Area and Perimeter of Plane Figures Class 9 ICSE — Ch 20
The Area and Perimeter of Plane Figures Class 9 ICSE chapter is the mensuration foundation of the Class 9 Mathematics book. It defines perimeter and area, then applies those definitions to triangles, quadrilaterals and circles. Every section states a formula, explains why it holds, and finishes with a worked example pattern you can imitate. The chapter runs from basic definitions in section 20.1 to the area of a circle in section 20.7.
Download Area and Perimeter of Plane Figures Class 9 ICSE Mathematics Chapter 20 PDF
Download the Area and Perimeter of Plane Figures Class 9 ICSE Mathematics chapter PDF to revise offline or print for practice. It is the full chapter as printed in the book — 22 pages, 7 sections, 78 questions.
What is inside Chapter 20
22 pages · 7 sections · 29 worked examples · 78 exercise questions
| Section | Title |
|---|---|
| 20.1 | INTRODUCTION |
| 20.2 | AREA AND PERIMETER OF TRIANGLES |
| 20.3 | SOME SPECIAL TYPES OF TRIANGLES |
| 20.4 | AREA AND PERIMETER OF QUADRILATERALS |
| 20.5 | SOME SPECIAL TYPES OF QUADRILATERALS |
| 20.6 | CIRCUMFERENCE OF A CIRCLE |
| 20.7 | AREA OF A CIRCLE |
- EXERCISE 20(A): 15 questions
- EXERCISE 20(B): 36 questions
- EXERCISE 20(C): 27 questions
20.1 INTRODUCTION
The perimeter of a plane figure is the length of its boundary. The area of a plane figure is the measure of the surface enclosed by that boundary. The two quantities use different units, so the chapter separates them from the start. Perimeter keeps the same unit as length — centimetre, metre and so on. Area uses square units such as \mathrm{cm}^2 and \mathrm{m}^2.
The chapter also draws a sharp line between ‘square metre’ and ‘metre square’. The phrase x square metre names an area of x square metres. The phrase x metre square describes a square whose side is x metres long, so its area is x \times x = x^2 square metres.
Take a 3 metre square. Its side is 3 m, so its area is 3 \times 3 = 9 square metres. Calling that area ‘3 square metres’ is a mistake, because 3 square metres names a much smaller region. Common mistakes to avoid:
- Confusing 3\ \mathrm{m}^2 with a 3 metre square: the first is a small area; the second has area 9\ \mathrm{m}^2.
- Mixing perimeter and area units: perimeter is measured in cm or m; area in \mathrm{cm}^2 or \mathrm{m}^2.
20.2 AREA AND PERIMETER OF TRIANGLES
The core formula is \text{Area} = \frac{1}{2} \times \text{base} \times \text{corresponding height}. The corresponding height is the perpendicular dropped from the opposite vertex onto the base you picked. Any of the three sides may serve as the base, and each side has its own corresponding height.
Why the \frac{1}{2} appears: two identical triangles joined along any side form a parallelogram, and a parallelogram’s area is base times height. Half of that parallelogram is one triangle.
When no height is known but all three sides are, Heron’s formula takes over. With sides a, b, c, the semi-perimeter is s = \frac{a+b+c}{2}, and the area is \sqrt{s(s-a)(s-b)(s-c)}.
Original pattern: sides 13 cm, 14 cm, 15 cm. First, s = \frac{13+14+15}{2} = 21 cm. Then area = \sqrt{21 \times 8 \times 7 \times 6} = 84\ \mathrm{cm}^2. The largest side is 15 cm, so \frac{1}{2} \times 15 \times \text{height} = 84, giving height = \frac{168}{15} = 11.2 cm.
The unit table that follows the formula is short: 1\ \mathrm{cm} = \frac{1}{100}\ \mathrm{m}, so 1\ \mathrm{cm}^2 = \frac{1}{10000}\ \mathrm{m}^2. Convert units before you multiply, not after. One common mistake deserves a warning: the height must be perpendicular to the base you chose. Pairing a base with a height from a different side gives a wrong area even when both numbers are right.
20.3 SOME SPECIAL TYPES OF TRIANGLES
An equilateral triangle has three equal sides. Side a gives perimeter 3a and area \frac{\sqrt{3}}{4}a^2. One special case is worth knowing: if the area equals the perimeter numerically, then \frac{\sqrt{3}}{4}a^2 = 3a, which forces a = \frac{12}{\sqrt{3}} = 4\sqrt{3} \approx 6.92 units, using \sqrt{3} \approx 1.73 as the chapter does.
For an isosceles triangle, the perpendicular from the vertex to the base bisects the base. That turns the figure into two right-angled triangles, which unlocks three routes to the area. Worked pattern: equal sides 10 cm, base 12 cm.
