Volume & Surface Area of Solids

(i) Given,

Length = 8.5 m

Breadth = 6.4 m

Height = 50 cm = 0.5 m.

Volume of rectangular solid = l × b × h

= 8.5 × 6.4 × 0.5

= 27.2 m³.

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

= 2(8.5 × 6.4 + 6.4 × 0.5 + 0.5 × 8.5)

= 2(54.4 + 3.2 + 4.25)

= 2(61.85)

= 123.70 m².

Lateral surface area of the solid = [2(l + b)h]

= [2(8.5 + 6.4) × 0.5]

= [2(14.9) × 0.5]

= 14.9 m².

Hence, volume = 27.2 m³, TSA = 123.7 m² and LSA = 14.9 m².

(ii) Given,

Length = 5.6 dm

Breadth = 2.5 dm

Height = 1 m = 10 dm

Volume of rectangular solid = l × b × h

= 5.6 × 2.5 × 10

= 140 dm³.

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

= 2(5.6 × 2.5 + 2.5 × 10 + 10 × 5.6)

= 2(14 + 25 + 56)

= 2(95)

= 190 dm².

Lateral surface area of the solid = [2(l + b)h]

= [2(5.6 + 2.5) × 10]

= [2(8.1) × 10]

= 162 dm².

Hence, volume = 140 dm³, TSA = 190 dm² and LSA = 162 dm².

Given,

Volume = 33 m³

Length = 16.5 m

Height = 8 m

Volume of rectangular solid = l × b × h

⇒ 33 = 16.5 × b × 8

⇒ 33 = 132 × b

⇒ b = $\frac{33}{132}$

⇒ b = 0.25 m

⇒ b = 0.25 × 100 cm = 25 cm.

Hence, width = 25 cm.

Given,

Dimensions of wall :

Length = 6 m = 600 cm.

Height = 5 m = 500 cm

Thickness = 50 cm.

Dimension of brick :

Length = 25 cm

Breadth = 12.5 cm

Height = 7.5 cm

Calculating the volume of the wall,

Total volume of the wall = l × b × h

= 600 × 500 × 50

= 1,50,00,000 cm³.

Calculating the volume of bricks,

Volume of bricks = Volume of the wall - $\frac{1}{20}$ th volume of the wall (occupied by sand and cement)

= 1,50,00,000 - $\frac{1}{20}$ × 1,50,00,000

= 1,50,00,000 - 7,50,000

= 1,42,50,000 cm³.

Calculating volume of one brick,

Volume of one brick = l × b × h

= 25 × 12.5 × 7.5

= 2343.75 cm³.

Number of bricks = $\frac{\text{Total volume of the bricks}}{\text{Volume of one brick}}$

= $\frac{1,42,50,000}{2343.75}$

= 6080.

Hence, number of bricks = 6080.

Given,

Room dimensions :

Length = 12.5 m

Breadth = 6.4 m

Height = 5 m

Students per floor area = 1.6 m²

Calculating the floor area,

Floor area = l × b = 12.5 × 6.4 = 80 m².

Number of students = $\frac{\text{Total floor area}}{\text{Area per student}}$

= $\frac{80}{1.6}$

= 50.

Calculating the Volume of the room,

Volume = l × b × h

= 12.5 × 6.4 × 5

= 400 m³.

Air per student = $\frac{\text{Total volume}}{\text{Number of students}}$

= $\frac{400}{50}$

= 8 m³.

Hence, number of students = 50 and air per student = 8 m³.

The dimensions of room :

Length = 12 m

Breadth = 9 m

Height = 8 m

Calculating the Diagonal of room,

$$\begin{aligned}\text{Diagonal} = \sqrt{l^2 + b^2 + h^2} \\[1em] = \sqrt{12^2 + 9^2 + 8^2} \\[1em] = \sqrt{144 + 81 + 64} \\[1em] = \sqrt{289} \\[1em] = 17 \text{ m}.\end{aligned}$$

Length of the longest rod that can be placed in a room = Length of the diagonal of the room.

∴ Length of the rod = 17 m.

Hence, length of the rod = 17 m.

Given,

Volume of cuboid = 14400 cm³

Height = 15 cm

Given,

The cross-section of the cuboid is a rectangle having its sides in the ratio 5 : 3.

Let length = 5x and breadth = 3x.

We know that,

Volume of cuboid = l × b × h

⇒ 14400 = 5x × 3x × 15

⇒ 15x² = $\frac{14400}{15}$

⇒ 15x² = 960

⇒ x² = $\frac{960}{15}$

⇒ x² = 64

⇒ x = $\sqrt{64}$

⇒ x = 8.

∴ Sides of the rectangle :

⇒ Length = 5x = 5 × 8 = 40 cm.

⇒ Breadth = 3x = 3 × 8 = 24 cm

Perimeter of cross-section = 2(l + b)

= 2(40 + 24)

= 2(64)

= 128 cm.

Hence, perimeter of cross section = 128 cm.

Given,

Area of path = 6500 m².

Rate of gravelling = ₹ 5.60 per m³.

Depth = 14 cm = 0.14 m.

Calculating the volume of the gravel,

Volume = Area × Depth

= 6500 × 0.14

= 910 m³.

Calculating total cost of the gravel,

Total cost = Volume × Rate

= 910 × 5.60

= ₹ 5,096.

Hence, cost of covering the path with gravel = ₹ 5,096.

Given,

Length = 12 m

Cost of papering (4 walls) = ₹ 1,638 at ₹ 6.50 per m².

