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WBB • Class 7 • Mathematics (গণিত প্রভা) • Ch 17

Estimated Time: 55 Minutes

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Area of Rectangle and Square

Welcome to Chapter 17 "Area of Rectangle and Square" of the West Bengal Board of Secondary Education (WBBSE) Class 7 Mathematics (Ganit Prabha) curriculum. This guide covers the distinction between perimeter and area, metric area unit conversions ($1\text{ m}^2 = 10,000\text{ cm}^2$, hectares, ares), diagonal theorems ($d = \sqrt{l^2+b^2}$ and $d = a\sqrt{2}$), external and internal perimeter pathway problems, central crossing pathways ($w(l+b-w)$), and commercial cost estimation for fencing vs paving using TargetExams Gold-Standard 5-step pedagogy.

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When designing a 2-meter wide walking pathway around a 50-meter by 30-meter rectangular garden, what will the total area of the pathway be?

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Why This Chapter Matters

Before You Begin (Prerequisites)

Fundamental properties of rectangles and squares
Distinction between perimeter (linear boundary) and area (enclosed space)
Linear metric unit conversions ($1\text{ m} = 100\text{ cm}$)
Basic binomial expansion and arithmetic simplification

What You Will Learn (Core Objectives)

Calculate perimeter, area, and diagonals for rectangles and squares
Convert between $\text{m}^2, \text{cm}^2$, and hectares without errors
Compute areas of external and internal pathways surrounding rectangular plots
Calculate the area of central crossing roads by subtracting the intersection square ($w^2$)
Determine costs for boundary fencing (per meter) and ground flooring/paving (per sq meter)

Chapter Roadmap & Progression

1 Concept 1: Foundations of Perimeter...

2 Concept 2: Diagonal Relationships f...

3 Concept 3: External & Internal Path...

4 Concept 4: Central Cross-Roads Para...

5 Concept 5: Commercial Applications...

Complete Concept Guide (100% Curriculum Coverage)

Concept 1: Foundations of Perimeter & Area and Unit Conversion

Step 1

Definitions

Perimeter: Total linear boundary distance enclosing a 2D shape (units: m, cm).
Area: Amount of surface enclosed by the boundary (units: $\text{m}^2, \text{cm}^2$).

Step 2

Metric Area Conversions

• $1\text{ m} = 100\text{ cm} \implies 1\text{ m}^2 = 10,000\text{ cm}^2$.
• $1\text{ hectare} = 10,000\text{ m}^2$.
• $1\text{ are} = 100\text{ m}^2$.

Step 3

Worked Example

A square room of side 4 m has area $4 \times 4 = 16\text{ m}^2 = 160,000\text{ cm}^2$.

Step 4

Examiner Trap

Never assume $1\text{ m}^2 = 100\text{ cm}^2$. Area conversion squares the linear factor ($100^2 = 10,000$).

Step 5

Land Surveying

Agricultural Registry: Official land deeds and records specify agricultural holdings in hectares.

Concept 2: Diagonal Relationships for Rectangles & Squares

Step 1

Pythagorean Diagonal

With $90^\circ$ corners, diagonal $d$ acts as hypotenuse: $d = \sqrt{l^2 + b^2}$.

Step 2

Square Diagonal & Area

For a square of side $a$: $d = a\sqrt{2}$. Given diagonal $d$, Area $= \frac{1}{2}d^2$.

Step 3

Worked Example

If diagonal $= 10\sqrt{2}\text{ m}$, Area $= \frac{1}{2} \times (10\sqrt{2})^2 = \frac{1}{2} \times 200 = 100\text{ m}^2$.

Step 4

Examiner Caution

To find a side from perimeter, divide by 4; to find a side from area, take the square root.

Step 5

Display Technology

Electronics: Monitor and TV screen sizes (e.g. 55-inch) refer strictly to the diagonal measurement.

Concept 3: External & Internal Pathways Around Plots

Step 1

External Path Dynamics

Outer length $= l + 2w$, Outer breadth $= b + 2w$. Path Area $= (l+2w)(b+2w) - lb$.

Step 2

Internal Path Dynamics

Inner length $= l - 2w$, Inner breadth $= b - 2w$. Path Area $= lb - (l-2w)(b-2w)$.

Step 3

Practical Example

A 2 m path inside a 40 m × 25 m field leaves an inner lawn of $36 \times 21 = 756\text{ m}^2$. Path Area $= 1000 - 756 = 244\text{ m}^2$.

Step 4

Examiner Warning

Carefully note whether the question specifies "outside" (+2w) or "inside" (-2w).

Step 5

Urban Parks

Jogging Tracks: Municipal parks install perimeter jogging tracks based on these equations.

Concept 4: Central Cross-Roads Parallel to Sides

Step 1

Geometric Configuration

Two perpendicular roads bisecting a field parallel to length and breadth.

Step 2

Formula Derivation

$$\text{Cross Roads Area} = lw + bw - w^2 = w(l + b - w)$$

Step 3

Remaining Lawns

Area of remaining 4 corner quadrants $= (l - w)(b - w)$.

Step 4

Critical Trap

Never forget to deduct the intersection square $w^2$.

