Measuring Space: Perimeter and Area Class 9 Notes, Formulas and Solved Questions

What Measuring Space: Perimeter and Area is about

Weightage: This chapter is CBSE's Area and Perimeter, part of Unit V, Mensuration, which carries 14 of the 80 marks in the school annual exam, shared with Surface Area and Volume.Source: CBSE Mathematics syllabus, Class IX 2026-27, course structure, Unit V Mensuration

Key concepts

Circumference and the number π

NCERT section 6.2 Perimeter of a Circle — The C/D Ratio

The perimeter is the total length around a shape. For a square it is $4a$; for a rectangle $2(a+b)$. For every square, perimeter ÷ side is 4, whatever its size.

The same is true of circles: circumference ÷ diameter is the same for all circles. This constant is π (from the Greek *perimetros*), so
$C=\pi d=2\pi r$

π is irrational (Lambert proved it in 1761), so no fraction equals it. Useful approximations: $\frac{22}{7}\approx 3.1429$, 3.14, Āryabhaṭa's 3.1416 (which he called *asanna*, approximate) and Zu Chongzhi's $\frac{355}{113}$. Mādhava gave the first exact formula as an infinite series. Write $\pi \approx \frac{22}{7}$, never $\pi =\frac{22}{7}$.

Worked example

The circumference of a circular garden is 88 m. Find its radius (use $\pi \approx \frac{22}{7}$). By how much would the circumference grow if the radius increased by 7 m?

Try it first, then show the solutionSolution

1. $2\pi r=88$, so $r=\frac{88}{2}\times \frac{7}{22}=14$ m.
2. New circumference $=2\pi (14+7)=2\times \frac{22}{7}\times 21=132$ m.
3. Increase $=132-88=44$ m, which is $2\pi \times 7$.

Answer: Radius 14 m; the circumference grows by 44 m.

Length of an arc and perimeter of a sector

NCERT section 6.4 Length of an Arc of a Circle

By symmetry, a semicircle is half the circumference ($\pi r$) and a quarter circle is a quarter ($\frac{\pi r}{2}$). In general, an arc that subtends $\theta °$ at the centre is that fraction of the full turn:
$l=2\pi r\times \frac{\theta }{360°}$

A sector is the region between an arc and the two radii to its ends. Its perimeter is the arc plus the two radii: $l+2r$. Do not forget the radii.

Worked example

Find the length of an arc of a circle of radius 21 cm that subtends 60° at the centre, and the perimeter of the sector.

Try it first, then show the solutionSolution

1. $l=2\times \frac{22}{7}\times 21\times \frac{60}{360}=132\times \frac{1}{6}=22$ cm.
2. Perimeter of sector $=22+2\times 21=64$ cm.

Answer: Arc 22 cm; sector perimeter 64 cm.

Why running tracks need a stagger

NCERT section 6.4 Length of an Arc of a Circle

On a standard 400 m track, the two bends together make a full circle. A runner in an outer lane runs on a circle of bigger radius, so her bends are longer while the straights are the same.

If the lane width is w, the outer runner's radius is w more, and her full circle is longer by
$2\pi (r+w)-2\pi r=2\pi w$
The difference does not depend on r. So every lane needs the same extra start (stagger) over the lane inside it: about $2\pi \times 1.22\approx 7.67$ m for a full lap with lanes 1.22 m wide, and half of that for one bend.

Worked example

Lanes are 1.22 m wide. Using $\pi \approx 3.1416$, find the stagger needed between lane 2 and lane 1 for (i) one full lap (two bends) (ii) a 200 m race with one bend.

Try it first, then show the solutionSolution

1. Extra distance for two bends (a full circle) $=2\pi \times 1.22=2\times 3.1416\times 1.22\approx 7.67$ m.
2. Extra for one bend (a semicircle) $=\pi \times 1.22\approx 3.83$ m.

Answer: (i) about 7.67 m (ii) about 3.83 m; the same between any two neighbouring lanes.

