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WBB • Class X • Mathematics • Ch 4

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Rectangular Parallelopiped or Cuboid

Rectangular Parallelopiped or Cuboid forms Chapter 4 of the WBBSE Class 10 Mathematics curriculum Ganit Prakash, introducing students to three-dimensional solid geometry and mensuration. A solid body bounded by six rectangular plane faces, with opposite faces being congruent and parallel, is defined as a rectangular parallelopiped or cuboid. It possesses 6 faces, 12 straight edges, and 8 vertices. Its spatial size is determined by three mutually perpendicular dimensions: length l, breadth b, and height h. The Total Surface Area is given by 2(lb + bh + hl), while the Lateral Surface Area, representing the four vertical walls of a room, equals 2(l + b)h. The interior space or Volume is the product of its three dimensions, V = l * b * h. The longest straight rod that can be placed inside a cuboid corresponds to its body diagonal, calculated as the square root of l^2 + b^2 + h^2. When all three dimensions are equal, the solid becomes a cube of edge a, with total surface area 6a^2, volume a^3, and diagonal a*sqrt(3). The chapter connects geometry with physics via the relation Mass = Volume * Density and establishes crucial metric conversions: 1 cubic decimeter equals 1 liter, and 1 cubic meter equals 1,000 liters. Students tackle diverse practical engineering problems including earthwork excavation, brickwork estimation with mortar deduction, fluid displacement under Archimedes' principle, and open wooden boxes with uniform wall thickness.

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Have You Ever Wondered?

From the fired bricks of ancient Indus Valley cities to modern shipping containers, residential rooms, and water reservoirs, the cuboid is the primary three-dimensional shape of human civilization. How do length, breadth, and height combine to determine total surface area, volume, the longest rod that can fit inside, and fluid storage capacity?

Why This Chapter Matters

Mensuration of solid objects is a major scoring component of the WBBSE Madhyamik examination, carrying compulsory 4-mark and multiple-choice questions. In everyday life and technical professions, cuboids are ubiquitous across civil engineering, architecture, packaging logistics, interior design, and hydraulics. Estimating how many bricks are needed to construct a room, calculating the capacity of an overhead water reservoir, determining the weight of cast iron ingots, or optimizing packaging dimensions all depend entirely on mastering cuboid mensuration.

Before You Begin (Prerequisites)

Area of two-dimensional rectangles (Area = Length * Breadth) and squares (Area = Side^2).
Pythagorean theorem in 2D and its extension to 3D space.
Unit conversions between linear, square, and cubic units (1 m = 100 cm, 1 m^2 = 10,000 cm^2, 1 m^3 = 1,000,000 cm^3).
Metric fluid capacity relations (1 cubic decimeter = 1 liter, 1 cubic meter = 1,000 liters).

What You Will Learn (Core Objectives)

Identify and describe the geometric anatomy of a cuboid and cube: 6 rectangular/square faces, 12 edges, and 8 vertices.
Derive and apply formulas for Total Surface Area: 2(lb + bh + hl) and Lateral Surface Area (4 walls): 2(l + b)h.
Derive and calculate Cuboid Volume: V = l * b * h and body diagonal: D = sqrt(l^2 + b^2 + h^2).
Master Cube formulas: TSA = 6a^2, LSA = 4a^2, Volume = a^3, and body diagonal = a*sqrt(3).
Apply physical relations connecting volume, density, and mass: Mass = Volume * Density.
Solve practical board problems: digging trenches and spreading earth, constructing brick walls, calculating open box capacities with wall thickness, and melting/recasting solids.

Chapter Roadmap & Progression

1 Module 1: Geometric Anatomy & Prope...

2 Module 2: Complete Derivations of S...

3 Module 3: Capacity, Density & Mass...

4 Module 4: Practical Board Applicati...

Complete Concept Guide (100% Curriculum Coverage)

Module 1: Geometric Anatomy & Properties of Cuboid and Cube

1.1 Anatomy of a Cuboid (আয়তঘনকের বৈশিষ্ট্য)

A rectangular parallelopiped or cuboid (আয়তঘন) is a three-dimensional solid bounded by six rectangular plane surfaces where every pair of adjacent faces meets at right angles ($90^\circ$).

Geometric Element	Count	Description & Relationships
Faces (তল)	6	All 6 faces are rectangles. Opposite faces are identical (congruent) and parallel.
Edges (ধার বা প্রান্তরেখা)	12	Consists of 4 lengths ($l$), 4 breadths ($b$), and 4 heights ($h$). Total edge length $= 4(l + b + h)$.
Vertices (শীর্ষবিন্দু)	8	Points where three mutually perpendicular edges intersect.
Diagonals (কর্ণ)	4	Body diagonals connecting opposite opposite vertices through the center of the solid.