Cost of matting (Floor) = ₹ 378 at ₹ 3.50 per m².

Calculating the floor area,

Floor area = $\frac{\text{Total cost of matting}}{\text{Rate}}$

= $\frac{378}{3.50}$

= 108 m².

Calculating width of the room,

Floor area = l × b

⇒ 108 = 12 × b

⇒ b = $\frac{108}{12}$ = 9 m

Calculating the area of four walls,

Area of four walls = $\frac{\text{Total cost of papering}}{\text{Rate}}$

= $\frac{1638}{6.5}$

= 252 m².

Calculating the height of the room,

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

⇒ 252 = [2(12 + 9) × h]

⇒ 252 = 2(21) × h

⇒ 252 = 42 × h

⇒ h = $\frac{252}{42}$

⇒ h = 6 m.

Hence, height of the room = 6 m.

Given,

Field dimension = 15 m × 12 m

Pit dimension = 7.5 m × 6 m × 1.5 m

Calculating the volume of the earth dug out,

Volume = 7.5 × 6 × 1.5

= 67.5 m³.

Calculating the remaining area,

Total area of the field = 15 × 12 = 180 m².

Area of pit opening = 7.5 × 6 = 45 m².

Remaining area = 180 - 45 = 135 m².

Calculating the rise in field level,

Rise in level = $\frac{\text{Volume of earth dug out}}{\text{Remaining area}}$

= $\frac{67.5}{135}$

= 0.50 m

= 0.50 × 100 = 50 cm.

Hence, rise in field level = 50 cm.

Given,

l + b + h = 19 cm ........(1)

Diagonal = 11 cm.

By formula,

Diagonal of a cuboid = $\sqrt{l^2 + b^2 + h^2}$

⇒ 11 = $\sqrt{l^2 + b^2 + h^2}$

⇒ 11² = l² + b² + h²

⇒ 121 = l² + b² + h² .........(2)

By formula,

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

Substituting the values from equation (1) and (2) in above equation, we get :

⇒ (19)² = 121 + 2(lb + bh + hl)

⇒ 361 = 121 + 2(lb + bh + hl)

⇒ 2(lb + bh + hl) = 361 - 121

⇒ 2(lb + bh + hl) = 240

We know that,

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

Hence, surface area of the cuboid = 240 cm².

Given,

Side = 6 cm

When three identical cubes are joined end to end they form a cuboid.

∴ New length = 6 + 6 + 6 = 18 cm.

Breadth = 6 cm.

Height = 6 cm.

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

= 2(18 × 6 + 6 × 6 + 6 × 18)

= 2(108 + 36 + 108)

= 2(252)

= 504 cm².

Hence, surface area of cuboid = 504 cm².

Let 'a' be the side of a cube.

(i) Volume of cube = (a)³

= (10)³

= 1000 cm³.

Hence, volume = 1000 cm³.

(ii) By formula,

Total surface area of cube = 6a²

= 6 × 10²

= 600 cm².

Hence, total surface area of cube = 600 cm².

(iii) By formula,

Lateral surface area of cube = 4a²

= 4 × 10²

= 400 cm².

Hence, lateral surface area of cube = 400 cm².

(iv) By formula,

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

= $10\sqrt{3}$

= 10 × 1.732

= 17.32 cm.

Hence, diagonal of cube = 17.32 cm.

Let 'a' be the length of the edge (side of a cube).

(i) Total surface area of cube = 6a²

⇒ 1536 = 6a²

⇒ a² = $\frac{1536}{6}$

⇒ a² = 256

⇒ a = $\sqrt{256}$

⇒ a = 16 cm.

Hence, length of the edge = 16 cm.

(ii) By formula,

Volume of cube = a³

= 16³

= 4096 cm³.

Hence, volume of cube = 4096 cm³.

(iii) Material thickness = 5 mm = 0.5 cm

Outer edge = 16 cm

Inner edge = 16 - (0.5 + 0.5) = 16 - 1 = 15 cm.

Volume of the material = (outer edge)³ - (inner edge)³

= (16)³ - (15)³

= 4096 - 3375

= 721 cm³.

Hence, volume of the material = 721 cm³.

Let length of each side of cube be a cm.

Given,

Volume of each cube = 512 cm³.

Calculating the side of cube,

Volume of cube = a³

512 = a³

a = $\sqrt[3]{512}$

a = 8 cm.

Two cubes are joined end to end to form a cuboid.

∴ Length (l) = a + a = 8 + 8 = 16 cm

Breadth (b) = a = 8 cm

Height (h) = a = 8 cm

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

= 2(16 × 8 + 8 × 8 + 8 × 16)

= 2(128 + 64 + 128)

= 2(320)

= 640 cm².

Hence, total surface area of cuboid = 640 cm².

(i) Given,

External dimensions :

Length (L) = 36 cm

Breadth (B) = 25 cm

Height (H) = 16.5 cm

Internal dimensions:

Since the thickness = 1.5 cm

∴ Length (l) = 36 - (1.5 + 1.5) = 36 - 3 = 33 cm

Breadth (b) = 25 - (1.5 + 1.5) = 25 - 3 = 22 cm

Height (h) = 16.5 - 1.5 = 15 cm (Since, box is open)

Calculating the external volume,

Volume = L × B × H

= 36 × 25 × 16.5

= 14,850 cm³.

Calculating internal volume,

Volume = l × b × h

= 33 × 22 × 15

= 10,890 cm³

Calculating volume of iron,

Volume of iron = External volume - Internal volume

= 14,850 - 10,890

= 3,960 cm³.

