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ICSE • Class 8 • Mathematics • Ch 22

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Area of Rectilinear Figures

In ICSE Class 8 Mathematics, "Area of Rectilinear Figures" provides an authoritative, mathematically rigorous master study guide investigating the perimeter and area of composite polygonal plane figures, triangles, quadrilaterals, paths, and field surveyor cross-staff maps. This comprehensive chapter explores Area of Triangles (1. Base-Height Formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$, 2. Right-Angled Triangle: $\text{Area} = \frac{1}{2} \times \text{product of legs}$, 3. Equilateral Triangle Formula: $\text{Area} = \frac{\sqrt{3}}{4} a^2$ with altitude $h = \frac{\sqrt{3}}{2}a$, 4. Isosceles Triangle: $\text{Area} = \frac{b}{4}\sqrt{4a^2 - b^2}$, 5. Heron's Master Area Formula for Scalene Triangles: $\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}$ where semi-perimeter $s = \frac{a + b + c}{2}$), Area of Quadrilaterals (1. General Quadrilateral given one diagonal $d$ and two perpendicular offsets $h_1, h_2$: $\text{Area} = \frac{1}{2} d (h_1 + h_2)$, 2. Trapezium: $\text{Area} = \frac{1}{2} \times (a + b) \times h$, 3. Parallelogram: $\text{Area} = \text{base} \times \text{height}$, 4. Rhombus: $\text{Area} = \frac{1}{2} \times d_1 \times d_2 = \text{base} \times \text{height}$), Area of Paths, Verandahs & Borders (Running inside or outside rectangular lawns; Cross-paths running through the middle of parks), and Surveying & Area of Irregular Polygons (Cross-staff surveyor field book method: dividing irregular land acreage into triangles and trapeziums along a central baseline) aligned with the 2026–27 CISCE ICSE curriculum.

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How Did an Ancient Greek Engineer in Alexandria Calculate the Area of a Scalene Triangle Without Ever Measuring Its Altitude?

Imagine an irregular triangular plot of land bounded by three steep ravines measuring $13\text{ meters}$, $14\text{ meters}$, and $15\text{ meters}$. You want to buy it, but to know the fair price, you need its exact area. The standard formula says $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$—but how do you drop a plumb line and measure the vertical height inside a rocky ravine? In 60 CE in Roman Egypt, Hero (Heron) of Alexandria solved this ancient crisis forever! He created a mind-blowing formula that calculates the exact area using ONLY THE THREE SIDE LENGTHS ($a, b, c$)—no height required! First calculate semi-perimeter $s = \frac{13 + 14 + 15}{2} = 21$. Then multiply: $\sqrt{21 \times (21-13) \times (21-14) \times (21-15)} = \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = \mathbf{84\text{ m}^2}$! Pure genius! How do modern land surveyors map entire irregular farming estates using a Cross-Staff Baseline? What is the secret area formula for a trapezium? Let's master the area of rectilinear figures.

Why This Chapter Matters

Planar mensuration of rectilinear figures is essential for civil construction (concrete slabs, tiling layouts), real estate cadastral land deeds, agriculture crop yield optimization, urban road planning, and materials manufacturing. Mastering Heron's formula and composite polygon surveyor decomposition is a core high-scoring ICSE mathematics topic.

Before You Begin (Prerequisites)

Perimeter and area of rectangles and squares from Class 7.
Pythagoras theorem ($a^2 + b^2 = c^2$).
Square roots and decimal arithmetic from Chapter 4.

What You Will Learn (Core Objectives)

Calculate triangle areas using standard base-height and Heron's formula.
Derive and apply the equilateral triangle area formula $\frac{\sqrt{3}}{4} a^2$.
Calculate the area of a trapezium using $\frac{1}{2}(a + b)h$.
Calculate the area of a rhombus using half the product of its diagonals: $\frac{1}{2} d_1 d_2$.
Calculate the area of internal and external running paths and crossing paths.
Calculate the total acreage of irregular land polygons using the Surveyor Field Book method.

Chapter Roadmap & Progression

1 1. Triangle Area Formulas & Heron's...

2 2. Area of Quadrilaterals: Trapeziu...

3 3. Area of Paths, Borders & Cross-R...

4 4. Surveyor's Field Book Decomposit...

Complete Concept Guide (100% Curriculum Coverage)

1. Triangle Area Formulas & Heron's Theorem

2. Area of Quadrilaterals: Trapezium, Parallelogram & Rhombus

3. Area of Paths, Borders & Cross-Roads

4. Surveyor's Field Book Decomposition

Key Formulas, Identities & Theorems

Heron's Area Formula

$$\text{Area} = \sqrt{s(s - a)(s - b)(s - c)} \quad [s = (a+b+c)/2]$$

Area of any triangle from three side lengths.

Trapezium Area Formula

$$\text{Area} = \frac{1}{2} (a + b) h$$

Half the sum of parallel sides times perpendicular height.