1.2 The Cube (ঘনক)

A cube is a special rectangular parallelopiped whose length, breadth, and height are all equal ($l = b = h = a$). All six faces are congruent squares of side length $a$.

Module 2: Complete Derivations of Surface Area, Volume & Diagonal

2.1 Total Surface Area and Lateral Surface Area

A cuboid has 3 pairs of identical rectangular faces:

Top and bottom faces: Area $= 2 imes (l imes b) = 2lb$
Front and back faces: Area $= 2 imes (l imes h) = 2lh$
Left and right side faces: Area $= 2 imes (b imes h) = 2bh$

Total Surface Area (সমগ্রতলের ক্ষেত্রফল): $ ext{TSA} = 2(lb + bh + hl) ext{ sq units}$

Lateral Surface Area / Area of 4 Walls (চার দেওয়ালের ক্ষেত্রফল): $ ext{LSA} = 2(l + b)h = ext{Perimeter of base} imes ext{Height}$

2.2 Volume and Derivation of Body Diagonal

The volume represents the total three-dimensional space enclosed:

Volume (আয়তন): $V = ext{Base Area} imes ext{Height} = l \cdot b \cdot h ext{ cubic units}$

Derivation of the Body Diagonal (কর্ণের দৈর্ঘ্য):
Let the base rectangle have vertices on the floor. By Pythagoras' theorem, the diagonal of the floor base is:
$$d_{ ext{base}} = \sqrt{l^2 + b^2}$$
The vertical height $h$ stands perpendicular to the base diagonal $d_{ ext{base}}$, forming a vertical right-angled triangle. Applying Pythagoras' theorem again to find the body diagonal $D$:
$$D^2 = d_{ ext{base}}^2 + h^2 = (\sqrt{l^2 + b^2})^2 + h^2 = l^2 + b^2 + h^2$$

Body Diagonal: $D = \sqrt{l^2 + b^2 + h^2} ext{ units}$

Summary of Cube Formulas (Edge = a):

• Total Surface Area $= 6a^2 ext{ sq units}$
• Lateral Surface Area (4 walls) $= 4a^2 ext{ sq units}$
• Volume $= a^3 ext{ cubic units}$
• Body Diagonal $= \sqrt{a^2 + a^2 + a^2} = a\sqrt{3} ext{ units}$

Module 3: Capacity, Density & Mass Relations

3.1 Fluid Capacity & Unit Equivalences

When measuring liquids stored in cuboidal tanks, water reservoirs, or cisterns, volume is converted to capacity (liters) using these standard conversions:

Cubic Measure	Capacity in Liters	Practical Rule of Thumb
$1 ext{ dm}^3$ (cubic decimeter)	$1 ext{ liter}$	A cube of edge $10 ext{ cm}$ holds exactly $1 ext{ liter}$.
$1 ext{ m}^3$ (cubic meter)	$1,000 ext{ liters} = 1 ext{ kiloliter}$	A water tank of $1 ext{ m} imes 1 ext{ m} imes 1 ext{ m}$ holds $1,000 ext{ kg}$ of water.
$1,000 ext{ cm}^3$ (cubic centimeters)	$1 ext{ liter}$	$1 ext{ cm}^3 = 1 ext{ milliliter (mL)}$.

3.2 Density and Mass Relations

For solid objects composed of metal, wood, or stone, the total weight or mass is computed using density:

$$ ext{Mass} = ext{Volume} imes ext{Density} \quad (M = V \cdot d)$$

Module 4: Practical Board Applications (Earthwork, Brickwork & Open Boxes)

4.1 Earth Excavation & Rise in Level

When a pit is dug in a corner of a field and the dug-out earth is spread over the remaining area:

$ ext{Volume of dug-out earth} = l_{ ext{pit}} imes b_{ ext{pit}} imes h_{ ext{pit}}$.
$ ext{Remaining area of the field} = ( ext{Total area of field}) - ( ext{Area of pit}) = (L imes B) - (l_{ ext{pit}} imes b_{ ext{pit}})$.
$ ext{Rise in level of field} = rac{ ext{Volume of dug-out earth}}{ ext{Remaining area of field}}$.

4.2 Brick Construction Problems

To calculate the number of bricks required to construct a wall:

$$ ext{Number of Bricks} = rac{ ext{Effective Volume of Wall (after deducting mortar/doors)}}{ ext{Volume of 1 Brick}}$$

4.3 Boxes with Wall Thickness

For a box made of wood of uniform thickness $t$:

Closed Box: Internal dimensions are $(l - 2t)$, $(b - 2t)$, and $(h - 2t)$.
Open Box (open at top): Internal dimensions are $(l - 2t)$, $(b - 2t)$, and $(h - t)$ (thickness deducted once from height).
Volume of material: $ ext{External Volume} - ext{Internal Volume}$.