Hence, volume of iron = 3,960 cm³.

(ii) Given,

Weight of iron = 15 g

Total weight of the box = Volume of iron × weight of iron.

= 3,960 × 15 g

= 59,400 g.

1000 g = 1 kg

∴ 59,400 g = $\frac{59,400}{1000}$ kg

= 59.4 kg

Hence, weight of the box = 59.4 kg.

Given,

Edge of largest cube = 12 cm

Edge of 1st smaller cube = 6 cm

Edge of 2nd smaller cube = 8 cm.

Let the edge of 3rd smaller cube = x cm

Calculating the volume of larger cube,

Volume = a³

= (12)³

= 1728 cm³.

Calculating the volume of smaller cubes,

Volume of 1st smaller cube = 6³ = 216 cm³.

Volume of 2nd smaller cube = 8³ = 512 cm³.

Volume of 3rd smaller cube = x³.

Since larger cube melts to form a smaller cubes,

∴ Volume of larger cube = Total volume of smaller cubes.

⇒ 1728 = 216 + 512 + x³

⇒ 1728 = 728 + x³

⇒ x³ = 1728 - 728

⇒ x³ = 1000

⇒ x = $\sqrt[3]{1000}$

⇒ x = 10 cm.

Hence, edge of third smaller cube = 10 cm.

(i) Given,

Length = 100 cm

Breadth = 80 cm

Height = 64 cm

Calculating the volume of the cube,

Volume of cube = l × b × h

= 100 × 80 × 64

= 512000 cm³.

Let the edge of cube be a cm,

⇒ a³ = 512000

⇒ a = $\sqrt[3]{512000}$

⇒ a = 80 cm.

Hence, edge of the cube = 80 cm.

(ii) Surface area of the cube = 6a².

= 6 × (80)²

= 6 × 6400

= 38400 cm².

Hence, surface area of the cube = 38400 cm².

(i) Given,

Area = 1875 cm².

We know that,

Diagonal of cube (d) = $\sqrt{3}$ a

The square on the diagonal means a square whose side is equal to the diagonal of the cube.

∴ Area of square = (diagonal)²

⇒ 1875 = (a $\sqrt{3})$²

⇒ 1875 = 3a²

⇒ a² = $\frac{1875}{3}$

⇒ a² = 625

⇒ a = $\sqrt{625}$

⇒ a = 25 cm.

Hence, side of cube = 25 cm.

(ii) Given,

Total surface area of cube = 6a².

= 6 × (25)²

= 6 × 625

= 3750 cm².

Hence, surface area of the cube = 3750 cm².

Let the dimensions of cuboid be,

Length = l

Breadth = b

Height = h

We know that,

Volume of cuboid (V) = l × b × h

Areas of three adjacent faces :

x = l × b

y = b × h

z = h × l

Multiplying these three areas :

xyz = (l × b) × (b × h) × (h × l)

xyz = l² × b² × h²

xyz = (l × b × h)²

Substituting the value of V = l × b × h

∴ xyz = V².

Hence, proved that V² = xyz.

Let the side of a cube = a cm and diagonal = d cm.

Given,

Diagonal of cube = $16\sqrt{3}$ cm.

By formula,

Diagonal of cube (d) = $\sqrt{3}a$

∴ $\sqrt{3}a = 16\sqrt{3}$

a = 16 cm.

Calculating the total surface area of cube,

Total surface area of cube = 6a².

= 6 × 16²

= 6 × 256

= 1536 cm².

Calculating the volume of the cube,

Volume of a cube = a³

= 16³

= 4096 cm³.

Hence, surface area = 1536 cm² and volume = 4096 cm³.

Given,

Tank dimension :

Length = 150 m

Breadth = 100 m

Depth = 7 m

Cross section of pipe = 5 dm × 3.5 dm

= 0.5 m × 0.35 m.

Speed of water = 15 km/hr = 15000 m/hr.

Let time taken by pipe to fill the tank be x hours.

⇒ Volume of tank = Time × Volume of water discharged by pipe (per hour)

⇒ 150 × 100 × 7 = x × (Cross section of pipe × Speed of water)

⇒ 150 × 100 × 7 = x × 0.5 × 0.35 × 15000

⇒ 105000 = x × 2625

⇒ x = $\frac{105000}{2625}$

⇒ x = 40 hours.

Hence, time taken = 40 hours.

Given,

Tank dimension:

Length (l) = 25 m

Depth (h) = 7.5 m

Drop in level = 1.8 m

Volume removed = 540 m³.

(i) Volume removed = length × width × drop in level

⇒ 540 = 25 × width × 1.8

⇒ 540 = width × 45

⇒ Width = $\frac{540}{45}$ = 12 m.

Hence, width of the tank = 12 m.

(ii) Calculating the capacity of the tank,

Capacity of the tank = Length × Width × Depth

= 25 × 12 × 7.5

= 300 × 7.5

= 2250 m³.

Hence, capacity of tank = 2250 m³.

From figure,

The swimming pool cross-section is in the shape of trapezium.

Area of cross-section = $\frac{1}{2} \times \text{ (Sum of // sides)} \times \text{ (Distance between them)}$

= $\frac{1}{2} \times (1.5 + 3.5) \times 30$

= 75 m².

Volume of pool = Area of cross-section × breadth

= 75 × 12

= 900 m³

= 900 × 1000 litres

= 900000 litres.