Areas of rectangles, parallelograms and triangles

NCERT section 6.8 Area of a Triangle

• Rectangle: $A=ab$; a square is the special case $a^2$.
• Parallelogram: cut a right triangle off one end and move it to the other to make a rectangle with the same base and height: $A=bh$. Knowing only the sides is not enough; the angle (or height) matters.
• Triangle: two copies make a parallelogram, so $A=\frac{1}{2}bh$.
• Median theorem: a median joins a vertex to the midpoint of the opposite side. The two triangles it makes have equal bases and the same height, so a median divides a triangle into two triangles of equal area, even though they are usually not congruent.

The height is always the perpendicular distance to the line of the base.

Worked example

A parallelogram ABCD has base AB = 12 cm and height 7.5 cm. P is any point on CD. Find the area of the parallelogram and of $△PAB$. If M is the midpoint of AB, find the area of $△PAM$.

Try it first, then show the solutionSolution

1. Area of ABCD $=12\times 7.5=90$ cm².
2. $△PAB$ has base AB and height 7.5 cm (P is on the opposite side): $\frac{1}{2}\times 12\times 7.5=45$ cm².
3. PM is a median of $△PAB$, so $△PAM$ has half its area: 22.5 cm².

Answer: 90 cm², 45 cm² and 22.5 cm².

Heron's formula

NCERT section 6.8 Area of a Triangle

If a triangle has sides a, b, c, first find the semi-perimeter $s=\frac{a+b+c}{2}$. Then
$\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}$

It needs no height, so it is the tool when only the three sides are known (land plots, signboards). Check it on a 3, 4, 5 triangle: $s=6$, area $=\sqrt{6\times 3\times 2\times 1}=6$, the same as $\frac{1}{2}\times 3\times 4$.

For an equilateral triangle of side a it gives $\frac{\sqrt{3}}{4}{a}^2$. The chapter also mentions $\text{Area}=\frac{abc}{4R}$ (R the circumradius) and $\frac{1}{2}r(a+b+c)$ (r the inradius).

Worked example

Find the area of a triangle with sides 13 cm, 14 cm and 15 cm, and the height on the 14 cm side.

Try it first, then show the solutionSolution

1. $s=\frac{13+14+15}{2}=21$.
2. Area $=\sqrt{21\times 8\times 7\times 6}=\sqrt{7056}=84$ cm².
3. Height on the 14 cm side: $\frac{1}{2}\times 14\times h=84$, so $h=12$ cm.

Answer: Area 84 cm²; height 12 cm.

Brahmagupta's formula for a cyclic quadrilateral

NCERT section 6.8 Area of a Triangle

The four sides alone do not fix the area of a quadrilateral (a rhombus of side 3 can be squashed flatter and flatter). But if the quadrilateral is cyclic, Brahmagupta (628 CE) showed that with $s=\frac{a+b+c+d}{2}$:
$\text{Area}=\sqrt{(s-a)(s-b)(s-c)(s-d)}$

• Every rectangle is cyclic: the formula gives $ab$.
• Every isosceles trapezium is cyclic: it agrees with $\frac{1}{2}(\text{sum of parallel sides})\times h$.
• Put $d=0$ (the fourth side shrinks to a point) and it becomes Heron's formula. So Brahmagupta's formula is a generalisation of Heron's, and Heron's is a special case.

Worked example

An isosceles trapezium has parallel sides 10 cm and 20 cm and each slant side 13 cm. Find its area by Brahmagupta's formula and check it with the trapezium formula.

Try it first, then show the solutionSolution

1. Isosceles trapezia are cyclic. $s=\frac{10+13+20+13}{2}=28$.
2. Area $=\sqrt{(28-10)(28-13)(28-20)(28-13)}=\sqrt{18\times 15\times 8\times 15}=\sqrt{32400}=180$ cm².
3. Check: each overhang is $\frac{20-10}{2}=5$, so height $=\sqrt{13^2-5^2}=12$.
4. $\frac{1}{2}(10+20)\times 12=180$ cm². The answers agree.