Key Formulas, Identities & Theorems

Total Surface Area of Cuboid

TSA = 2(lb + bh + hl)

Area of 4 Walls (Lateral Surface Area)

LSA = 2(l + b)h

Volume of Cuboid

V = l * b * h

Body Diagonal of Cuboid

$$D = sqrt(l^2 + b^2 + h^2)$$

Total Surface Area of Cube

$$TSA = 6 * a^2$$

Volume of Cube

$$V = a^3$$

Body Diagonal of Cube

D = a * sqrt(3)

Fluid Capacity Conversion

$$1 m^3 = 1000 L; 1 dm^3 = 1 L$$

Mass and Density Relation

Mass = Volume * Density

Excavation Rise in Level

Rise = Volume of pit / Remaining area

Conceptual Solved Examples & Case Studies

Example 1

Find the length of the longest pole that can be placed in a room 10 meters long, 8 meters broad, and 6 meters high.

Step-by-Step Solution:

Step 1: Identify dimensions: Length l = 10 m, Breadth b = 8 m, Height h = 6 m. Step 2: The longest pole that can fit inside the room corresponds to its body diagonal D. Step 3: Apply the body diagonal formula: D = sqrt(l^2 + b^2 + h^2) D = sqrt(10^2 + 8^2 + 6^2) D = sqrt(100 + 64 + 36) D = sqrt(200) Step 4: Simplify the radical: D = sqrt(100 * 2) = 10*sqrt(2) m. Taking sqrt(2) approx 1.414: D approx 10 * 1.414 = 14.14 m. Hence, the length of the longest pole is 10*sqrt(2) meters (approx 14.14 m).

Example 2

The ratio of the dimensions of a cuboid is 5 : 3 : 2 and its total surface area is 558 sq cm. Find its volume and body diagonal.

Step-by-Step Solution:

Step 1: Assign variables using the common ratio x: Let Length l = 5x, Breadth b = 3x, Height h = 2x. Step 2: Apply the total surface area formula: TSA = 2(lb + bh + hl) = 558 2[(5x)(3x) + (3x)(2x) + (2x)(5x)] = 558 2[15x^2 + 6x^2 + 10x^2] = 558 2[31x^2] = 558 62x^2 = 558 Step 3: Solve for x: x^2 = 558 / 62 = 9 => x = 3 cm. Step 4: Determine actual dimensions: l = 5 * 3 = 15 cm, b = 3 * 3 = 9 cm, h = 2 * 3 = 6 cm. Step 5: Compute Volume: V = l * b * h = 15 * 9 * 6 = 810 cubic cm. Step 6: Compute Body Diagonal: D = sqrt(15^2 + 9^2 + 6^2) = sqrt(225 + 81 + 36) = sqrt(342) = 3*sqrt(38) cm. Hence, Volume = 810 cm^3 and Diagonal = 3*sqrt(38) cm.

Example 3

A field is 20 m long and 14 m wide. In one corner of the field, a pit 8 m long, 5 m wide, and 3 m deep is dug, and the dug-out earth is spread evenly over the remaining part of the field. Find the rise in the level of the field.

Step-by-Step Solution:

Step 1: Calculate the volume of earth dug out from the pit: Volume of pit = l * b * h = 8 * 5 * 3 = 120 m^3. Step 2: Calculate the total area of the field and the area occupied by the pit: Total area of field = 20 * 14 = 280 m^2. Area of pit = 8 * 5 = 40 m^2. Step 3: Compute the remaining area of the field on which the earth is spread: Remaining area = Total Area - Area of Pit = 280 - 40 = 240 m^2. Step 4: Compute the rise in the level of the field: Rise in level = Volume of dug-out earth / Remaining area Rise = 120 / 240 = 0.5 meters. Converting to centimeters: 0.5 * 100 = 50 cm. Hence, the rise in the level of the field is 0.5 m or 50 cm.

Example 4

How many bricks each measuring 25 cm x 12.5 cm x 8 cm will be needed to build a wall 15 m long, 2 m high, and 0.5 m thick, if mortar occupies 5% of the total volume of the wall?

Step-by-Step Solution:

Step 1: Calculate the gross volume of the wall in cm^3: Length = 15 m = 1500 cm. Height = 2 m = 200 cm. Thickness = 0.5 m = 50 cm. Gross volume of wall = 1500 * 200 * 50 = 15,000,000 cm^3. Step 2: Deduct volume occupied by mortar (5%): Mortar volume = 5% of 15,000,000 = 0.05 * 15,000,000 = 750,000 cm^3. Effective volume of bricks = 15,000,000 - 750,000 = 14,250,000 cm^3. Step 3: Calculate the volume of 1 brick: Volume of 1 brick = 25 * 12.5 * 8 = 25 * 100 = 2,500 cm^3. Step 4: Compute the number of bricks required: Number of bricks = Effective Volume / Volume of 1 brick Number of bricks = 14,250,000 / 2,500 = 5,700 bricks. Hence, 5,700 bricks are needed.

