Volume & Surface area of Solids

Q1: Find the volume, the total surface area and the lateral surface area of the cuboid having:

i. length = 24 cm, breadth = 16 cm and height = 7.5 cm

Step 1: Write the given dimensions.
Length (l) = 24 cm
Breadth (b) = 16 cm
Height (h) = 7.5 cm
Step 2: Find the volume of the cuboid.
Volume = l × b × h
= 24 × 16 × 7.5
= 384 × 7.5
= 2880 cm³
Step 3: Find the total surface area of the cuboid.
Total Surface Area = 2(lb + bh + hl)
= 2(24×16 + 16×7.5 + 7.5×24)
= 2(384 + 120 + 180)
= 2 × 684
= 1368 cm²
Step 4: Find the lateral surface area of the cuboid.
Lateral Surface Area = 2h(l + b)
= 2 × 7.5 (24 + 16)
= 15 × 40
= 600 cm²
Answer: Volume = 2880 cm³, Total Surface Area = 1368 cm², Lateral Surface Area = 600 cm²

ii. length = 10 m, breadth = 35 cm and height = 1.2 m

Step 1: Convert all dimensions into the same unit (meters).
Breadth = 35 cm = 0.35 m
Length (l) = 10 m
Breadth (b) = 0.35 m
Height (h) = 1.2 m
Step 2: Find the volume of the cuboid.
Volume = l × b × h
= 10 × 0.35 × 1.2
= 4.2 m³
Step 3: Find the total surface area of the cuboid.
Total Surface Area = 2(lb + bh + hl)
= 2(10×0.35 + 0.35×1.2 + 1.2×10)
= 2(3.5 + 0.42 + 12)
= 2 × 15.92
= 31.84 m²
Step 4: Find the lateral surface area of the cuboid.
Lateral Surface Area = 2h(l + b)
= 2 × 1.2 (10 + 0.35)
= 2.4 × 10.35
= 24.84 m²
Answer: Volume = 4.2 m³, Total Surface Area = 31.84 m², Lateral Surface Area = 24.84 m²

Q2: Find the capacity of a rectangular cistern whose length is 6 m, breadth 2.5 m and depth 1.4 m. Also find the area of the iron sheet required to make the cistern.

i. Capacity of the Cistern

Step 1: Given dimensions of the cistern (cuboid):
Length \(l = 6\, m\)
Breadth \(b = 2.5\, m\)
Depth \(h = 1.4\, m\)
Step 2: Volume of a cuboid:\[ V = l \times b \times h \]Substitute the values:\[ V = 6 \times 2.5 \times 1.4 \\ V = 21\, m^3 \]Step 3: Convert cubic metres into litres:\[ 1\, m^3 = 1000 \text{ litres} \\ 21 \times 1000 = 21000 \text{ litres} \]Answer: Capacity of the cistern = 21 m³ = 21,000 litres

ii. Area of Iron Sheet Required

Step 4: A closed cistern requires iron sheet equal to the **Total Surface Area (TSA)** of a cuboid.\[ TSA = 2(lb + bh + lh) \]Step 5: Substitute the values:\[ lb = 6 \times 2.5 = 15 \\ bh = 2.5 \times 1.4 = 3.5 \\ lh = 6 \times 1.4 = 8.4 \\ TSA = 2(15 + 3.5 + 8.4) \\ TSA = 2(26.9) \\ TSA = 53.8\, m^2 \]Answer: Area of iron sheet required = 53.8 m²

Q3: A wall of length 13.5 m, width 60 cm and height 1.6 m is to be constructed by using bricks, each of dimensions 22.5 cm by 12 cm by 8 cm. How many bricks will be needed?