Hence, 900000 litres of water will fill the pool.

Given,

Length of the metal piece = 2 m = 200 cm.

(i) From figure,

Rectangle ABCD :

Length (AB = CD) = 12 cm

Breadth (BC = AD) = 13 cm.

Area of rectangle ABCD = length × breadth

= 12 × 13 = 156 cm².

Triangle DEF :

Height (DF) = DC - FC = 12 - 8 = 4 cm.

Base (DE) = AE - AD = 16 - 13 = 3 cm.

Area of triangle DEF = $\frac{1}{2}$ × Base × Height

= $\frac{1}{2}$ × 3 × 4

= 6 cm².

Total cross-section area = 156 + 6 = 162 cm².

Hence, area of cross-section = 162 cm².

(ii) Calculating the volume of metal piece,

Volume of metal piece = Area of cross section × Length

= 162 × 200

= 32400 cm³.

Hence, volume of metal piece = 32400 cm³.

(iii) Given,

1 cm³ of metal = 6.5 g.

Calculating the weight of the metal piece,

Total weight of the metal piece = Volume × Weight of metal at 1 cm³

= 32400 × 6.5

= 210600 g.

1000 g = 1 kg

∴ 210600 g = $\frac{210600}{1000}$ kg

= 210.6 kg ≈ 211 kg.

Hence, weight of the piece of the metal = 211 kg.

Let the bottom width be 'a' and top width be 'b'.

Given,

Bottom width (a) = 1.9 m

Top width (b) = 90 cm = 0.9 m

Height (h) = 5 m

Length = 20 m

(i) Cross-sectional area :

$$\begin{aligned}\text{Area of trapezium} = \frac{1}{2} \times \text{ (sum of // sides)} \times \text{ (distance between them)} \\[1em] = \frac{1}{2} \times (1.9 + 0.9) \times 5 \\[1em] = \frac{1}{2} \times 2.8 \times 5 \\[1em] = 1.4 \times 5 \\[1em] = 7 \text{ m}^2.\end{aligned}$$

Hence, area of cross-section = 7 m².

(ii) Volume of concrete :

Volume of concrete = Area of cross-section × Length of the wall.

= 7 × 20

= 140 m³.

Hence, volume of concrete = 140 m³.

From figure,

AD = BC = 8 cm

AB = CD = 6.5 cm

Triangle sides = AE = 5 cm & DE = 5 cm

Triangle base (AD) = 8 cm.

Length of metal piece = 6 m = 600 cm.

(i) Area of cross section :

Calculating the area of rectangle ABCD,

Area of rectangle ABCD = length × breadth

= 8 × 6.5

= 52 cm².

EAD is an isosceles triangle,

In an isosceles triangle, the perpendicular drawn from common vertex to the base, bisects the base.

AF = FD = $\frac{AD}{2} = \frac{8}{2}$ = 4 cm.

By using pythagoras theorem for the triangle EFD,

⇒ Hypotenuse² = Base² + Height²

⇒ ED² = FD² + EF²

⇒ 5² = 4² + EF²

⇒ 25 = 16 + EF²

⇒ EF² = 25 - 16

⇒ EF² = 9

⇒ EF = $\sqrt{9}$

⇒ EF = 3 cm.

∴ Height (EF) = 3 cm

Area of △ EAD = $\frac{1}{2}$ × Base × Height

= $\frac{1}{2}$ × AD × EF

= $\frac{1}{2}$ × 8 × 3

= 4 × 3

= 12 cm².

Total area = Area of rectangle ABCD + Area of triangle EAD

= 52 + 12 = 64 cm².

Hence, area of cross-section = 64 cm².

(ii) Volume of metal:

Volume of metal = Area of cross-section × Length of the metal piece

= 64 × 600

= 38400 cm³.

Hence, volume of metal = 38400 cm³.

Given,

Side of square = x cm.

Thickness of brass = 1 mm = 0.1 cm

Total weight = 5.44 kg = 5440 g

1 cm³ of brass weighs 8.5 g

Density = 8.5 g/cm³

Calculating the volume of the brass,

Volume of the brass = $\frac{\text{Total weight of brass}}{\text{Density of the brass}}$

= $\frac{5440}{8.5}$

= 640 cm³.

Calculating the area of the plate,

Area of the plate = length × breadth

= x × x = x²

We know that,

Volume = Area of the plate × Thickness of the brass.

⇒ 640 = x² × 0.1

⇒ x² = $\frac{640}{0.1}$

⇒ x² = 6400

⇒ x = $\sqrt{6400}$

⇒ x = 80 cm.

Hence, x = 80 cm.

Given,

Area of cross-section of rectangular pipe = 5.4 cm².

Speed of water = 27 km/hr = 2700000 cm/hr

1 hr = 60 minutes

∴ Speed of water = $\frac{2700000}{60}$ = 45000 cm/min.

Calculating the distance traveled by water in 1 minute,

Distance = Speed × time

= 45000 × 1 = 45000 cm.

Calculating the volume of water,

Volume of water which flows out of pipe in 1 min = Area of cross-section × Distance traveled by water in 1 min

= 5.4 × 45000

= 243000 cm³.

1000 cm³ = 1 litre

∴ 243000 cm³ = $\frac{243000}{1000}$ liters = 243 litres.

Hence, 243 litres of water flows out of the pipe in 1 minute.

Given solid figure consists of two cuboids.

