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ICSE • Class 9 • Mathematics • Ch 20
Estimated Time: 45 Mins
Study Progress: In Progress

Mensuration II

In ICSE Class 9 Mathematics, "Mensuration II" transitions spatial geometry into three dimensions, providing the rigorous analytical formulas and computational algorithms for evaluating the Surface Area (Lateral and Total) and Volume (Capacity) of right prisms, specifically Cuboids and Cubes. A cuboid is a three-dimensional solid bounded by six rectangular faces meeting at right angles, defined by length ($l$), breadth ($b$), and height ($h$). Its fundamental metrics include: Volume $V = l \times b \times h$; Total Surface Area $\text{TSA} = 2(lb + bh + hl)$; Lateral Surface Area (Area of 4 vertical walls) $\text{LSA} = 2(l + b)h$; and Space Diagonal (the maximum length of a rod that can fit inside) $d = \sqrt{l^2 + b^2 + h^2}$. For a cube of side $a$: $V = a^3$, $\text{TSA} = 6a^2$, $\text{LSA} = 4a^2$, and $d = a\sqrt{3}$. The chapter rigorously investigates advanced real-world engineering problems: (1) Open-topped rectangular tanks; (2) Hollow wooden or metallic rectangular boxes with internal and external dimensions governed by uniform wall thickness $t$ ($l_{\text{ext}} = l_{\text{int}} + 2t$, Volume of material $= V_{\text{ext}} - V_{\text{int}}$); (3) Melting, recasting, and conservation of volume (e.g., how many small cubes of side $x$ can be recast from a solid block); (4) Uniform rectangular cross-sections and rate of fluid flow through rectangular pipes (Volume $=$ Cross-sectional Area $\times$ Speed $\times$ Time); and (5) Unit conversions ($1\text{ m}^3 = 1000\text{ liters} = 1,000,000\text{ cm}^3$).

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The Delian Plague and the Riddle of Apollo: Why Doubling a Cube's Volume Requires Higher Mathematics

In 430 BCE, a devastating plague struck the Greek city-state of Athens. Desperate for relief, the citizens sent an emissary to the Oracle of Apollo at the sacred island of Delos. The Oracle replied with a bizarre mathematical command: "To stop the plague, you must double the size of Apollo's cubical altar." The Athenian stonemasons immediately rushed to the altar, measured the sides of the golden cubical block, and chiseled a new altar with double the length, double the breadth, and double the height. But to their horror, the plague grew ten times worse! Why? The Oracle had asked to double the Volume, but by doubling each edge from $a$ to $2a$, the volume had exploded by $2^3 = \mathbf{8\text{ times}}$! Apollo's wrath deepened because the Athenians failed the mathematics of 3D mensuration! How does scaling the side of a cube affect its area versus its volume? How do naval architects calculate the water capacity of massive ship tanks? Let us master the geometry of 3D solids!

Why This Chapter Matters

3D mensuration governs architecture, HVAC room ventilation, packaging design, warehouse storage optimization, civil reservoir water capacity, and industrial casting.

Before You Begin (Prerequisites)

  • Plane mensuration of rectangles and squares from Chapter 19.
  • Pythagoras theorem applied in 2D and 3D space.

What You Will Learn (Core Objectives)

  • Calculate the volume, total surface area, lateral surface area, and diagonal of cuboids and cubes.
  • Solve problems involving open boxes, four-wall room plastering, and whitewashing costs.
  • Compute material volume, capacity, and mass of hollow rectangular boxes with specified wall thickness.
  • Model fluid flow problems through rectangular channels and pipes using volumetric rates.
  • Handle volume conservation during melting, recasting, and reshaping of metal ingots.

Chapter Roadmap & Progression

1 1. Metrics of a Cuboid & Cube
2 2. Hollow Boxes, Thickness & Unit C...
3 3. Worked ICSE Problem Archetypes

Complete Concept Guide (100% Curriculum Coverage)