Step 1: Convert all dimensions of the wall into the same unit (cm).
Length of wall = 13.5 m = 1350 cm
Width of wall = 60 cm
Height of wall = 1.6 m = 160 cm
Step 2: Find the volume of the wall.
Volume of wall = l × b × h
= 1350 × 60 × 160
= 12,960,000 cm³
Step 3: Write the dimensions of one brick.
Length of brick = 22.5 cm
Breadth of brick = 12 cm
Height of brick = 8 cm
Step 4: Find the volume of one brick.
Volume of one brick = 22.5 × 12 × 8
= 2160 cm³
Step 5: Find the number of bricks required.
Number of bricks = (Volume of wall) ÷ (Volume of one brick)
= 12,960,000 ÷ 2160
= 6000
Answer: Number of bricks required = 6000

Q4: How many planks each measuring 5 m by 24 cm by 10 cm can be stored in a place 15 m long, 4 m wide and 60 cm deep?

Step 1: Convert all dimensions into the same unit (cm).
Length of place = 15 m = 1500 cm
Width of place = 4 m = 400 cm
Depth of place = 60 cm
Step 2: Find the volume of the place.
Volume of place = l × b × h
= 1500 × 400 × 60
= 36,000,000 cm³
Step 3: Write the dimensions of one plank.
Length of plank = 5 m = 500 cm
Breadth of plank = 24 cm
Thickness of plank = 10 cm
Step 4: Find the volume of one plank.
Volume of one plank = 500 × 24 × 10
= 120,000 cm³
Step 5: Find the number of planks that can be stored.
Number of planks = (Volume of place) ÷ (Volume of one plank)
= 36,000,000 ÷ 120,000
= 300
Answer: Number of planks that can be stored = 300

Q5: A classroom is 10 m long, 6.4 m broad and 5 m high. If each student be given 1.6 m² of the floor area, how many students can be accommodated in the room? How many cubic metres of air would each student get?

Step 1: Write the given dimensions of the classroom.
Length (l) = 10 m
Breadth (b) = 6.4 m
Height (h) = 5 m
Step 2: Find the floor area of the classroom.
Floor area = l × b
= 10 × 6.4
= 64 m²
Step 3: Find the number of students that can be accommodated.
Floor area required for each student = 1.6 m²
Number of students = Total floor area ÷ Area per student
= 64 ÷ 1.6
= 40
Step 4: Find the volume of the classroom.
Volume of classroom = l × b × h
= 10 × 6.4 × 5
= 320 m³
Step 5: Find the volume of air available to each student.
Volume of air per student = Total volume ÷ Number of students
= 320 ÷ 40
= 8 m³
Answer: Number of students accommodated = 40 and volume of air per student = 8 m³

Q6: Find the length of the longest pole that can be placed in a room 12 m long, 8 m broad and 9 m high.

Step 1: Write the given dimensions of the room.
Length (l) = 12 m
Breadth (b) = 8 m
Height (h) = 9 m
Step 2: Understand the concept used.
The longest pole that can be placed inside the room is equal to the diagonal of the cuboid.
Step 3: Write the formula for the diagonal of a cuboid.
Diagonal (d) = √(l² + b² + h²)
Step 4: Substitute the given values in the formula.
d = √(12² + 8² + 9²)
= √(144 + 64 + 81)
= √289
Step 5: Find the value of the square root.
√289 = 17
Answer: Length of the longest pole = 17 m

Q7: The volume of a cuboid is 972 m³. If its length and breadth be 16 m and 13.5 m respectively, find its height.

Step 1: Write the given data.
Volume of cuboid = 972 m³
Length (l) = 16 m
Breadth (b) = 13.5 m
Step 2: Write the formula for volume of a cuboid.
Volume = l × b × h
Step 3: Substitute the given values in the formula.
972 = 16 × 13.5 × h
Step 4: Simplify the equation.
16 × 13.5 = 216
972 = 216h
Step 5: Find the value of height.
h = 972 ÷ 216
h = 4.5 m
Answer: Height of the cuboid = 4.5 m

Q8: The volume of a cuboid is 1296 m³. Its length is 24 m and its breadth and height are in the ratio 3 : 2. Find the breadth and height of the cuboid.