Cuboid 1:

Length (l) = 4 cm

Breadth (b) = 6 cm

Height (h) = 3 cm

Cuboid 2:

Length (l) = 4 cm

Breadth (b) = 3 cm

Height (h) = 9 cm

Calculating the volume of cuboid 1,

Volume of cuboid 1 = length × breadth × height

= 4 × 6 × 3

= 72 cm³.

Calculating the volume of cuboid 2,

Volume of cuboid 2 = length × breadth × height

= 4 × 3 × 9

= 108 cm³.

∴ Total volume = 72 + 108

= 180 cm³.

Hence, volume of solid = 180 cm³.

Given,

AB = 8 m

DC = 6 m

Height = 2.4 m

Length of the tunnel = 40 m

AL = BM

From figure, LM = DC = 6 m (since perpendiculars drop from D and C).

From figure,

⇒ AB = AL + LM + MB

⇒ 8 = AL + 6 + BM

Since, AL = BM

∴ 8 = 2AL + 6

⇒ 2AL = 8 - 6

⇒ 2AL = 2

⇒ AL = $\frac{2}{2}$ = 1 m

∴ AL = BM = 1 m.

Calculating the lengths of the sloping sides AD and BC,

In right triangle ALD,

AL = 1 m

DL = 2.4 m

Using pythagoras theorem for the triangle ALD,

⇒ Hypotenuse² = Base² + Height²

⇒ AD² = AL² + DL²

⇒ AD² = 1² + (2.4)²

⇒ AD² = 1 + 5.76

⇒ AD² = 6.76

⇒ AD = $\sqrt{6.76}$

⇒ AD = 2.6 m.

∴ AD = BC = 2.6 m.

(i) Cost of paving the floor:

Area of tunnel = AB × length

= 8 × 40

= 320 m².

Total cost = Area × cost per m²

= 320 × 16

= ₹ 5,120.

Hence, cost of paving the floor = ₹ 5,120.

(ii) Cost of painting internal surface (excluding floor)

Area to be painted are :

Roof DC and Two sloping sides AD and BC.

Calculating the area of DC,

Area = length × DC

= 40 × 6 = 240 m².

Calculating the area of slope AD,

Area = length × AD

= 40 × 2.6 = 104 m².

Calculating the area of slope BC,

Area = Length × BC

= 40 × 2.6 = 104 m².

Total area = 240 + 104 + 104

= 448 m².

Cost rate = ₹ 5 per m²

Calculating the cost of painting the internal surface of the tunnel,

Total cost = Total area × Cost

= 448 × 5

= ₹ 2,240.

Hence, cost = ₹ 2,240.

Let the original side of cube be a units and side of cube after reduction be a'.

According to the question,

a' = a - 25% of a = a - 0.25a = 0.75a = $\frac{3}{4}a$.

We know that,

Volume of cube = (side)³.

Calculating original volume,

V₁ = a³.

Calculating the volume after reduction,

V₂ = (a')³

= $\Big(\frac{3}{4}a\Big)$³

= $\frac{27}{64}$ a³

Ratio of original volume to the new volume

V₁ : V₂

a³ : $\frac{27}{64}$ a³

$\frac{a^3}{\frac{27}{64}a^3}$

$\frac{64}{27}$

64 : 27.

Hence, option 3 is the correct option.

Let the original side be a units and new side be a' units.

Reduction of 50% :

a' = a - 50% of a = a - $\frac{1}{2}a = \frac{a}{2}$ units.

We know that,

Total surface area of cube (TSA) = 6(side)².

Calculating the original total surface area of cube,

original TSA = 6a²

Calculating the new total surface area of a cube,

new TSA = 6(a')²

= 6 $\Big(\frac{a}{2}\Big)^2$

= 6 × $\frac{a^2}{4}$

= $\frac{3a^2}{2}$.

Ratio of TSA of original cube to the new cube:

original TSA : new TSA

6a² : $\frac{3a^2}{2}$

$\frac{6a^2}{\frac{3a^2}{2}}$

$\frac{4}{1}$

4 : 1.

Hence, option 2 is the correct option.

Given,

Length = 60 cm

Breadth = 50 cm

Total surface area = 14800 cm².

By formula,

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

⇒ 14800 = 2(60 × 50 + 50 × h + h × 60)

⇒ 14800 = 2(3000 + 50h + 60h)

⇒ 3000 + 50h + 60h = $\frac{14800}{2}$

⇒ 3000 + 110h = 7400

⇒ 110h = 7400 - 3000

⇒ 110h = 4400

⇒ h = $\frac{4400}{110}$ = 40 cm.

Hence, option 1 is the correct option.

The base is a square having length and breadth both equal to x cm.

Given,

Height = 4 cm

Volume = 200 cm³.

We know that,

Volume = length × breadth × height

⇒ 200 = x × x × 4

⇒ 200 = x² × 4

⇒ x² = $\frac{200}{4}$

⇒ x² = 50

⇒ x = $\sqrt{50}$

⇒ x = $\sqrt{25 × 2}$

⇒ x = $5\sqrt{2}$ cm.

Hence, option 4 is the correct option.

Given,

Diagonal of cube (d) = $16\sqrt{3} m$.

Let side of cube be a meters.

Diagonal (d) = $a\sqrt{3}$

⇒ $16\sqrt{3} = a\sqrt{3}$

⇒ a = 16 m.

Calculating the volume of cube,

Volume of cube = a³

= 16³

= 4096 m³.

Hence, option 1 is the correct option.

Given,

Total surface area of cube = 96 cm².

