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Class 9 · Chapter 12 · Mensuration unit (14 of 80 marks)

Area and Perimeter Class 9: notes and important questions

Revision notes, 17 board-style questions with the step marks shown, and a four-step route from the basics to 95+, each step ending in a short checkpoint.

  • 17 questions
  • 5 multiple choice, 1 assertion–reason, 3 very short answer, 4 short answer, 2 long answer, 2 case study
  • About 9 hours to master

Mensuration unit: 14 of 80 theory marks (Area and Perimeter, Surface Area and Volume).

In the NCERT book Ganita Manjari: Part I, Chapter 6, Measuring Space: Perimeter and Area.

Revision notes

Area and Perimeter: revision notes

1. Perimeter and arc length

  • Perimeter of a polygon: the sum of its sides. Circumference of a circle: \(C = 2\pi r = \pi d\).
  • An arc of a circle that subtends \(\theta^\circ\) at the centre has length \(l = \dfrac{\theta}{360} \times 2\pi r\). The perimeter of the sector is \(l + 2r\).
  • A semicircle of radius \(r\) has perimeter \(\pi r + 2r\).

2. Areas of standard shapes

Triangle \(\dfrac12 \times \text{base} \times \text{height}\); rectangle \(lb\); parallelogram \(bh\); trapezium \(\dfrac12(a + b)h\); rhombus \(\dfrac12 d_1d_2\); equilateral triangle of side \(a\): \(\dfrac{\sqrt3}{4}a^2\). Circle \(\pi r^2\); sector of angle \(\theta^\circ\): \(\dfrac{\theta}{360}\pi r^2\).

3. Heron's formula

For a triangle with sides \(a, b, c\) and semi-perimeter \(s = \dfrac{a + b + c}{2}\):

\[\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}\]

It needs only the three sides, so it works when no height is known. A quadrilateral with a known diagonal splits into two triangles.

4. Incircle

The incircle touches all three sides; its centre is where the angle bisectors meet. Joining the incentre to the vertices splits the triangle into three triangles of height \(r\), so \(\text{Area} = rs\) and \(r = \dfrac{\text{Area}}{s}\).

5. Brahmagupta's formula

For a cyclic quadrilateral with sides \(a, b, c, d\) and \(s = \dfrac{a + b + c + d}{2}\): \(\text{Area} = \sqrt{(s - a)(s - b)(s - c)(s - d)}\). Heron's formula is the case \(d = 0\).

Worked example 1

Find the area of a triangle with sides \(13\), \(14\) and \(15\) cm.

\(s = 21\); area \(= \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84\) cm².

Worked example 2

Find the length of an arc of angle \(90^\circ\) in a circle of radius \(14\) cm (take \(\pi = \dfrac{22}{7}\)).

\(l = \dfrac14 \times 2 \times \dfrac{22}{7} \times 14 = 22\) cm.

Worked example 3

Find the inradius of the triangle with sides \(6\), \(8\) and \(10\) cm.

Area \(= 24\) cm², \(s = 12\), so \(r = 2\) cm.

Common errors

  • Using the perimeter instead of the semi-perimeter \(s\) in Heron's formula.
  • Forgetting the two radii in the perimeter of a sector or the diameter in the perimeter of a semicircle.
  • Using Brahmagupta's formula for a quadrilateral that is not cyclic.
  • Mixing units (cm and m) or writing an area in cm instead of cm².

Exam tips

  • Write \(s\), then \(s - a\), \(s - b\), \(s - c\) on separate lines before multiplying; factorise under the root instead of multiplying out.
  • Use the value of \(\pi\) the question gives; otherwise leave the answer in terms of \(\pi\).

Topics in this chapter: Heron's formula · Perimeter and arc length · Areas of standard shapes · Brahmagupta's formula · Incircle.

Route to 95: four steps

Work through the steps in order. Take each checkpoint closed book, about 1.5 minutes per mark; pass at 80% to move on. Two misses in a row means going back one step.

Step 1

Secure the basics

You can use Heron's formula, circle and arc formulas and the standard area formulas in one step.

Read first: 1. Perimeter and arc length; 2. Areas of standard shapes; 3. Heron's formula 5 practice questions · checkpoint: 3 questions, 3 marks, pass 80%
Practise step 1
Step 2

Exam standard

You can find missing sides from a perimeter, an inradius from Area = rs, and areas of rhombuses and isosceles triangles.

Read first: 3. Heron's formula; 4. Incircle; Worked examples 1-3 6 practice questions · checkpoint: 3 questions, 8 marks, pass 80%
Practise step 2
Step 3

Full marks on long answers

You can find the area of a quadrilateral in two triangles and answer multi-part questions on tracks, sectors and signboards.

Read first: 1. Arc length; 3. Heron's formula (quadrilaterals) 3 practice questions · checkpoint: 2 questions, 9 marks, pass 80%
Practise step 3
Step 4

95+ stretch (HOTS)

You can use Brahmagupta's formula, altitudes from areas, and composite shapes with circles.

Read first: 4. Incircle; 5. Brahmagupta's formula; Common errors 3 practice questions · checkpoint: 3 questions, 9 marks, pass 80%
Practise step 4

Practice questions

Original questions in the board's styles. Multiple-choice answers are checked as you go; for written answers, compare your working with the step mark scheme and record your marks. 3 of the 17 are competency-based (case studies and questions set in a real-life situation): the "Competency-based" button shows just those.

Q1·1 mark·Multiple choiceHeron's formula

The area of a triangle with sides \(13\) cm, \(14\) cm and \(15\) cm is

  1. (a)\(105\) cm²
  2. (b)\(91\) cm²
  3. (c)\(42\) cm²
  4. (d)\(84\) cm²
Q2·1 mark·Multiple choicePerimeter and arc length

The perimeter of a semicircular protractor of radius \(7\) cm is (take \(\pi = \dfrac{22}{7}\))

  1. (a)\(22\) cm
  2. (b)\(44\) cm
  3. (c)\(36\) cm
  4. (d)\(29\) cm
Q3·1 mark·Multiple choiceAreas of standard shapes

The area of an equilateral triangle of side \(8\) cm is

  1. (a)\(32\sqrt3\) cm²
  2. (b)\(16\sqrt3\) cm²
  3. (c)\(64\) cm²
  4. (d)\(8\sqrt3\) cm²

Stretch yourself: original geometry problems at Intermediate level (ages 13 to 16), with hints and full solutions: Problem I99 · Problem I17 · Problem I85 · Problem I92. All 28 →

Where marks are lost in Area and Perimeter

  • Perimeter used in place of the semi-perimeter in Heron's formula. Fix: s = (a + b + c)/2, then s − a, s − b, s − c.
  • Sector or semicircle perimeter without the straight edges. Fix: arc + 2r for a sector, πr + 2r for a semicircle.
  • Arithmetic slips multiplying large numbers under the root. Fix: factorise into primes first (21 × 8 × 7 × 6 = 2⁴ × 3² × 7²).
  • Brahmagupta's formula used on a quadrilateral not known to be cyclic. Fix: otherwise split along a diagonal and use Heron twice.
  • Wrong or missing units, or π = 22/7 used when the question did not give it. Fix: square units for area; exact π unless a value is given.

Test Area and Perimeter against the clock

Want it against the clock? Take a timed 30-mark chapter test on Area and Perimeter, new questions each time, or climb the Difficulty ladder, levels 1 to 10, three questions a level (CBSE Essentials or the free trial).