Step 1: Write the given data.
Volume of cuboid = 1296 m³
Length (l) = 24 m
Ratio of breadth : height = 3 : 2
Step 2: Assume the breadth and height using the given ratio.
Let breadth = 3x m
Let height = 2x m
Step 3: Write the formula for volume of a cuboid.
Volume = l × b × h
Step 4: Substitute the given and assumed values.
1296 = 24 × 3x × 2x
1296 = 144x²
Step 5: Solve for x.
x² = 1296 ÷ 144
x² = 9
x = 3
Step 6: Find the breadth and height.
Breadth = 3x = 3 × 3 = 9 m
Height = 2x = 2 × 3 = 6 m
Answer: Breadth of the cuboid = 9 m and Height of the cuboid = 6 m

Q9: The surface area of a cuboid is 468 cm². Its length and breadth are 12 cm and 9 cm respectively. Find its height.

Step 1: Write the given data.
Total surface area of cuboid = 468 cm²
Length (l) = 12 cm
Breadth (b) = 9 cm
Step 2: Write the formula for total surface area of a cuboid.
Total Surface Area = 2(lb + bh + hl)
Step 3: Substitute the given values in the formula.
468 = 2(12×9 + 9h + 12h)
Step 4: Simplify the expression.
12 × 9 = 108
468 = 2(108 + 21h)
Step 5: Solve the equation to find height.
468 ÷ 2 = 108 + 21h
234 = 108 + 21h
21h = 234 − 108
21h = 126
h = 126 ÷ 21
h = 6 cm
Answer: Height of the cuboid = 6 cm

Q10: The length, breadth and height of a room are 8 m, 6.5 m and 3.5 m respectively. Find:

i. the area of 4 walls of the room.

Step 1: Write the given dimensions:
Length \(l = 8\, m\), Breadth \(b = 6.5\, m\), Height \(h = 3.5\, m\)
Step 2: Recall the formula for the area of 4 walls (also called the lateral surface area of the cuboid):
Area of 4 walls \(= 2h(l + b)\)
Step 3: Substitute the given values:
\(= 2 \times 3.5 \times (8 + 6.5)\)
\(= 7 \times 14.5 = 101.5\, m^2\)
Answer: Area of 4 walls = \(101.5\, m^2\)

ii. the area of the floor of the room.

Step 1: Area of floor is length \(\times\) breadth:
Area of floor \(= l \times b = 8 \times 6.5 = 52\, m^2\)
Answer: Area of the floor = \(52\, m^2\)

Q11: A room 9 m long, 6 m broad and 3.6 m high has one door 1.4 m by 2 m and two windows, each 1.6 m by 75 cm. Find:

i. the area of 4 walls, excluding doors and windows.

Step 1: Write the given dimensions:
Length \(l = 9\, m\), Breadth \(b = 6\, m\), Height \(h = 3.6\, m\)
Door dimensions = \(1.4\, m \times 2\, m\)
Window dimensions = \(1.6\, m \times 0.75\, m\) (since 75 cm = 0.75 m)
Step 2: Calculate the total area of 4 walls:
Area of 4 walls \(= 2h(l + b)\)
\(= 2 \times 3.6 \times (9 + 6)\)
\(= 7.2 \times 15 = 108\, m^2\)
Step 3: Calculate the total area of the door and windows:
Area of door = \(1.4 \times 2 = 2.8\, m^2\)
Area of one window = \(1.6 \times 0.75 = 1.2\, m^2\)
Area of two windows = \(2 \times 1.2 = 2.4\, m^2\)
Total area of door and windows = \(2.8 + 2.4 = 5.2\, m^2\)
Step 4: Calculate area of walls excluding door and windows:
\(= 108 – 5.2 = 102.8\, m^2\)
Answer: Area of 4 walls excluding doors and windows = \(102.8\, m^2\)

ii. the cost of white-washing the walls from inside at the rate of ₹22.50 per m².

Step 1: Cost of white-washing = Area \(\times\) rate per m²
\(= 102.8 \times 22.50 = ₹2313\)
Answer: Cost of white-washing = ₹2313

iii. the cost of painting its ceiling at ₹25 per m².