- Drop the perpendicular: the half-base is 6 cm, height \sqrt{10^2-6^2} = 8 cm, area \frac{1}{2} \times 12 \times 8 = 48\ \mathrm{cm}^2.
- Heron’s formula: s = \frac{10+10+12}{2} = 16, area \sqrt{16 \times 6 \times 6 \times 4} = 48\ \mathrm{cm}^2.
- Direct formula: \frac{1}{4} \times b \times \sqrt{4a^2-b^2} = \frac{1}{4} \times 12 \times \sqrt{400-144} = 48\ \mathrm{cm}^2.
A right-angled triangle’s area is half the product of the sides containing the right angle. When those sides are written in terms of x, the area equation becomes a quadratic. Pattern: sides 2x and 3x-2 with area 40\ \mathrm{cm}^2. Then \frac{1}{2}(2x)(3x-2) = 40, so 3x^2 - 2x - 40 = 0. Solving gives x = 4 or x = -\frac{10}{3}.
The negative root is rejected, because a side length can never be negative. The sides are 8 cm and 10 cm.
20.4 AREA AND PERIMETER OF QUADRILATERALS
Any quadrilateral that is not a rectangle or square falls back on two general methods. The first method needs one diagonal and the perpendiculars to it from the remaining two vertices. That diagonal splits the quadrilateral into two triangles sharing the same base.
So \text{Area} = \frac{1}{2} \times \text{diagonal} \times (\text{sum of the two perpendiculars}). Pattern: a diagonal of 12 cm with perpendiculars 5 cm and 3 cm gives \frac{1}{2} \times 12 \times (5+3) = 48\ \mathrm{cm}^2.
The second method applies when the two diagonals cut each other at right angles. The same triangle split produces \text{Area} = \frac{1}{2} \times d_1 \times d_2. Pattern: diagonals 10 cm and 6 cm meeting at right angles give \frac{1}{2} \times 10 \times 6 = 30\ \mathrm{cm}^2.
Keep both rules in mind for section 20.5. Every special quadrilateral’s formula is a shortcut for one of them. A rhombus, for instance, has diagonals that always meet at right angles.
20.5 SOME SPECIAL TYPES OF QUADRILATERALS
Five shapes get their own formulas. The table collects them for quick reference.
| Shape | Area | Perimeter | Diagonal / note |
|---|---|---|---|
| Rectangle, sides l, b | lb | 2(l+b) | \sqrt{l^2+b^2} |
| Square, side a | a^2 | 4a | a\sqrt{2}, also \sqrt{2 \times \text{area}} |
| Parallelogram | \text{base} \times \text{height} | — | height is the distance between base and opposite side |
| Rhombus, diagonals d_1, d_2 | \frac{1}{2}d_1 d_2 | 4a | (\text{side})^2 = \left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2; also \text{base} \times \text{height} |
| Trapezium, parallel sides a, b, height h | \frac{1}{2}(a+b)h | sum of the four sides | mid-segment = \frac{a+b}{2}, so area = \text{mid-segment} \times h |
Two details in the table earn attention. A rhombus is also a parallelogram, so base times height works there too. A trapezium’s mid-segment equals \frac{a+b}{2}, which is why area equals mid-segment times height. These five formulas stay in use across the rest of the Class 9 Maths book hub.
Wire problems keep the perimeter fixed. A wire bent into a square of area 144\ \mathrm{cm}^2 has side 12 cm, so the wire is 48 cm long. Bent next into a rectangle of length 18 cm, 2(18+b) = 48 gives b = 6, so the rectangle encloses 108\ \mathrm{cm}^2.
Footpath problems subtract the inner rectangle from the outer one. Pattern: a field 34 m by 26 m has a uniform inner path occupying 224\ \mathrm{m}^2. With width x, 34 \times 26 - (34-2x)(26-2x) = 224, which simplifies to x^2 - 30x + 56 = 0.
So x = 2 or x = 28; a 28 m path cannot fit inside a 26 m wide field, so the width is 2 m.
20.6 CIRCUMFERENCE OF A CIRCLE
The circumference of a circle is the length of its boundary. For every circle, the ratio of circumference to diameter is the same constant, called \pi (pi). This gives C = \pi d = 2\pi r, since the diameter is twice the radius. The chapter takes \pi = \frac{22}{7} = 3.14 approximately.
Why the ratio is constant: every circle is a scaled copy of every other. Double the radius, and both circumference and diameter double, so their quotient never changes.
Pattern: a circle whose circumference exceeds its diameter by 90 cm. Write 2\pi r - 2r = 90, factor to 2r(\pi - 1) = 90. With \pi = \frac{22}{7}, \pi - 1 = \frac{15}{7}, so 2r \times \frac{15}{7} = 90 and r = \frac{90 \times 7}{2 \times 15} = 21 cm. The diameter is 42 cm.
