CG Board Class 9 Mathematics Chapter 15 Solutions: Cube and Cuboid (घन और घनाभ) – 2026-27

Activity Step:**

1. Draw a 2D net consisting of 6 congruent squares, each of side 8 cm.
2. Cut out the net along its outer border.
3. Fold along the inner edges to form a 3D box and glue the margins together.
4. This forms a cube of side 8 cm with total surface area = 6 × 8² = 384 cm².

Let initial edge of cube, a₁ = 4 cm.

Initial Total Surface Area (TSA₁) = 6 × a₁² = 6 × (4)² = 6 × 16 = 96 cm².

New edge of cube, a₂ = 8 cm.

New Total Surface Area (TSA₂) = 6 × a₂² = 6 × (8)² = 6 × 64 = 384 cm².

Increase in TSA = TSA₂ / TSA₁ = 384 / 96 = 4 times.

Hence, the total surface area will increase by 4 times.

Total Surface Area of a Cuboid = 2(lb + bh + hl)

For a cube, length = breadth = height = a (i.e., l = b = h = a).

Substituting l = a, b = a, and h = a in the cuboid formula:

Total Surface Area of Cube = 2(a × a + a × a + a × a) = 2(a² + a² + a²) = 2(3a²) = 6a².

Given dimensions: Length l = 6 cm, Breadth b = 3 cm, Height h = 2 cm.

Total Surface Area of Cuboid = 2(lb + bh + hl)

= 2(6 × 3 + 3 × 2 + 2 × 6)

= 2(18 + 6 + 12)

= 2 × 36 = 72 cm².

Let dimensions of the mathematics book be: Length l, Breadth b, Height (thickness) h.

• Position 1: Resting flat on the table (Base = l × b).
• Position 2: Standing upright on its longer edge (Base = l × h).
• Position 3: Standing upright on its shorter edge (Base = b × h).

In all three positions, the total surface area remains constant and equal to 2(lb + bh + hl) because changing the orientation of a solid object does not change its outer surface dimensions or total area.

Let assumed dimensions of classroom be: Length l = 8 m, Breadth b = 6 m, Height h = 4 m.

Area of doors and windows = 10 m².

(i) Area of 4 walls (Lateral Surface Area) = 2h(l + b) = 2 × 4 × (8 + 6) = 8 × 14 = 112 m².

Area excluding doors and windows = 112 – 10 = 102 m².

(ii) Area to be white washed = Area of 4 walls (excluding doors/windows) + Area of ceiling

= 102 + (l × b) = 102 + (8 × 6) = 102 + 48 = 150 m².

(iii) Assuming rate of white washing = Rs. 10 per m²,

Total expenditure = 150 × 10 = Rs. 1500.

1. Let length of page l = 25 cm, breadth b = 18 cm.

Area of rectangular sheet = 25 × 18 = 450 cm².

1. Thickness (height) of book h = 2 cm.

Volume of book = Base Area × Height = 450 × 2 = 900 cm³.

1. Let inner volume of almirah shelf = 45,000 cm³.

Number of books that can fit = 45000 / 900 = 50 books.

1. Newspaper estimation: Volume of 1 folded newspaper ≈ 30 cm³.

Total newspapers to fill shelf = 45000 / 30 = 1500.

If 1 newspaper arrives daily, it will fill the shelf in 1500 days.

Given: Total surface area of cube = 1350 m².

Formula: Total surface area of cube = 6a² (where a is the side of the cube)

6a² = 1350

⇒ a² = 1350 / 6 = 225

⇒ a = √225 = 15 m.

Now, Volume of cube = a³

Volume = (15)³ = 15 × 15 × 15 = 3375 m³ (or 3375 cubic meters).

Given: Volume of cuboid = 1200 cm³, Length l = 15 cm, Breadth b = 10 cm.