Step 1: Area of ceiling = length \(\times\) breadth
\(= 9 \times 6 = 54\, m^2\)
Step 2: Cost of painting ceiling = Area \(\times\) rate per m²
\(= 54 \times 25 = ₹1350\)
Answer: Cost of painting the ceiling = ₹1350

Q12: An assembly hall is 45 m long, 30 m broad and 16 m high. It has five doors, each measuring 4 m by 3.5 m and four windows, each 2.5 m by 1.6 m. Find:

i. the cost of papering its walls at the rate of ₹35 per m²;

Step 1: Write the given dimensions:
Length \(l = 45\, m\), Breadth \(b = 30\, m\), Height \(h = 16\, m\)
Number of doors = 5, each \(4\, m \times 3.5\, m\)
Number of windows = 4, each \(2.5\, m \times 1.6\, m\)
Step 2: Calculate the total area of 4 walls:
Area of 4 walls \(= 2h(l + b)\)
\(= 2 \times 16 \times (45 + 30) = 32 \times 75 = 2400\, m^2\)
Step 3: Calculate total area of doors:
Area of one door = \(4 \times 3.5 = 14\, m^2\)
Area of 5 doors = \(5 \times 14 = 70\, m^2\)
Step 4: Calculate total area of windows:
Area of one window = \(2.5 \times 1.6 = 4\, m^2\)
Area of 4 windows = \(4 \times 4 = 16\, m^2\)
Step 5: Calculate area of walls to be papered (excluding doors and windows):
\(= 2400 – (70 + 16) = 2400 – 86 = 2314\, m^2\)
Step 6: Calculate cost of papering:
Cost \(= 2314 \times 35 = ₹80990\)
Answer: Cost of papering the walls = ₹80990

ii. the cost of carpeting its floor at the rate of ₹154 per m².

Step 1: Calculate the area of the floor:
Area \(= l \times b = 45 \times 30 = 1350\, m^2\)
Step 2: Calculate the cost of carpeting:
Cost \(= 1350 \times 154 = ₹207900\)
Answer: Cost of carpeting the floor = ₹207900

Q13: The length, breadth and height of a cuboid are in the ratio 7 : 6 : 5. If the surface area of the cuboid is 1926 cm², find its dimensions. Also find the volume of the cuboid.

Step 1: Let the common ratio be \(x\). Then,
Length \(l = 7x\), Breadth \(b = 6x\), Height \(h = 5x\)
Step 2: Surface area (SA) formula of cuboid is:
\(SA = 2(lb + bh + hl)\)
Given \(SA = 1926\, cm^2\)
Substitute values:
\(2[(7x)(6x) + (6x)(5x) + (5x)(7x)] = 1926\)
\(2[42x^2 + 30x^2 + 35x^2] = 1926\)
\(2 \times 107x^2 = 1926\)
\(214x^2 = 1926\)
Step 3: Solve for \(x^2\):
\(x^2 = \frac{1926}{214} = 9\)
\(x = \sqrt{9} = 3\)
Step 4: Find the dimensions:
Length \(l = 7 \times 3 = 21\, cm\)
Breadth \(b = 6 \times 3 = 18\, cm\)
Height \(h = 5 \times 3 = 15\, cm\)
Step 5: Calculate the volume \(V\) of the cuboid:
\(V = l \times b \times h = 21 \times 18 \times 15\)
\(= 5670\, cm^3\)
Answer: Dimensions are Length = 21 cm, Breadth = 18 cm, Height = 15 cm
Volume = \(5670\, cm^3\)

Q14: If the areas of the three adjacent faces of a cuboidal box are 120 cm², 72 cm² and 60 cm² respectively, then find the volume of the box.