Mensuration: Area of Trapezium & Heron's Triangle Decomposition

² TRAPEZIUM & RHOMBUS FORMULAS Parallel side a Parallel side b h Trapezium: Area = ½ (a + b) × h Rhombus: Area = ½ × d1 × d2 • General Quad: ½ d (h1 + h2) • Cross-roads: w(L+B) - w² HERON: √[s(s-a)(s-b)(s-c)] • TRAPEZIUM: ½(a+b)h • RHOMBUS: ½ d1*d2 • SUBTRACT OVERLAPS

Chapter Summary & 10 Key Takeaways

Takeaway 1

Standard triangle area: Area = 1/2 * base * height.

Takeaway 2

Heron's formula computes triangle area from 3 sides: Area = sqrt[s(s - a)(s - b)(s - c)], where s = (a + b + c)/2.

Takeaway 3

Area of an equilateral triangle of side a: Area = (sqrt(3)/4) * a^2.

Takeaway 4

Area of a general quadrilateral with diagonal d and offsets h1, h2: Area = 1/2 * d * (h1 + h2).

Takeaway 5

Area of a trapezium: Area = 1/2 * (sum of parallel sides) * height = 1/2 * (a + b) * h.

Takeaway 6

Area of a rhombus: Area = 1/2 * d1 * d2.

Takeaway 7

Outer path area: Area = (L + 2w)(B + 2w) - LB.

Takeaway 8

Inner path area: Area = LB - (L - 2w)(B - 2w).

Takeaway 9

Central cross-roads area: Area = w(L + B) - w^2 (subtract the central square overlap).

Takeaway 10

Surveyor field book method calculates total acreage by summing decomposed right triangles and trapeziums.

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.

Find the area of a triangle whose sides are $13\text{ cm}, 14\text{ cm},$ and $15\text{ cm}$ using Heron's formula.

Reveal Answer & Explanation

Answer: Step 1: Calculate semi-perimeter $s$:
$$s = \frac{a + b + c}{2} = \frac{13 + 14 + 15}{2} = \frac{42}{2} = \mathbf{21\text{ cm}}$$
Step 2: Calculate $(s - a), (s - b),$ and $(s - c)$:
• $s - a = 21 - 13 = 8\text{ cm}$
• $s - b = 21 - 14 = 7\text{ cm}$
• $s - c = 21 - 15 = 6\text{ cm}$

Step 3: Apply Heron's formula:
$$\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}$$
$$\text{Area} = \sqrt{21 \times 8 \times 7 \times 6}$$
Factorize under square root:
$$= \sqrt{(7 \times 3) \times (4 \times 2) \times 7 \times (3 \times 2)}$$
$$= \sqrt{7^2 \times 3^2 \times 4 \times 4} = 7 \times 3 \times 4 = \mathbf{84\text{ cm}^2}$$.
$s = 21$. $\text{Area} = \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84\text{ cm}^2$.

The parallel sides of a trapezium are $25\text{ cm}$ and $13\text{ cm}$, and its non-parallel sides are equal, each being $10\text{ cm}$. Find the area of the trapezium.

Reveal Answer & Explanation

Answer: Step 1: In an isosceles trapezium, drop perpendicular heights $h$ from the ends of the shorter base ($13\text{ cm}$) to the longer base ($25\text{ cm}$).
The longer base is split into $13\text{ cm}$ in the middle and two equal segments on the sides:
$$\text{Each side segment} = \frac{25 - 13}{2} = \frac{12}{2} = 6\text{ cm}$$
Step 2: By Pythagoras theorem, find height $h$:
$$h^2 + 6^2 = 10^2 \implies h^2 + 36 = 100 \implies h^2 = 64 \implies \mathbf{h = 8\text{ cm}}$$
Step 3: Apply the Trapezium Area Formula:
$$\text{Area} = \frac{1}{2} (a + b) \times h = \frac{1}{2} (25 + 13) \times 8 = \frac{1}{2} \times 38 \times 8 = 38 \times 4 = \mathbf{152\text{ cm}^2}$$.
Base segment is $(25 - 13)/2 = 6\text{ cm}$. Height $h = \sqrt{10^2 - 6^2} = 8\text{ cm}$. $\text{Area} = \frac{1}{2}(38)(8) = 152\text{ cm}^2$.

The area of a rhombus is $216\text{ cm}^2$. If one of its diagonals measures $24\text{ cm}$, find: (a) the length of the other diagonal, (b) the length of each side of the rhombus.

Reveal Answer & Explanation

Answer:

• (a) Length of other diagonal ($d_2$):

$$\text{Area} = \frac{1}{2} d_1 d_2$$

$$216 = \frac{1}{2} \times 24 \times d_2$$

$$216 = 12 d_2 \implies d_2 = \frac{216}{12} = \mathbf{18\text{ cm}}$$

• (b) Length of side of the rhombus:
Diagonals bisect each other at $90^{\circ}$. Half-diagonals are:
• $\frac{d_1}{2} = \frac{24}{2} = 12\text{ cm}$
• $\frac{d_2}{2} = \frac{18}{2} = 9\text{ cm}$
By Pythagoras theorem:

$$\text{Side} = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = \mathbf{15\text{ cm}}$$

(a) $d_2 = (2 \times 216) / 24 = 18\text{ cm}$. (b) $\text{Side} = \sqrt{12^2 + 9^2} = 15\text{ cm}$.