By formula,

Total surface of a cube = 6a²

⇒ 96 = 6a²

⇒ a² = $\frac{96}{6}$

⇒ a² = 16

⇒ a = $\sqrt{16}$

⇒ a = 4 cm.

Calculating the length of diagonal of a cube,

Diagonal of a cube = $a\sqrt{3} = 4\sqrt{3}$ cm.

Hence, option 4 is the correct option.

Given,

Length of the diagonal of a cube = $6\sqrt{6}$ cm

Calculating the side of a cube,

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

⇒ $6\sqrt{6} = a\sqrt{3}$

⇒ a = $\frac{6\sqrt{6}}{\sqrt{3}}$

⇒ a = $6\sqrt{2}$ cm.

Calculating the total surface area of a cube,

Total surface area of a cube = 6a²

= 6 × $(6\sqrt{2})^2$

= 6 × 36 × 2

= 432 cm².

Hence, option 2 is the correct option.

Given,

Volume of cube = 125 m³

Calculating the side of a cube,

Volume of cube = a³

⇒ 125 = a³

⇒ a = $\sqrt[3]{125}$

⇒ a = 5 m.

Calculating the total surface area of the cube,

Total surface area of the cube = 6a²

= 6 × 5²

= 6 × 25

= 150 m².

Hence, option 1 is the correct option.

Given,

Length(l) = 12 cm

Height(h) = 5 cm

Total surface area of sheet required to make a closed box = 460 cm²

Let breadth of the box be b cm.

Calculating the breadth of the box,

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

⇒ 460 = 2(12 × b + b × 5 + 5 × 12)

⇒ 460 = 2(12b + 5b + 60)

⇒ 460 = 2(17b + 60)

⇒ 17b + 60 = $\frac{460}{2}$

⇒ 17b + 60 = 230

⇒ 17b = 230 - 60

⇒ 17b = 170

⇒ b = $\frac{170}{17}$ = 10 cm.

Hence, option 1 is the correct option.

Given,

Length (l) = 5 m

Breadth (b) = 4 m

Height (h) = 3 m

Length of the largest rod that can fit in the room is the diagonal of the room.

Calculating the length of the diagonal of the room (cuboid),

$$\begin{aligned}\text{Diagonal of room (d)} = \sqrt{l^2 + b^2 + h^2} \\[1em] = \sqrt{5^2 + 4^2 + 3^2} \\[1em] = \sqrt{25 + 16 + 9} \\[1em] = \sqrt{50} \\[1em] = \sqrt{25 × 2} \\[1em] = 5 \sqrt{2} \text{ m}.\end{aligned}$$

Hence, option 2 is the correct option.

Given,

Sum of dimensions : l + b + h = 24

Length of diagonal (d) = 15 cm.

We know that,

Diagonal of cuboid (d) = $\sqrt{l^2 + b^2 + h^2}$

⇒ 15 = $\sqrt{l^2 + b^2 + h^2}$

Squaring on both sides,

⇒ 15² = $(\sqrt{l^2 + b^2 + h^2})^2$

⇒ 225 = l² + b² + h²

By formula,

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

By substituting the values we get,

⇒ (24)² = 225 + 2(lb + bh + hl)

⇒ 576 = 225 + 2(lb + bh + hl)

⇒ 2(lb + bh + hl) = 576 - 225

⇒ 2(lb + bh + hl) = 351

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

∴ Total surface area of cuboid = 351 cm².

Hence, option 4 is the correct option.

Given,

Length (l) = 8 cm

Breadth (b) = 5 cm

Height (h) = 3 cm

Calculating the volume of cuboid,

Volume of cuboid = l × b × h

= 8 × 5 × 3

= 120 cm³.

Hence, option 1 is the correct option.

Given,

Area of cross-section = 3 cm².

Speed of water = 50 cm/sec.

1 minute = 60 seconds

Calculating the distance traveled by water in 1 minute,

Distance = Speed × Time

Distance = 50 cm/sec × 60 sec

= 3000 cm.

Calculating the volume of water,

Volume = Area of cross-section × Distance by water in 1 minute

= 3 × 3000

= 9000 cm³.

1000 cm³ = 1 litre

∴ 9000 cm³ = $\frac{9000}{1000}$ litres

= 9 litres.

Hence, option 3 is the correct option.

Given,

Sum of sides : l + b + h = 41 cm.

Length of diagonal (d) = 25 cm.

We know that,

Diagonal of cuboid (d) = $\sqrt{l^2 + b^2 + h^2}$

⇒ 25 = $\sqrt{l^2 + b^2 + h^2}$

Squaring on both sides,

⇒ 25² = $(\sqrt{l^2 + b^2 + h^2})^2$

⇒ 625 = l² + b² + h²

By formula,

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

By substituting the values we get,

⇒ (41)² = 625 + 2(lb + bh + hl)

⇒ 1681 = 625 + 2(lb + bh + hl)

⇒ 2(lb + bh + hl) = 1681 - 625

⇒ 2(lb + bh + hl) = 1056

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

∴ Total surface area of cuboid = 1056 cm².

Hence, option 4 is the correct option.

Given,

External length of rectangular box = 16 dm

External breadth of rectangular box = 8.5 dm

Thickness = 5 cm = 0.5 dm

Density of water = 1.2 kg/litre

Weight of box with lid = 60 kg

Weight of box filled with water = 600 kg

Calculating the weight of the water,

Weight of the water = Weight of filled box - Weight of empty box.

= 600 - 60 = 540 kg.