Step 1: Let the dimensions of the cuboid be \(l\), \(b\) and \(h\).
Given areas of three adjacent faces:
\(lb = 120\)
\(bh = 72\)
\(hl = 60\)
Step 2: Multiply all three equations:
\((lb) \times (bh) \times (hl) = 120 \times 72 \times 60\)
\(\Rightarrow (l^2 b^2 h^2) = 120 \times 72 \times 60\)
\(\Rightarrow (lbh)^2 = 518400\)
Step 3: Take square root on both sides to find volume \(V = lbh\):
\(V = \sqrt{518400} = 720\, cm^3\)
Answer: Volume of the cuboidal box = \(720\, cm^3\)

Q15: A river 2 m deep and 40 m wide is flowing at the rate of 4.5 km per hour. How many cubic metres of water runs into the sea per minute?

Step 1: Write the given data:
Depth \(d = 2\, m\)
Width \(w = 40\, m\)
Speed of river flow \(v = 4.5\, km/h\)
Step 2: Convert speed from km/h to m/min:
\(4.5\, km/h = 4.5 \times 1000\, m/h = 4500\, m/h\)
Convert hours to minutes:
\(4500\, m/h = \frac{4500}{60} = 75\, m/min\)
Step 3: Calculate the volume of water flowing per minute:
Volume \(= \text{Area of cross-section} \times \text{Speed}\)
Area of cross-section = Depth \(\times\) Width = \(2 \times 40 = 80\, m^2\)
Volume per minute = \(80 \times 75 = 6000\, m^3\)
Answer: Volume of water flowing into the sea per minute = \(6000\, m^3\)

Q16: A closed wooden box 80 cm long, 65 cm wide and 45 cm high, is made up 2.5 cm thick. Find:

i. the capacity of the box

Step 1: Write the external dimensions:
Length \(L = 80\, cm\), Breadth \(B = 65\, cm\), Height \(H = 45\, cm\)
Thickness \(t = 2.5\, cm\)
Step 2: Calculate internal dimensions by subtracting thickness from both sides:
Internal length \(l = 80 – 2 \times 2.5 = 80 – 5 = 75\, cm\)
Internal breadth \(b = 65 – 5 = 60\, cm\)
Internal height \(h = 45 – 5 = 40\, cm\)
Step 3: Calculate capacity (volume of inside space):
Capacity \(= l \times b \times h = 75 \times 60 \times 40 = 180000\, cm^3\)
Answer: Capacity of the box = \(180000\, cm^3\)

ii. the weight of the box, if 100 cm³ of wood weighs 8 grams.

Step 1: Calculate volume of the wood used:
Volume of wood \(= \text{External volume} – \text{Internal volume}\)
External volume \(= 80 \times 65 \times 45 = 234000\, cm^3\)
Internal volume (from step 3) \(= 180000\, cm^3\)
Volume of wood \(= 234000 – 180000 = 54000\, cm^3\)
Step 2: Calculate weight:
Given, 100 cm³ weighs 8 grams
So, weight of 1 cm³ = \(\frac{8}{100} = 0.08\) grams
Weight of wood \(= 54000 \times 0.08 = 4320\) grams
Convert grams to kilograms:
\(4320\, g = \frac{4320}{1000} = 4.32\, kg\)
Answer: Weight of the box = \(4.32\, kg\)

Q17: The external dimensions of a wooden box, open at the top are 54 cm by 30 cm by 16 cm. It is made of wood 2 cm thick. Calculate:

i. the capacity of the box

Step 1: Write the external dimensions:
Length \(L = 54\, cm\), Breadth \(B = 30\, cm\), Height \(H = 16\, cm\)
Thickness \(t = 2\, cm\)
Step 2: Calculate internal dimensions by subtracting thickness from length and breadth, and height:
Internal length \(l = 54 – 2 \times 2 = 54 – 4 = 50\, cm\)
Internal breadth \(b = 30 – 4 = 26\, cm\)
Internal height \(h = 16 – 2 = 14\, cm\) (since open at top, thickness subtracted only once from height)
Step 3: Calculate capacity (volume of internal space):
Capacity \(= l \times b \times h = 50 \times 26 \times 14 = 18200\, cm^3\)
Answer: Capacity of the box = \(18200\, cm^3\)

ii. the volume of wood

Step 1: Calculate external volume:
External volume \(= L \times B \times H = 54 \times 30 \times 16 = 25920\, cm^3\)
Step 2: Calculate internal volume (from step 3):
Internal volume \(= 18200\, cm^3\)
Step 3: Volume of wood used:
\(= \text{External volume} – \text{Internal volume} = 25920 – 18200 = 7720\, cm^3\)
Answer: Volume of wood = \(7720\, cm^3\)

Q18: The internal dimensions of a closed box, made up of iron 1 cm thick, are 24 cm by 18 cm by 12 cm. Find the volume of iron in the box.