The direct method handles combined circles. If one circle’s circumference equals the sum of several others, 2\pi R = 2\pi r_1 + 2\pi r_2 + \cdots, then dividing by 2\pi leaves R = r_1 + r_2 + \cdots. Pattern: radii 4 cm, 7 cm and 9 cm combine to a circle of radius 20 cm, whose diameter is 40 cm.
20.7 AREA OF A CIRCLE
The area of a circle of radius r is \pi r^2. A neat check: if the area equals the circumference numerically, \pi r^2 = 2\pi r, so r = 2; the circle then has area 4\pi square units.
A running track or ring is the region between two concentric circles, and its area is outer minus inner: \pi R^2 - \pi r^2 = \pi(R^2 - r^2). Factoring gives \pi(R+r)(R-r), which becomes useful when the width R-r is known. Pattern: two circles of radii 13 cm and 5 cm enclose \pi(169-25) = 144\pi \approx 452.6\ \mathrm{cm}^2.
A wheel covers exactly one circumference per rotation. So distance equals 2\pi r \times number of rotations. Pattern: a wheel of radius 42 cm makes 1500 rotations, covering 2 \times \frac{22}{7} \times 42 \times 1500 = 396000\ \mathrm{cm} = 3.96 km. To turn revolutions per minute into km/h, convert m/s and multiply by \frac{18}{5}.
The largest circle cut from a sheet uses the smaller dimension as its diameter. A sheet 42 cm by 30 cm gives radius 15 cm and area 225\pi \approx 707\ \mathrm{cm}^2. When a square is inscribed in a circle, the square’s diagonal equals the circle’s diameter: for side x, 2r = x\sqrt{2}. The ratio of circle area to square area works out to \pi : 2.
What to practise from Area and Perimeter of Plane Figures Class 9 ICSE
The chapter’s practice is grouped into three exercise blocks. Each block targets a specific pair of sections from the chapter.
- Exercise 20(A) — 15 questions on triangles, testing sections 20.2 and 20.3.
- Exercise 20(B) — 36 questions on quadrilaterals, testing sections 20.4 and 20.5.
- Exercise 20(C) — 27 questions on circles, testing sections 20.6 and 20.7.
| Exercise | Sections it tests | Core content |
|---|---|---|
| Exercise 20(A) | 20.2, 20.3 | triangles: base-height formula, Heron’s formula, special triangles |
| Exercise 20(B) | 20.4, 20.4 | quadrilaterals: the two general methods, then the five special shapes |
| Exercise 20(C) | 20.6, 20.7 | circles: circumference, area, rings, wheel rotations |
You can work the exercises in book order, or jump straight to the block you need. Every chapter guide for this year group, including this one, sits on the Class 9 hub; the chapter itself follows the Class 9 Mathematics syllabus published on the official CISCE website.
Area and Perimeter of Plane Figures Class 9 ICSE frequently asked questions
What does area and perimeter of plane figures class 9 ICSE cover in Chapter 20?
Chapter 20 of the Class 9 ICSE Mathematics book covers perimeter and area for plane figures. Sections 20.1 to 20.3 treat triangles, sections 20.4 and 20.5 treat quadrilaterals, and sections 20.6 and 20.7 treat circles. The three exercises attached to the chapter test each block in turn.
What is the difference between square metre and metre square?
‘Square metre’ is a unit of area, written \mathrm{m}^2. ‘Metre square’ describes a square whose side is one metre long, and that square’s area is one square metre. The two diverge with bigger numbers: 3 square metres is a small area, but a 3 metre square has area 9\ \mathrm{m}^2.
When should I use Heron’s formula instead of half base times height?
Use \frac{1}{2} \times \text{base} \times \text{height} when you know a base and its corresponding perpendicular height. Use Heron’s formula when you know all three sides and no height. Section 20.2 demonstrates the switch with sides 13 cm, 14 cm and 15 cm.
How many methods are there to find the area of an isosceles triangle?
This ICSE Class 9 chapter shows three methods. Drop the perpendicular from the vertex and apply Pythagoras, apply Heron’s formula, or use \frac{1}{4}b\sqrt{4a^2-b^2}. All three return the same area — 48 cm² for the worked pattern with equal sides 10 cm and base 12 cm.
How do I find the number of rotations a wheel makes over a given distance?
Divide the total distance by the wheel’s circumference, after converting both to the same unit. One rotation covers exactly one circumference, so rotations = distance \div 2\pi r. The worked pattern in section 20.7 gives 1500 rotations for 3.96 km with a 42 cm radius wheel.
The syllabus this chapter follows is set by the CISCE board.
Reference: ICSE Class 9 Mathematics textbook material, chapter 20: Area and Perimeter of Plane Figures — CISCE-aligned.