Answer: 180 cm²

Squaring a rectangle (Baudhāyana)

NCERT section 6.9 Squaring a Rectangle

To square a shape is to construct a square of the same area. Baudhāyana's Śulbasūtra (c. 800 BCE) squares a rectangle of sides a and b ($a>b$). The construction rests on the identity
${\left(\frac{a+b}{2}\right)}^2-{\left(\frac{a-b}{2}\right)}^2=ab$
In the figure, a right triangle is built with hypotenuse $\frac{a+b}{2}$ and one leg $\frac{a-b}{2}$; by the Baudhāyana–Pythagoras theorem its other leg is $\sqrt{ab}$, the side of the required square.

So the square equal to an $a\times b$ rectangle has side $\sqrt{ab}$.

Worked example

A rectangle is 9 cm by 4 cm. Use Baudhāyana's idea to find the side of the square with the same area, and check.

Try it first, then show the solutionSolution

1. Hypotenuse $=\frac{9+4}{2}=6.5$ cm; one leg $=\frac{9-4}{2}=2.5$ cm.
2. Other leg${}^2=6.5^2-2.5^2=42.25-6.25=36$, so the leg is 6 cm.
3. Check: $6^2=36=9\times 4$.

Answer: A square of side 6 cm.

Area of a circle

NCERT section 6.10 Area of a Circle

$A=\pi r^2$

Why (Nīlakaṇṭha's picture): cut the circle into many thin equal slices and lay them alternately point-up and point-down. The arrangement looks more and more like a parallelogram with base half the circumference ($\pi r$) and height r, so the area is $\pi r\times r$.

Archimedes put it as: the area equals a right triangle with legs r and the circumference, $\frac{1}{2}\times 2\pi r\times r$. Earlier cultures had approximations, such as the Egyptian rule (also in Baudhāyana) $A\approx {\left(\frac{8}{9}d\right)}^2$.

For a ring (annulus) between radii R and r: $A=\pi (R^2-r^2)$.

Worked example

A circle has circumference 88 cm. Find its area.

Try it first, then show the solutionSolution

1. $2\pi r=88$ gives $r=14$ cm.
2. $A=\frac{22}{7}\times 14\times 14=616$ cm².

Answer: 616 cm²

Area of a sector and a segment

NCERT section 6.10 Area of a Circle

A sector with angle $\theta °$ is that fraction of the disc:
$A_{\text{sector}}=\pi r^2\times \frac{\theta }{360°}$
A semicircular disc is $\frac{1}{2}\pi r^2$; a quadrant is $\frac{1}{4}\pi r^2$.

A segment is the region between an arc and its chord:

• minor segment $=$ sector $-$ triangle formed by the two radii and the chord
• major segment $=$ circle $-$ minor segment

For $\theta =90°$ the triangle is right-angled with area $\frac{1}{2}{r}^2$; for $\theta =60°$ it is equilateral with area $\frac{\sqrt{3}}{4}{r}^2$.

Worked example

A chord of a circle of radius 14 cm subtends 90° at the centre. Find the areas of the minor sector and the minor segment.

Try it first, then show the solutionSolution

1. Sector $=\frac{22}{7}\times 196\times \frac{90}{360}=154$ cm².
2. Triangle (right-angled, legs 14 and 14) $=\frac{1}{2}\times 14\times 14=98$ cm².
3. Minor segment $=154-98=56$ cm².

Answer: Sector 154 cm²; segment 56 cm².

Maths in real life: Measuring Space: Perimeter and Area

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Circumference, Arc Length, Sector and Segment
• Heron's Formula and Areas of Quadrilaterals

Measuring Space: Perimeter and Area formula sheet

Perimeters

$\text{square}=4a,\text{rectangle}=2(a+b),\text{equilateral triangle}=3a$

Circumference

$C=\pi d=2\pi r$

π is irrational; $\pi \approx \frac{22}{7}$ or 3.14.