Formula: Volume of cuboid = l × b × h

1200 = 15 × 10 × h

⇒ 1200 = 150 × h

⇒ h = 1200 / 150 = 8 cm.

Hence, the height of the cuboid is 8 cm.

Let the original side of the cubical box be ‘a’.

Original Total Surface Area (TSA) = 6a².

(i) When side is doubled (new side = 2a):

New TSA = 6(2a)² = 6(4a²) = 4 × (6a²) = 4 times the original surface area.

(ii) When side is tripled (new side = 3a):

New TSA = 6(3a)² = 6(9a²) = 9 × (6a²) = 9 times the original surface area.

(iii) When side is made n times (new side = na):

New TSA = 6(na)² = 6(n²a²) = n² × (6a²) = n² times (or n × n times) the original surface area.

Given:

Perimeter of the floor of the room = 2(l + b) = 250 m.

Rate of white washing = Rs. 10 per sq m.

Total cost of white washing four walls = Rs. 15000.

Area of four walls = Total Cost / Rate

⇒ Area of four walls = 15000 / 10 = 1500 m².

Formula for area of four walls (Lateral Surface Area) = 2(l + b) × h

⇒ Perimeter × Height = 1500

⇒ 250 × h = 1500

⇒ h = 1500 / 250 = 6 m.

Hence, the height of the room is 6 meters.

For Cubical Box:

Side a = 10 cm.

• Lateral Surface Area (LSA) = 4a² = 4 × (10)² = 400 cm².
• Total Surface Area (TSA) = 6a² = 6 × (10)² = 600 cm².

For Cuboidal Box:

Length l = 12.5 cm, Breadth b = 10 cm, Height h = 8 cm.

• Lateral Surface Area (LSA) = 2h(l + b) = 2 × 8 × (12.5 + 10) = 16 × 22.5 = 360 cm².
• Total Surface Area (TSA) = 2(lb + bh + hl) = 2(12.5×10 + 10×8 + 8×12.5) = 2(125 + 80 + 100) = 2(305) = 610 cm².

(i) Comparing LSA:

Cubical box LSA (400 cm²) > Cuboidal box LSA (360 cm²).

Difference = 400 – 360 = 40 cm².

The cubical box has more lateral surface area by 40 cm².

(ii) Comparing TSA:

Cubical box TSA (600 cm²) < Cuboidal box TSA (610 cm²).

Difference = 610 – 600 = 10 cm².

The cubical box has less total surface area by 10 cm².

Population of village = 4000.

Water required per person per day = 150 litres.

Total daily water requirement of village = 4000 × 150 = 6,00,000 litres.

Dimensions of water tank: Length l = 20 m, Breadth b = 15 m, Height h = 6 m.

Capacity (Volume) of tank = l × b × h = 20 × 15 × 6 = 1800 m³.

Since 1 m³ = 1000 litres,

Capacity of tank in litres = 1800 × 1000 = 18,00,000 litres.

Number of days water will last = Capacity of tank / Daily water requirement

= 18,00,000 / 6,00,000 = 3 days.

Given: Depth h = 3 m, Width b = 40 m.

Flow rate of water = 2 km/hour.

Distance covered by water in 1 hour (60 minutes) = 2 km = 2000 m.

Distance covered by water in 1 minute (Length l) = 2000 / 60 = 100 / 3 m.

Volume of water flowing into the sea per minute = l × b × h

= (100 / 3) × 40 × 3 = 100 × 40 = 4000 m³ (or 4000 cubic meters per minute).

Let the length l = 4x, breadth b = 3x, and height h = 2x.

Total Surface Area of Cuboid = 2(lb + bh + hl) = 3328

⇒ 2(4x × 3x + 3x × 2x + 2x × 4x) = 3328

⇒ 2(12x² + 6x² + 8x²) = 3328

⇒ 2(26x²) = 3328

⇒ 52x² = 3328

⇒ x² = 3328 / 52 = 64

⇒ x = √64 = 8 m.