Step 5

Heritage Architecture

Charbagh Gardens: The gardens surrounding the Taj Mahal embody this cross-road geometry.

Concept 5: Commercial Applications (Fencing vs Tiling Costs)

Step 1

Cost Principles

• Boundary work (fencing): $\text{Cost} = \text{Perimeter} \times \text{Rate/m}$.
• Surface work (paving, turfing): $\text{Cost} = \text{Area} \times \text{Rate/m}^2$.

Step 2

Tile Count Formula

$$\text{Number of Tiles} = \frac{\text{Area of Floor}}{\text{Area of One Tile}}$$ Always harmonize units before division.

Step 3

Worked Example

Tiling a 6 m × 4 m floor with 20 cm × 20 cm tiles requires $\frac{600 \times 400}{20 \times 20} = 600\text{ tiles}$.

Step 4

Caution

Do not divide meters by centimeters without converting units first.

Step 5

Construction

Building Estimations: Civil engineers compute flooring tiles and paint budgets using these exact formulas.

Key Formulas, Identities & Theorems

Perimeter & Area of Rectangle

$$P = 2(l + b), \quad A = l \times b$$

Where $l$ = length and $b$ = breadth.

Diagonal of Rectangle

$$d = \sqrt{l^2 + b^2}$$

Derived via the Pythagorean theorem.

Square Formulas

$$P = 4a, \quad A = a^2, \quad d = a\sqrt{2}$$

Where $a$ = side length. Area in terms of diagonal is $A = \frac{1}{2}d^2$.

External Pathway Area

$$A_{\text{path}} = (l + 2w)(b + 2w) - lb = 2w(l + b + 2w)$$

Where $w$ = path width. Increases dimensions on both sides by $2w$.

Internal Pathway Area

$$A_{\text{path}} = lb - (l - 2w)(b - 2w) = 2w(l + b - 2w)$$

Area of outer plot minus inner remaining plot.

Central Crossing Roads Area

$$A_{\text{cross}} = lw + bw - w^2 = w(l + b - w)$$

Central square ($w^2$) subtracted once to avoid double counting.

Conceptual Solved Examples & Case Studies

Example 1

A rectangular garden is $50\text{ m}$ long and $30\text{ m}$ wide. A pathway $2\text{ m}$ wide is built around its outside. Find the area of the pathway.

Step-by-Step Solution:

Area of garden $= 50 \times 30 = 1500\text{ m}^2$.
Length including path $= 50 + 2(2) = 54\text{ m}$.
Breadth including path $= 30 + 2(2) = 34\text{ m}$.
Total area including path $= 54 \times 34 = 1836\text{ m}^2$.
Area of pathway $= 1836 - 1500 = 336\text{ m}^2$.

Example 2

The area of a square field is equal to that of a rectangular field measuring $18\text{ m} \times 8\text{ m}$. Find the side length and perimeter of the square field.

Step-by-Step Solution:

Area of rectangle $= 18 \times 8 = 144\text{ m}^2$.
Area of square ($a^2$) $= 144\text{ m}^2 \implies a = \sqrt{144} = 12\text{ m}$.
Perimeter of square $= 4a = 4 \times 12 = 48\text{ m}$.

Example 3

Two central perpendicular roads of width $2\text{ m}$ run through a playground measuring $60\text{ m} \times 40\text{ m}$. Calculate the cost of paving these roads at ₹50 per $\text{m}^2$.

Step-by-Step Solution:

Road area $= w(l + b - w) = 2(60 + 40 - 2) = 2 \times 98 = 196\text{ m}^2$.
Total cost $= 196 \times 50 = ₹9,800$.

Common Misconceptions & Examiner Traps

Common Misconception

Adding $w$ instead of $2w$ to length and breadth when a path runs around the entire perimeter.

Scientific Reality & Correction

Paths surround both sides, so dimension expansion is $2w$ on each axis ($l+2w, b+2w$).

Common Misconception

Forgetting to subtract the common intersection square ($w^2$) in central cross-roads.

Scientific Reality & Correction

The cross section is counted in both length and breadth strips; subtract $w^2$ once.

Common Misconception

Using area instead of perimeter for fencing problems, or perimeter for flooring.

Scientific Reality & Correction

Fencing is along the linear boundary (perimeter, m); paving is over the surface (area, $\text{m}^2$).

Rectangular Pathway & Central Crossing Roads Geometry Model

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All Class 7 Mathematics (গণিত প্রভা) Chapters

Ch 1: Revision of Previous Lessons Ch 2: Ratio Ch 3: Proportion Ch 4: Addition, Subtraction, Multiplication and Division of Integers Ch 5: Concept of Indices Ch 6: Algebraic Operations Ch 7: Construction of Specific Angles with Compass Ch 8: Construction of Triangles Ch 9: Concept of Congruence Ch 10: Approximation Ch 11: Square Root of Fractions Ch 12: Algebraic Formulae Ch 13: Parallel Lines and Transversals Ch 14: Properties of Triangles Ch 15: Time and Distance Ch 16: Double Bar Graph Ch 17: Area of Rectangle and Square Ch 18: Symmetry Ch 19: Factorization Ch 20: Classification of Quadrilaterals Ch 21: Construction of Quadrilaterals Ch 22: Formation and Solution of Equations

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