A rectangular grassy lawn measures $50\text{ m} \times 40\text{ m}$. A path $2.5\text{ m}$ wide is constructed all around on the outside of the lawn. Find the area of the path and the cost of gravelling it at $\text{Rs } 20\text{ per m}^2$.

Reveal Answer & Explanation

Answer: Step 1: Area of inner lawn $= 50 \times 40 = \mathbf{2000\text{ m}^2}$.
Step 2: Outer dimensions with $w = 2.5\text{ m}$:
• Outer Length $= 50 + 2(2.5) = 50 + 5 = 55\text{ m}$
• Outer Breadth $= 40 + 2(2.5) = 40 + 5 = 45\text{ m}$
Outer Area $= 55 \times 45 = \mathbf{2475\text{ m}^2}$.
Step 3: Area of path $= \text{Outer Area} - \text{Inner Area}$:
$$\text{Area of Path} = 2475 - 2000 = \mathbf{475\text{ m}^2}$$
Step 4: Cost of gravelling:
$$\text{Cost} = 475 \times 20 = \mathbf{\text{Rs } 9,500}$$.
Outer dimensions: $55 \times 45 = 2475\text{ m}^2$. Path area: $2475 - 2000 = 475\text{ m}^2$. Cost: $475 \times 20 = \text{Rs } 9500$.

Two cross-roads, each of width $5\text{ meters}$, run at right angles through the center of a rectangular park of dimensions $70\text{ m} \times 50\text{ m}$. Find the area of the cross-roads.

Reveal Answer & Explanation

Answer: Step 1: Area of road parallel to length ($70\text{ m}$):
$$\text{Area}_1 = 70 \times 5 = 350\text{ m}^2$$
Step 2: Area of road parallel to breadth ($50\text{ m}$):
$$\text{Area}_2 = 50 \times 5 = 250\text{ m}^2$$
Step 3: Area of common overlapping square in the middle:
$$\text{Common Area} = 5 \times 5 = 25\text{ m}^2$$
Step 4: Net Area of cross-roads:
$$\text{Net Area} = \text{Area}_1 + \text{Area}_2 - \text{Common Area}$$
$$= 350 + 250 - 25 = 600 - 25 = \mathbf{575\text{ m}^2}$$.
Area $= (70 \times 5) + (50 \times 5) - (5 \times 5) = 350 + 250 - 25 = 575\text{ m}^2$.

Find the area of an equilateral triangle whose perimeter is $36\text{ cm}$. Take $\sqrt{3} \approx 1.732$.

Reveal Answer & Explanation

Answer: Step 1: Find the side length $a$:
$$3a = 36 \implies a = \frac{36}{3} = 12\text{ cm}$$
Step 2: Apply the Equilateral Triangle Area Formula:
$$\text{Area} = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \times 12^2 = \frac{\sqrt{3}}{4} \times 144 = 36\sqrt{3}\text{ cm}^2$$
Step 3: Substitute $\sqrt{3} \approx 1.732$:
$$\text{Area} = 36 \times 1.732 = \mathbf{62.352\text{ cm}^2}$$.
Side $a = 12\text{ cm}$. $\text{Area} = (\sqrt{3}/4)(144) = 36\sqrt{3} \approx 62.352\text{ cm}^2$.

A diagonal of a quadrilateral is $26\text{ cm}$ long and the perpendicular offsets drawn to it from the opposite vertices are $8.5\text{ cm}$ and $6.5\text{ cm}$. Find the area of the quadrilateral.

Reveal Answer & Explanation

Answer: Apply the general quadrilateral formula: $\text{Area} = \frac{1}{2} d (h_1 + h_2)$.
Here $d = 26\text{ cm}, h_1 = 8.5\text{ cm}, h_2 = 6.5\text{ cm}$:
$$\text{Area} = \frac{1}{2} \times 26 \times (8.5 + 6.5)$$
$$= 13 \times 15 = \mathbf{195\text{ cm}^2}$$.
$\text{Area} = \frac{1}{2} \times 26 \times (8.5 + 6.5) = 13 \times 15 = 195\text{ cm}^2$.

Explain how a field surveyor uses a cross-staff and baseline to find the total area of an irregular polygonal plot.

Reveal Answer & Explanation

Answer:

• The surveyor establishes a straight central baseline between two fixed boundary stations (e.g., $A$ and $B$).
• As the surveyor walks along the baseline, a cross-staff is used to sight right angles ($90^{\circ}$) to all outer corner vertices of the property, recording the chainage distance along the baseline and the perpendicular offset lengths to the left and right.
• This procedure partitions the irregular estate into a sequence of right-angled triangles and right-angled trapeziums.
• The area of each individual geometric strip is computed using $\frac{1}{2} \times \text{base} \times \text{offset}$ or $\frac{1}{2}(h_1 + h_2) \times \text{interval}$, and their sum gives the exact total land area.

A central baseline with perpendicular cross-staff offsets divides irregular land into right triangles and trapeziums whose areas sum to the total.

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