Dimensions of cuboid:

Length l = 4 × 8 = 32 m,

Breadth b = 3 × 8 = 24 m,

Height h = 2 × 8 = 16 m.

Volume of cuboid = l × b × h = 32 × 24 × 16 = 12288 m³ (or 12288 cubic meters).

Volume of 1 cube = a³ = 125 cm³

⇒ Side of each cube a = ∛125 = 5 cm.

When 3 identical cubes are joined end-to-end:

Length of resulting cuboid L = 5 + 5 + 5 = 15 cm,

Breadth B = 5 cm,

Height H = 5 cm.

Total Surface Area of Cuboid = 2(LB + BH + HL)

= 2(15 × 5 + 5 × 5 + 5 × 15)

= 2(75 + 25 + 75) = 2(175) = 350 cm² (or 350 sq cm).

Note: Here “18 l.” in the textbook signifies 18 kilolitres (18 kL = 18,000 litres = 18 m³).

Given: Length l = 20 m, Drop in water level h = 15 cm = 0.15 m.

Volume of water removed = 18 m³.

Volume of removed water = l × b × h

⇒ 18 = 20 × b × 0.15

⇒ 18 = 3 × b

⇒ b = 18 / 3 = 6 m.

Hence, the breadth of the water body is 6 meters.

Dimensions of the wall:

Length L = 10 m = 1000 cm,

Height H = 4 m = 400 cm,

Thickness B = 24 cm.

Volume of wall = L × B × H = 1000 × 24 × 400 = 96,00,000 cm³.

Dimensions of one brick:

Length l = 24 cm, Breadth b = 12 cm, Height h = 8 cm.

Volume of 1 brick = l × b × h = 24 × 12 × 8 = 2304 cm³.

Number of bricks required = Volume of wall / Volume of 1 brick

= 96,00,000 / 2304 = 4166.67 ≈ 4167 bricks.

Given dimensions of pit: Length l = 8 m, Width b = 6 m, Depth h = 3 m.

Volume of pit = l × b × h = 8 × 6 × 3 = 144 m³.

Rate of digging = Rs. 30 per cubic meter.

Total expenditure on digging = Volume × Rate

= 144 × 30 = Rs. 4320.

Dimensions of godown: Length L = 60 m, Breadth B = 25 m, Height H = 10 m.

Volume of godown = L × B × H = 60 × 25 × 10 = 15,000 m³.

Dimensions of 1 wooden crate: l = 1.5 m, b = 1.25 m, h = 0.5 m.

Volume of 1 crate = l × b × h = 1.5 × 1.25 × 0.5 = 0.9375 m³.

Maximum number of crates = Volume of godown / Volume of 1 crate

= 15000 / 0.9375 = 16,000 crates.

Side of original large cube A = 12 cm.

Volume of large cube = A³ = 12³ = 1728 cm³.

Volume of each new small cube = Volume of large cube / 8

= 1728 / 8 = 216 cm³.

Let side of new small cube be ‘a’.

a³ = 216 ⇒ a = ∛216 = 6 cm.

Surface Area Calculations:

• Surface area of large cube = 6A² = 6 × (12)² = 6 × 144 = 864 cm².
• Surface area of small cube = 6a² = 6 × (6)² = 6 × 36 = 216 cm².
• Ratio of surface areas = 864 / 216 = 4 : 1.

Dimensions of the box: Length l = 80 cm, Breadth b = 40 cm, Height h = 20 cm.

Total Surface Area of box = 2(lb + bh + hl)

= 2(80 × 40 + 40 × 20 + 20 × 80)

= 2(3200 + 800 + 1600)

= 2(5600) = 11,200 cm².

Area of 1 square paper sheet = Side × Side = 40 × 40 = 1600 cm².

Number of paper sheets required = Total surface area of box / Area of 1 sheet

= 11,200 / 1600 = 7 sheets.