Step 1: Write the given data:
Internal length \(l = 24\, cm\), Internal breadth \(b = 18\, cm\), Internal height \(h = 12\, cm\)
Thickness \(t = 1\, cm\)
Step 2: Calculate the external dimensions by adding thickness on both sides:
External length \(L = 24 + 2 \times 1 = 26\, cm\)
External breadth \(B = 18 + 2 \times 1 = 20\, cm\)
External height \(H = 12 + 2 \times 1 = 14\, cm\)
Step 3: Calculate external volume:
\(V_{external} = L \times B \times H = 26 \times 20 \times 14 = 7280\, cm^3\)
Step 4: Calculate internal volume:
\(V_{internal} = l \times b \times h = 24 \times 18 \times 12 = 5184\, cm^3\)
Step 5: Calculate volume of iron used:
Volume of iron \(= V_{external} – V_{internal} = 7280 – 5184 = 2096\, cm^3\)
Answer: Volume of iron in the box = \(2096\, cm^3\)

Q19: Find the volume, the total surface area, the lateral surface area and the diagonal of a cube each of whose edges measures:

i. 8 m

Step 1: Edge length \(a = 8\, m\)
Step 2: Calculate volume \(V\):
\(V = a^3 = 8^3 = 512\, m^3\)
Step 3: Calculate total surface area (TSA):
\(TSA = 6a^2 = 6 \times 8^2 = 6 \times 64 = 384\, m^2\)
Step 4: Calculate lateral surface area (LSA):
\(LSA = 4a^2 = 4 \times 8^2 = 4 \times 64 = 256\, m^2\)
Step 5: Calculate diagonal \(d\):
\(d = a \sqrt{3} = 8 \times \sqrt{3} = 8 \times 1.73 = 13.84\, m\)
Answer: Volume = \(512\, m^3\)
Total Surface Area = \(384\, m^2\)
Lateral Surface Area = \(256\, m^2\)
Diagonal = \(13.84\, m\)

ii. 6.5 cm

Step 1: Edge length \(a = 6.5\, cm\)
Step 2: Calculate volume \(V\):
\(V = a^3 = 6.5^3 = 274.625\, cm^3\)
Step 3: Calculate total surface area (TSA):
\(TSA = 6a^2 = 6 \times 6.5^2 = 6 \times 42.25 = 253.5\, cm^2\)
Step 4: Calculate lateral surface area (LSA):
\(LSA = 4a^2 = 4 \times 42.25 = 169\, cm^2\)
Step 5: Calculate diagonal \(d\):
\(d = a \sqrt{3} = 6.5 \times 1.73 = 11.245\, cm\)
Answer: Volume = \(274.625\, cm^3\)
Total Surface Area = \(253.5\, cm^2\)
Lateral Surface Area = \(169\, cm^2\)
Diagonal = \(11.245\, cm\)

iii. 2 cm 6 mm

Step 1: Convert 2 cm 6 mm to cm:
\(2\, cm + 6\, mm = 2 + \frac{6}{10} = 2.6\, cm\)
Step 2: Edge length \(a = 2.6\, cm\)
Step 3: Calculate volume \(V\):
\(V = a^3 = 2.6^3 = 17.576\, cm^3\)
Step 4: Calculate total surface area (TSA):
\(TSA = 6a^2 = 6 \times 2.6^2 = 6 \times 6.76 = 40.56\, cm^2\)
Step 5: Calculate lateral surface area (LSA):
\(LSA = 4a^2 = 4 \times 6.76 = 27.04\, cm^2\)
Step 6: Calculate diagonal \(d\):
\(d = a \sqrt{3} = 2.6 \times 1.73 = 4.498\, cm\)
Answer: Volume = \(17.576\, cm^3\)
Total Surface Area = \(40.56\, cm^2\)
Lateral Surface Area = \(27.04\, cm^2\)
Diagonal = \(4.498\, cm\)

Q20: The surface area of a cube is 1176 cm². Find its volume.