1. Metrics of a Cuboid & Cube

Fundamental Solid Metrics
A. Cuboid (Length $l$, Breadth $b$, Height $h$):
  • Volume (Capacity): $$V = l \times b \times h = (\text{Base Area}) \times \text{Height}$$
  • Total Surface Area (TSA): $$\text{TSA} = 2(lb + bh + hl)$$
  • Lateral Surface Area (Area of 4 Walls): $$\text{LSA} = 2(l + b)h = (\text{Perimeter of Base}) \times \text{Height}$$
  • Area of an Open Box (without top lid): $$\text{Area} = lb + 2(l + b)h = 2(lh + bh) + lb$$
  • Longest Diagonal (Space Diagonal): $$d = \sqrt{l^2 + b^2 + h^2}$$
B. Cube (All edges equal to $a$):
  • $V = a^3$
  • $\text{TSA} = 6a^2$
  • $\text{LSA} = 4a^2$
  • $d = a\sqrt{3}$

2. Hollow Boxes, Thickness & Unit Conversions

Hollow Solids & Units
A. Hollow Box with Wall Thickness $t$:
  • Internal Dimensions: $l_{\text{int}} = l_{\text{ext}} - 2t, \; b_{\text{int}} = b_{\text{ext}} - 2t, \; h_{\text{int}} = h_{\text{ext}} - 2t$ (for closed box; for open top, $h_{\text{int}} = h_{\text{ext}} - t$).
  • Internal Volume (Capacity): $V_{\text{int}} = l_{\text{int}} \times b_{\text{int}} \times h_{\text{int}}$.
  • Volume of Material (Wood/Metal): $$V_{\text{material}} = V_{\text{ext}} - V_{\text{int}} = (l_{\text{ext}} b_{\text{ext}} h_{\text{ext}}) - (l_{\text{int}} b_{\text{int}} h_{\text{int}})$$
  • $\text{Mass} = \text{Volume of Material} \times \text{Density}$.
B. Mandatory Volumetric Conversions:
  • $1\text{ m}^3 = 1000\text{ litres}$
  • $1\text{ litre} = 1000\text{ cm}^3 = 1\text{ dm}^3$
  • $1\text{ m}^3 = 1,000,000\text{ cm}^3$

3. Worked ICSE Problem Archetypes

Exemplary Solutions
Problem 1: Find the length of the longest rod that can be placed in a room measuring $12\text{ m} \times 9\text{ m} \times 8\text{ m}$.

Solution:

The longest rod that can fit in a rectangular room corresponds to the space diagonal $d$:

$$d = \sqrt{l^2 + b^2 + h^2} = \sqrt{12^2 + 9^2 + 8^2} = \sqrt{144 + 81 + 64} = \sqrt{289} = \mathbf{17\text{ meters}}$$
Problem 2: A solid metal cuboid measuring $18\text{ cm} \times 12\text{ cm} \times 8\text{ cm}$ is melted and recast into identical small cubes of side $2\text{ cm}$. Find the number of cubes formed.

Solution:

By conservation of volume during melting:

$$\text{Total Volume of Cuboid} = 18 \times 12 \times 8 = 1728\text{ cm}^3$$ $$\text{Volume of one small cube} = a^3 = 2^3 = 8\text{ cm}^3$$ $$\text{Number of cubes formed} = \frac{\text{Volume of Cuboid}}{\text{Volume of one cube}} = \frac{1728}{8} = \mathbf{216\text{ cubes}}$$

Key Formulas, Identities & Theorems

Cuboid Volume
$$V = l \times b \times h$$
Capacity in cubic units.
Cuboid Total Surface Area
$$\text{TSA} = 2(lb + bh + hl)$$
Sum of areas of all 6 faces.
Four Walls Area
$$\text{LSA} = 2(l + b)h$$
Lateral surface area.
Space Diagonal
$$d = \sqrt{l^2 + b^2 + h^2}$$
Maximum straight rod length.
Hollow Solid Material Volume
$$V_{\text{material}} = V_{\text{external}} - V_{\text{internal}}$$
Volume of wood or metal.

Mathematics: 3D Cuboid Space Diagonal & Net Surface Layout

3D Mensuration: Cuboid Space Diagonal & Surface Anatomy 3D Cuboid & Space Diagonal d = √(l²+b²+h²) d (Longest Rod) Length l h b Volume = l × b × h  |  TSA = 2(lb + bh + hl) Hollow Box: V_material = V_ext - V_int Internal Cavity (Capacity) t t Formulas for Hollow Rectangular Box: • Internal l = l_ext - 2t, b = b_ext - 2t, h = h_ext - 2t • V_wood = V_external - V_internal • 1 m³ = 1,000 Litres = 1,000,000 cm³