Arc length

$l=2\pi r\times \frac{\theta }{360^{\circ }}$

Perimeter of a sector

$l+2r$

Track stagger (one full circle)

$2\pi (r+w)-2\pi r=2\pi w$

w = lane width; half of this for one bend.

Rectangle and parallelogram

$A=ab,A=bh$

Triangle

$A=\frac{1}{2}bh$

A median splits it into two equal areas.

Heron's formula

$A=\sqrt{s(s-a)(s-b)(s-c)},s=\frac{a+b+c}{2}$

Equilateral triangle

$A=\frac{\sqrt{3}}{4}{a}^2$

Brahmagupta's formula (cyclic 4-gon)

$A=\sqrt{(s-a)(s-b)(s-c)(s-d)},s=\frac{a+b+c+d}{2}$

Trapezium

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

Rhombus or kite

$A=\frac{1}{2}{d}_1{d}_2$

Squaring a rectangle

${\left(\frac{a+b}{2}\right)}^2-{\left(\frac{a-b}{2}\right)}^2=ab$

Side of the equal square $=\sqrt{ab}$.

Area of a circle

$A=\pi r^2$

Area of a sector

$A=\pi r^2\times \frac{\theta }{360^{\circ }}$

Area of a segment

$\text{minor}=\text{sector}-△;\text{major}=\pi r^2-\text{minor}$

Ring (annulus)

$A=\pi (R^2-r^2)$

How the school exam asks Measuring Space: Perimeter and Area

The question patterns to expect, and the method that works for each.

Tricks to remember Measuring Space: Perimeter and Area

Short memory aids. Say each one aloud once, then use it on a question the same day.

• How I wish I could recollect pi

  Count the letters: 3, 1, 4, 1, 5, 9, 2.

  Use it when: Recalling π ≈ 3.141592.

• Slice of the turn

  Arc length and sector area are just 'full circle × fraction of the turn', and the fraction is $\frac{\theta }{360°}$.

  Use it when: Writing arc and sector formulas without memorising separately.

• s minus each side

  Heron: s times (s minus each side). Brahmagupta: drop the lone s and use four sides.

  Use it when: Keeping the two formulas apart.

• Segment = slice minus triangle

  Cut a pizza slice straight across its crust ends: what is left near the crust is the segment.

  Use it when: Minor segment = sector − triangle.

• Stagger = 2π × lane width

  The radius cancels: only the lane width matters.

  Use it when: Track and ring problems.

Measuring Space: Perimeter and Area: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Measuring Space: Perimeter and Area

The slips examiners see most often, and what to write instead.

• ✗Mistake: Forgetting the two radii in a sector's perimeter

  Only the curved part is calculated.

  Wrong working

  Perimeter of sector ($r=21$, $60°$) $=22$ cm

  Right working

  $22+21+21=64$ cm

  ✓Do this: Trace the whole boundary with your finger before adding.

  Usually costs: 1 mark.

• ✗Mistake: Using the perimeter instead of the semi-perimeter in Heron's formula

  s is mistaken for the full perimeter.

  Wrong working

  Sides 13, 14, 15: $s=42$

  Right working

  $s=\frac{42}{2}=21$

  ✓Do this: Write 's = half the perimeter' as the first line.

  Usually costs: Most of the marks for the question.

• ✗Mistake: Mixing units in wheel problems

  Distance is in km and circumference in cm.

  Wrong working

  Revolutions $=\frac{11}{176}=0.0625$

  Right working

  $11$ km $=1100000$ cm; $\frac{1100000}{176}=6250$ revolutions

  ✓Do this: Convert to the same unit before dividing.

  Usually costs: The final answer mark.

• ✗Mistake: Using a slant side as the height

  The side looks like it goes 'up'.