Step 1: Let the edge of the cube be \(a\) cm.
Step 2: Surface area (SA) formula of a cube is: \[ SA = 6a^2 \] Given \(SA = 1176\, cm^2\)
So, \[ 6a^2 = 1176 \]Step 3: Solve for \(a^2\): \[ a^2 = \frac{1176}{6} = 196 \]Step 4: Find \(a\): \[ a = \sqrt{196} = 14\, cm \]Step 5: Calculate volume \(V\): \[ V = a^3 = 14^3 = 2744\, cm^3 \]Answer: Volume of the cube = \(2744\, cm^3\)

Q21: The volume of a cube is 216 cubic cm. Find its surface area.

Step 1: Let the edge of the cube be \(a\) cm.
Step 2: Volume (V) formula of a cube is: \[ V = a^3 \] Given \(V = 216\, cm^3\)
So, \[ a^3 = 216 \]Step 3: Find \(a\): \[ a = \sqrt[3]{216} = 6\, cm \]Step 4: Calculate surface area (SA) of the cube: \[ SA = 6a^2 = 6 \times 6^2 = 6 \times 36 = 216\, cm^2 \]Answer: Surface area of the cube = \(216\, cm^2\)

Q22: The volume of a cube is 343 cm³. Find its surface area.

Step 1: Let the edge of the cube be \(a\) cm.
Step 2: Volume (V) formula of a cube is: \[ V = a^3 \] Given \(V = 343\, cm^3\)
So, \[ a^3 = 343 \]Step 3: Find \(a\): \[ a = \sqrt[3]{343} = 7\, cm \]Step 4: Calculate surface area (SA) of the cube: \[ SA = 6a^2 = 6 \times 7^2 = 6 \times 49 = 294\, cm^2 \]Answer: Surface area of the cube = \(294\, cm^2\)

Q23: A solid piece of metal in the form of a cuboid of dimensions 24 cm × 18 cm × 4 cm is melted down and recasted into a cube. Find the length of each edge of the cube.

Step 1: Write the dimensions of the cuboid:
Length \(l = 24\, cm\), Breadth \(b = 18\, cm\), Height \(h = 4\, cm\)
Step 2: Calculate the volume of the cuboid: \[ V = l \times b \times h = 24 \times 18 \times 4 = 1728\, cm^3 \]Step 3: Since the metal is melted and recasted into a cube, volume remains the same.
Let the edge length of the cube be \(a\). Then, \[ a^3 = 1728 \]Step 4: Find \(a\): \[ a = \sqrt[3]{1728} = 12\, cm \]Answer: Length of each edge of the cube = \(12\, cm\)

Q24: Three cubes of metal with edges 5 cm, 4 cm and 3 cm respectively are melted to form a single cube. Find the lateral surface area of the new cube formed.

Step 1: Find the volumes of the three cubes: \[ V_1 = 5^3 = 125\, cm^3 \\ V_2 = 4^3 = 64\, cm^3 \\ V_3 = 3^3 = 27\, cm^3 \]Step 2: Calculate total volume of the new cube: \[ V = V_1 + V_2 + V_3 = 125 + 64 + 27 = 216\, cm^3 \]Step 3: Let the edge length of the new cube be \(a\). Then, \[ a^3 = 216 \]Step 4: Find \(a\): \[ a = \sqrt[3]{216} = 6\, cm \]Step 5: Calculate lateral surface area (LSA) of the new cube: \[ LSA = 4a^2 = 4 \times 6^2 = 4 \times 36 = 144\, cm^2 \]Answer: Lateral surface area of the new cube = \(144\, cm^2\)