Calculating the volume of water,

Volume of water = $\frac{\text{Weight of the water}}{\text{Density of the water}}$

= $\frac{540}{1.2}$

= 450 litres.

1 litre = 1 dm³

∴ 450 litres = 450 dm³.

Calculating internal dimensions,

Internal length = External length - 2 × Thickness

= 16 - (2 × 0.5)

= 16 - 1 = 15 dm.

Internal breadth = External breadth - 2 × Thickness

= 8.5 - (2 × 0.5)

= 8.5 - 1 = 7.5 dm.

Calculating the internal volume of rectangular box,

Volume of cuboid = l × b × h

⇒ 450 = 15 × 7.5 × h

⇒ 450 = 112.5 × h

⇒ h = $\frac{450}{112.5}$

⇒ h = 4

∴ Internal height = 4 dm.

Calculating the external height of the box,

Since the box has thickness at both the top and bottom,

So,

External height = Internal height + 2(Thickness)

= 4 + 2(0.5)

= 4 + 1

= 5 dm.

Hence, option 1 is the correct option.

Given,

Length (l) = 6 m

Breadth (b) = 4 m

Height (h) = 3 m

Dimension of each window = 1 m × 0.5 m

Dimension of door = 2 m × 1 m

1. Area of one window = 1 × 0.5 = 0.5 m²

∴ Area of two windows = 2 (1 × 0.5) = 1 m²

Area of one door = 2 × 1 = 2 m².

Total area = 2 + 1 = 3 m².

Hence, option (d) is the correct option.

2. Area to be whitewashed = Area of four walls - Area of two windows and a door.

Calculating the area of four walls,

We know that,

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

= 2 × 3 × (6 + 4)

= 6 × 10

= 60 m².

Area of two windows and a door = 3 m².

Area to be whitewashed = 60 - 3 = 57 m².

Hence, option (b) is the correct option.

3. Given,

Tile side = 50 cm = 0.5 m

Calculating the floor area,

Area of floor = Length × Breadth

= 6 × 4

= 24 m².

Area of square tile = (side)²

= (0.5)²

= 0.25 m²

Number of tiles = $\frac{\text{Area of floor}}{\text{Area of each tile}}$

= $\frac{24}{0.25}$

= 96.

Hence, option (c) is the correct option.

4. Volume of air contained in the room = l × b × h

= 6 × 4 × 3

= 72 m³.

Hence, option (a) is the correct option.

5. Length of the longest rod that can be placed in the room = Diagonal of room

$$\begin{aligned}\text{Diagonal (d)} = \sqrt{l^2 + b^2 + h^2} \\[1em] = \sqrt{6^2 + 4^2 + 3^2} \\[1em] = \sqrt{36 + 16 + 9} \\[1em] = \sqrt{61} \\[1em] = 7.81 \approx 8 \text{ m}.\end{aligned}$$

Hence, option (c) is the correct option.

Given,

Internal edge of cube = 18 cm

Thickness of plywood = 1 cm

1. Inside the open cubical box there are 5 faces (4 sides + 1 bottom)

Calculating area of one face,

Area = (side)²

= (18)²

= 324 cm².

Calculating total inner area,

Total inner area = 5 × 324 = 1620 cm².

Thus, the total area to be painted black is 1620 cm².

Hence, option (b) is the correct option.

2. Thickness 1 cm will be on both sides of the box.

∴ Outer edge = Internal edge of cube + 2(Thickness)

= 18 + 2(1)

= 18 + 2 = 20 cm.

Hence, option (a) is the correct option.

3. Outer edge = 20 cm

Lateral surface area refers to the 4 side walls only (not the bottom).

Outer lateral area = 4 × (outer edge)²

= 4 × (20)²

= 4 × 400

= 1600 cm².

Hence, option (d) is the correct option.

4. Total outer surface area of the box (excluding the top)

Total faces excluding the top = 5

Area of one face = (20)² = 400 cm².

Calculating the area of all the faces,

Area of all the faces = 5 × 400 = 2000 cm².

Hence, option (c) is the correct option.

5. The rod must lie completely inside the open box, so the maximum straight length inside the box equals the internal edge length = 18 cm.

Hence, option (a) is the correct option.

Given,

Room dimensions = 10 m × 10 m × 5 m

The longest rod that can be placed in a room = Diagonal of room.

We know that,

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

Diagonal of room = $\sqrt{10^2 + 10^2 + 5^2}$

= $\sqrt{100 + 100 + 25}$

= $\sqrt{225}$

= 15 m.

So length of the longest rod that can be placed in the room = 15 m.

∴ Assertion (A) is true.

By formula,

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

∴ Reason (R) is true.

Both Assertion (A) and Reason (R) are true.

Hence, option 3 is the correct option.

Given,

Perimeter = 20 cm

Each face of a cube is a square.

Let the side of square be a cm.

Perimeter of a square = 4 × side.

⇒ 20 = 4a

⇒ a = $\frac{20}{4}$

⇒ a = 5 cm.

Calculating the volume of a cube,

Volume of cube = a³

= 5³

= 125 cm³.

∴ Assertion (A) is false.

Given,

Edge of a cube = a

By formula,

Volume of a cube = a³

∴ Reason (R) is true.

Assertion (A) is false, Reason (R) is true.

Hence, option 2 is the correct option.

Given,

Volume of water (V) = 2.6 m³

Base area (A) = 6500 cm²

= $\frac{6500}{10000}$ = 0.65 m².

