions of warehouse = 88 m × 66 m × 5.5 m

(i) Largest cube size that can fit = HCF of 88, 66 and 5.5

HCF of 88, 66, 5.5 = 5.5.

Hence, largest cube size that can fit = 5.5 m.

(ii) Volume of warehouse = l × b × h

= 88 × 66 × 5.5

= 31944 m³.

Hence, volume of warehouse = 31944 m³.

(iii) Size of a cubical carton = 5.5 m

Volume of cube = (side)³

= (5.5)³

= 166.375 m³.

Hence, volume of a cubical carton = 166.375 m³.

(iv) Maximum number of cartons that can be stored = $\frac{\text{Warehouse volume}}{\text{Cube volume}}$

= $\frac{31944}{166.375}$

= 192.

Hence, maximum number of cartons that can be stored in warehouse = 192.

Given,

Each matchbox size = 6 cm × 4 cm × 1.5 cm

(i) Total number of match boxes used to make table are:

Left support (leg) = 2 boxes

Middle span (the top) = 1 box

Right support(drawers) = 3 boxes

Total = 2 + 1 + 3 = 6 boxes.

Hence, total number of match boxes used to make the table = 6 boxes.

(ii) Area of one base = 6 × 4 = 24 cm²

Since there are two bases:

The area covered by left section : 6 × 1.5 = 9 cm²

The area covered by right section : 6 × 4 = 24 cm²

Total floor area covered by the table = 9 + 24 = 33 cm².

Hence, total floor area covered by the table = 33 cm².

(iii) Area of one base = 6 × 4 = 24 cm².

Since three distinct sections (left, middle, right) make up the table.

So,

Total surface area of table = 3 × 24

= 72 cm².

Hence, total surface area of the top of the table = 72 cm².

(iv) Since matchbox is in cuboid shape.

So, Volume of cuboid = l × b × h

= 6 × 4 × 1.5

= 36 cm³.

Total matchboxes required to make a table = 6

∴ Volume of the table = 6 × 36 = 216 cm³.

Hence, volume of the table = 216 cm³.