Chapter Summary & 10 Key Takeaways

Takeaway 1
A cuboid is a 3D rectangular box with dimensions l, b, h.
Takeaway 2
Volume of a cuboid is V = l * b * h.
Takeaway 3
Total Surface Area of a cuboid is 2(lb + bh + hl).
Takeaway 4
Lateral Surface Area (Area of 4 walls) is 2(l + b)h.
Takeaway 5
The longest rod (space diagonal) in a cuboid has length d = √(l^2 + b^2 + h^2).
Takeaway 6
For a cube of edge a: V = a^3, TSA = 6a^2, LSA = 4a^2, diagonal d = a√3.
Takeaway 7
In a hollow box with wall thickness t, internal dimensions are reduced by 2t: l_int = l_ext - 2t.
Takeaway 8
Volume of material in a hollow box is V_external - V_internal.
Takeaway 9
1 cubic meter equals 1,000 liters and 1,000,000 cubic centimeters.
Takeaway 10
When a solid is melted and recast, total volume is strictly conserved.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
A water reservoir is $8\text{ m}$ long, $6\text{ m}$ wide, and $2.5\text{ m}$ deep. How many liters of water can it hold?
Reveal Answer & Explanation
Answer: • Volume of the reservoir:
$$V = l \times b \times h = 8 \times 6 \times 2.5 = 120\text{ m}^3$$
• Convert cubic meters to liters ($1\text{ m}^3 = 1000\text{ liters}$):
$$\text{Capacity} = 120 \times 1000 = \mathbf{120,000\text{ liters}}$$
V = 8 * 6 * 2.5 = 120 m^3. Multiply by 1000 to get liters.
2
The total surface area of a cube is $294\text{ cm}^2$. Find: (i) The length of its edge, (ii) Its volume, (iii) The length of its diagonal.
Reveal Answer & Explanation
Answer:

• (i) Length of edge ($a$):

$$\text{TSA} = 6a^2 = 294 \implies a^2 = \frac{294}{6} = 49 \implies a = \sqrt{49} = \mathbf{7\text{ cm}}$$


• (ii) Volume:

$$V = a^3 = 7^3 = \mathbf{343\text{ cm}^3}$$


• (iii) Length of diagonal ($d$):

$$d = a\sqrt{3} = \mathbf{7\sqrt{3}\text{ cm}} \approx 7 \times 1.732 = \mathbf{12.124\text{ cm}}$$


6a^2 = 294 -> a^2 = 49 -> a = 7 cm. V = 7^3 = 343 cm^3. Diagonal = 7√3 cm.
3
The dimensions of a room are $10\text{ m} \times 8\text{ m} \times 4\text{ m}$. Find the cost of whitewashing its four walls and ceiling at the rate of $\text{₹}20\text{ per m}^2$.
Reveal Answer & Explanation
Answer:

• Area of 4 walls (LSA):

$$\text{Area}_{\text{walls}} = 2(l + b)h = 2(10 + 8)(4) = 2(18)(4) = 144\text{ m}^2$$


• Area of ceiling:

$$\text{Area}_{\text{ceiling}} = l \times b = 10 \times 8 = 80\text{ m}^2$$


• Total area to be whitewashed:

$$\text{Total Area} = 144 + 80 = 224\text{ m}^2$$


• Cost:

$$\text{Cost} = 224 \times 20 = \mathbf{\text{₹}4,480}$$


Area = 2(l+b)h + lb = 144 + 80 = 224 m^2. Cost = 224 * 20 = ₹4,480.
4
An open rectangular tank is made of iron sheet. Its outer dimensions are $1.5\text{ m} \times 1.2\text{ m} \times 0.8\text{ m}$ and the iron sheet is $2\text{ cm}$ thick throughout. Find the volume of iron used to make the tank.
Reveal Answer & Explanation
Answer:

• Note the dimensions in centimeters ($1\text{ m} = 100\text{ cm}$):
$l_{\text{ext}} = 150\text{ cm}, \quad b_{\text{ext}} = 120\text{ cm}, \quad h_{\text{ext}} = 80\text{ cm}$

$$V_{\text{ext}} = 150 \times 120 \times 80 = 1,440,000\text{ cm}^3$$


• For an OPEN tank (no top lid), thickness $t = 2\text{ cm}$ is subtracted twice from length and breadth, but only once from height:
- $l_{\text{int}} = 150 - 2(2) = 146\text{ cm}$
- $b_{\text{int}} = 120 - 2(2) = 116\text{ cm}$
- $h_{\text{int}} = 80 - 2 = 78\text{ cm}$
• Internal volume (capacity):

$$V_{\text{int}} = 146 \times 116 \times 78 = 1,321,008\text{ cm}^3$$


• Volume of Iron Used:

$$V_{\text{iron}} = V_{\text{ext}} - V_{\text{int}} = 1,440,000 - 1,321,008 = \mathbf{118,992\text{ cm}^3} \quad (\text{or } \mathbf{0.118992\text{ m}^3})$$


For an open tank, deduct 2t from l and b, but only 1t from height. V_iron = V_ext - V_int.
5
Three solid metal cubes of edges $3\text{ cm}, 4\text{ cm}, 5\text{ cm}$ are melted together to form a single new cube. Find the edge and total surface area of the new cube.
Reveal Answer & Explanation
Answer:

• Sum of volumes of the three cubes:

$$V_{\text{new}} = 3^3 + 4^3 + 5^3 = 27 + 64 + 125 = 216\text{ cm}^3$$


• Let the edge of the new cube be $A$:

$$A^3 = 216 \implies A = \sqrt[3]{216} = \mathbf{6\text{ cm}}$$


• Total Surface Area of new cube:

$$\text{TSA} = 6A^2 = 6(6^2) = 6 \times 36 = \mathbf{216\text{ cm}^2}$$


Total volume = 27 + 64 + 125 = 216. Side A = ∛216 = 6 cm. TSA = 6 * 36 = 216 cm^2.
6
Water flows through a rectangular pipe of cross-section $5\text{ cm} \times 3\text{ cm}$ at the speed of $2\text{ meters per second}$. How many liters of water flow out in $10\text{ minutes}$?
Reveal Answer & Explanation
Answer: • Convert all quantities to centimeters and seconds:
- Area of cross-section $A = 5 \times 3 = 15\text{ cm}^2$
- Speed $v = 2\text{ m/s} = 200\text{ cm/s}$
- Time $t = 10\text{ minutes} = 600\text{ seconds}$
• Total length of water column flowing out in 10 minutes:
$$L = v \times t = 200 \times 600 = 120,000\text{ cm}$$
• Total volume of water:
$$V = A \times L = 15 \times 120,000 = 1,800,000\text{ cm}^3$$
• Convert cubic centimeters to liters ($1000\text{ cm}^3 = 1\text{ liter}$):
$$\text{Liters} = \frac{1,800,000}{1000} = \mathbf{1,800\text{ liters}}$$
Volume = Area * Speed * Time = 15 cm^2 * 200 cm/s * 600 s = 1,800,000 cm^3 = 1,800 liters.
7
If the length, breadth, and height of a cuboid are in the ratio $5 : 3 : 2$ and its volume is $240\text{ cm}^3$, find its dimensions and total surface area.
Reveal Answer & Explanation
Answer:

• Let the dimensions be $l = 5x, b = 3x, h = 2x$.

$$V = l \times b \times h = (5x)(3x)(2x) = 30x^3$$


• Given $V = 240$:

$$30x^3 = 240 \implies x^3 = \frac{240}{30} = 8 \implies x = 2$$


• Dimensions:
- $l = 5(2) = \mathbf{10\text{ cm}}$
- $b = 3(2) = \mathbf{6\text{ cm}}$
- $h = 2(2) = \mathbf{4\text{ cm}}$
• Total Surface Area:

$$\text{TSA} = 2(lb + bh + hl) = 2(10 \times 6 + 6 \times 4 + 4 \times 10) = 2(60 + 24 + 40) = 2(124) = \mathbf{248\text{ cm}^2}$$


30x^3 = 240 -> x = 2. Dimensions are 10, 6, 4 cm. TSA = 2(60 + 24 + 40) = 248 cm^2.
8
How does the volume of a cube change if each of its edges is tripled?
Reveal Answer & Explanation
Answer:

• Let the original edge be $a$. Original volume $V_1 = a^3$.
• If the edge is tripled, the new edge is $a' = 3a$.
• The new volume is:

$$V_2 = (a')^3 = (3a)^3 = 27a^3 = 27V_1$$


• The volume increases by a factor of $27$ times.


(3a)^3 = 27 a^3. Volume becomes 27 times the original.
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