Let the depth of the water be 'd'.

Calculating the depth of the water,

Volume of water = Base area × Depth of water

⇒ 2.6 = 0.65 × d

⇒ d = $\frac{2.6}{0.65}$ = 4 m.

Hence, option 2 is the correct option.

Given,

Volume of cuboid = V

Surface area of cuboid = S

Dimensions of cuboid = x, y, z

Volume of cuboid = length × breadth × height

V = xyz

Addition of the reciprocal values of the given dimensions,

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$

⇒ $\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ = $\frac{yz + xz + xy}{xyz}$

By substituting the value V = xyz, we get

⇒ $\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ = $\frac{yz + xz + xy}{V}$

⇒ V$\Big(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\Big)$ = yz + xz + xy ..........(1)

We know that,

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

S = 2(xy + yz + zx)

⇒ $\frac{S}{2}$ = xy + yz + zx

By substituting the value of xy + yz + zx in equation (1), we get :

$$\begin{aligned}\Rightarrow \frac{S}{2} = V\Big(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\Big) \\[1em] \Rightarrow \frac{1}{V} = \frac{2}{S}\Big(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\Big).\end{aligned}$$

Hence, option 1 is the correct option.

Let the side of the first cube be a units and volume be V cubic units.

We know that,

Surface area of the cube = 6a²

⇒ S = 6a²

⇒ a² = $\frac{S}{6}$

⇒ a = $\sqrt{\frac{S}{6}}$

Calculating the volume of first cube,

Volume of cube (V) = a³

Substituting the value of a, we get,

V = $\Big(\sqrt{\frac{S}{6}}\Big)^3$

V = $\Big(\frac{S}{6}\Big)^\frac{3}{2}$ .......(1)

Let the side of second cube be b units and volume be V' cubic units.

Surface area = 6b²

⇒ 2S = 6b²

⇒ S = 3b²

⇒ b² = $\frac{S}{3}$

⇒ b = $\sqrt{\frac{S}{3}}$

Volume of second cube (V') = b³

Substituting the value of b, we get

V' = $\Big(\sqrt{\frac{S}{3}}\Big)^3$

V' = $\Big(\frac{S}{3}\Big)^\frac{3}{2}$.......(2)

Taking the ratio of equation (1) and (2) we get,

$$\begin{aligned}\Rightarrow \frac{V}{V'} = \frac{\Big(\frac{S}{6}\Big)^\frac{3}{2}}{\Big(\frac{S}{3}\Big)^\frac{3}{2}} \\[1em] \Rightarrow \frac{V}{V'} = \Big(\frac{\frac{S}{6}}{\frac{S}{3}}\Big)^{\frac{3}{2}} \\[1em] \Rightarrow \frac{V}{V'} = \Big(\frac{3}{6}\Big)^\frac{3}{2} \\[1em] \Rightarrow \frac{V}{V'} = \Big(\frac{1}{2}\Big)^\frac{3}{2} \\[1em] \Rightarrow \frac{V}{V'} = \frac{1}{2\sqrt{2}} \\[1em] \Rightarrow V' = 2 \sqrt{2}V\end{aligned}$$

Hence, option 3 is the correct option.

Given,

Total height = 10 cm

Inner vertical step height = 6 cm

So lower height = 10 - 6 = 4 cm

Total front width = 10 cm

Top block width = 5 cm

Bottom block width = 5 cm.

Calculating the area of top face of the tall block,

Area of rectangle = length × breadth

= 5 × 10 = 50 cm².

Calculating the area of top face of the lower block,

Area of rectangle = 5 × 10 = 50 cm².

Calculating the vertical inner face (step)

Length = 6 cm

Breadth = 10 cm

Area = 6 × 10 = 60 cm².

Calculating the area of bottom side outer face,

Length = 10 cm

Breadth = 4 cm

Area = 4 × 10 = 40 cm².

Total shaded area = 50 + 50 + 60 + 40

= 200 cm².

Hence, option 3 is the correct option.

Given,

Number of notes = 50

Volume of bundle = 98 cm³

Length (l) = 14 cm

Breadth (b) = 7 cm

Let the thickness of the bundle be h cm.

Calculating the thickness of the bundle,

Volume of cuboid = length × breadth × height

⇒ 98 = 14 × 7 × h

⇒ 98 = 98 × h

⇒ h = $\frac{98}{98}$ = 1 cm.

So total thickness of 50 notes = 1 cm.

∴ Thickness of 1 note = $\frac{1}{50}$ cm = 0.02 cm.

1 cm = 10 mm

∴ 0.02 cm = 0.02 × 10 = 0.2 mm.

Hence, option 4 is the correct option.

Given,

Let the dimensions of the cuboid be :

Length = l

Breadth = b

Height = h

From the figure (i) the three faces have areas:

x = length × height = lh

y = breadth × height = bh

z = length × breadth = lb

When two identical cuboids, are stacked the height is doubled (2h), length (l) and breadth (b) will remain same.

In (ii) figure,

Surface area of top face = length × breadth = lb = z

Surface area of right vertical face = length × height = 2lh = 2x

Surface area of left vertical face = breadth × height = 2bh = 2y

Total surface area = 2x + 2y + z.

Hence, surface area of new cuboid = 2x + 2y + z.

Given,

Side lengths of each small cube = x units.

From figure,

At each corner of a Rubik's cube is a small cube of length x units.

Outer surface area of cube = 6x²

Hence, outer surface area of the cube at one of the corners = 6x² sq. units.