Example 5

A cuboidal water reservoir of dimensions 6 m x 4 m x 3 m is completely filled with water. If 36,000 liters of water are pumped out, find the fall in the water level.

Step-by-Step Solution:

Step 1: Identify reservoir base dimensions: Length l = 6 m, Breadth b = 4 m. Base Area = 6 * 4 = 24 m^2. Step 2: Convert the pumped-out water volume into cubic meters: Since 1,000 liters = 1 m^3: Volume of water removed = 36,000 / 1,000 = 36 m^3. Step 3: Relate volume of water removed to the fall in level h_fall: Volume removed = Base Area * Fall in level 36 = 24 * h_fall Step 4: Solve for h_fall: h_fall = 36 / 24 = 1.5 meters. Hence, the water level will fall by 1.5 meters.

Common Misconceptions & Examiner Traps

Common Misconception

Mixing units during calculation (e.g. multiplying meters by centimeters directly).

Scientific Reality & Correction

Convert all dimensions into a single uniform unit (either all meters or all centimeters) before computing.

Common Misconception

Confusing 1 m^3 with 100 liters instead of 1,000 liters.

Scientific Reality & Correction

Remember 1 m^3 = 1,000 liters (1 m = 10 dm, so 1 m^3 = 10^3 dm^3 = 1,000 dm^3 = 1,000 L).

Common Misconception

Forgetting to subtract the pit area from the field area when spreading excavated earth.

Scientific Reality & Correction

Always compute remaining area: Field Area - Pit Area.

Common Misconception

Using 2(l + b)h for total surface area instead of lateral surface area.

Scientific Reality & Correction

2(l + b)h is the area of 4 walls (LSA). Total surface area is 2(lb + bh + hl).

Common Misconception

Deducting thickness 2t from height in an OPEN box.

Scientific Reality & Correction

In an open box, deduct thickness once (h - t) from height, but twice (l - 2t, b - 2t) from length and breadth.

Common Misconception

Taking cube diagonal as a*sqrt(2) instead of a*sqrt(3).

Scientific Reality & Correction

Face diagonal of a square is a*sqrt(2); the 3D body diagonal of a cube is a*sqrt(3).

Concept Map: Rectangular Parallelopiped or Cuboid (WBBSE Class 10 Ganit Prakash)

Chapter Summary & 10 Key Takeaways

Takeaway 1

A cuboid has 6 rectangular faces, 12 edges, and 8 vertices; opposite faces are congruent and parallel.

Takeaway 2

Total Surface Area of cuboid: TSA = 2(lb + bh + hl); Area of 4 walls: LSA = 2(l + b)h.

Takeaway 3

Volume of cuboid: V = l * b * h; Body diagonal: D = sqrt(l^2 + b^2 + h^2).

Takeaway 4

A cube of edge a has TSA = 6a^2, LSA = 4a^2, Volume = a^3, and Diagonal = a*sqrt(3).

Takeaway 5

Fluid capacity: 1 cubic decimeter = 1 liter; 1 cubic meter = 1,000 liters.

Takeaway 6

Mass = Volume * Density.

Takeaway 7

When excavated earth from a pit is spread over a field: Rise in level = Pit Volume / (Field Area - Pit Area).

Takeaway 8

Number of bricks = Effective wall volume / Volume of 1 brick.

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.

What is the total length of all 12 edges of a cuboid with dimensions l, b, and h?

Reveal Answer & Explanation

Answer: Since there are 4 lengths, 4 breadths, and 4 heights, the total edge length is 4(l + b + h).

If the length of the body diagonal of a cube is 6*sqrt(3) cm, find its volume.

Reveal Answer & Explanation

Answer: Body diagonal = a*sqrt(3) = 6*sqrt(3) => a = 6 cm. Volume V = a^3 = 6^3 = 216 cm^3.

How many liters of water can be stored in a cubical tank of edge 2 meters?

Reveal Answer & Explanation

Answer: Volume = a^3 = 2^3 = 8 m^3. Since 1 m^3 = 1,000 liters, capacity = 8 * 1,000 = 8,000 liters.

If the total surface area of a cube is 96 sq cm, find its edge length.

Reveal Answer & Explanation

Answer: TSA = 6a^2 = 96 => a^2 = 16 => a = 4 cm.

If each edge of a cube is doubled, how many times does its volume increase?

Reveal Answer & Explanation

Answer: New edge = 2a. New volume = (2a)^3 = 8a^3 = 8 times the original volume.

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