  Wrong working

  Parallelogram with base 10 and slant side 6: area $=60$

  Right working

  Area $=$ base × perpendicular height; with height 5, area $=50$

  ✓Do this: The height must be perpendicular to the base.

  Usually costs: 1 to 2 marks.

• ✗Mistake: Writing π = 22/7

  The approximation is treated as exact.

  Wrong working

  π $=\frac{22}{7}$, so π is rational.

  Right working

  $\pi \approx \frac{22}{7}$; π is irrational.

  ✓Do this: Use '≈' and use the value of π the question specifies.

  Usually costs: 1-mark MCQ or assertion-reason.

• ✗Mistake: Applying Brahmagupta's formula to any quadrilateral

  The cyclic condition is forgotten.

  Wrong working

  Rhombus of side 3: area $=\sqrt{3^4}=9$ always.

  Right working

  Only the square (which is cyclic) has area 9; a non-square rhombus is not cyclic and has a smaller area.

  ✓Do this: Check the shape is cyclic before using the formula.

  Usually costs: The whole answer.

NCERT exercise guide: Measuring Space: Perimeter and Area

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Practice questions on Measuring Space: Perimeter and Area

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The area of a semicircle of diameter 14 cm is
   (a) 154 cm²
   (b) 77 cm²
   (c) 44 cm²
   (d) 22 cm²

   Hint

   Radius 7 cm; half of $\pi r^2$.

   Answer

   (b) 77 cm²

2. Q2 · 1 mark · MCQExam level

   For a triangle with sides 7 cm, 24 cm and 25 cm, the semi-perimeter s is
   (a) 56 cm
   (b) 28 cm
   (c) 25 cm
   (d) 14 cm

   Hint

   Half of the perimeter.

   Answer

   (b) 28 cm

3. Q3 · 1 mark · Assertion-reasonExam level

   Assertion (A): A median of a triangle divides it into two triangles of equal area.
   Reason (R): The two triangles have equal bases and the same height.
   (a) Both A and R are true and R is the correct explanation of A.
   (b) Both A and R are true but R is not the correct explanation of A.
   (c) A is true but R is false.
   (d) A is false but R is true.

   Hint

   Where does the median meet the opposite side?

   Answer

   (a)

4. Q4 · 2 marks · Very short answerExam level

   Find the length of an arc of a circle of radius 10.5 cm that subtends 120° at the centre.

   Hint

   $\frac{120}{360}=\frac{1}{3}$.

   Answer

   $2\times \frac{22}{7}\times 10.5\times \frac{1}{3}=22$ cm.

5. Q5 · 3 marks · Short answerHarder exam level

   Two sides of a triangle are 9 cm and 10 cm, and its perimeter is 36 cm. Find its area.

   Hint

   Find the third side, then use Heron's formula.

   Answer

   Third side 17 cm, $s=18$, area $=\sqrt{18\times 9\times 8\times 1}=36$ cm².

6. Q6 · 2 marks · Very short answerHarder exam level

   One diagonal of a rhombus is twice the other, and its area is 100 cm². Find both diagonals.

   Hint

   Area $=\frac{1}{2}{d}_1{d}_2$.

   Answer

   $\frac{1}{2}\times d\times 2d=d^2=100$, so 10 cm and 20 cm.

7. Q7 · 2 marks · Very short answerHarder exam level

   A car wheel has radius 35 cm. How many revolutions does it make in 2.2 km?

   Hint

   Circumference in cm first.

   Answer

   Circumference $=220$ cm; $\frac{220000}{220}=1000$ revolutions.

8. Q8 · 3 marks · Short answerHarder exam level

   In a circle of radius 10 cm, a chord makes a right angle at the centre. Work out the areas of the minor and major segments. (Use $\pi \approx 3.14$.)

   Hint

   The triangle is right-angled.

   Answer

   Sector $=78.5$ cm², triangle $=50$ cm², minor segment $=28.5$ cm², major segment $=314-28.5=285.5$ cm².

9. Q9 · 3 marks · Short answerHarder exam level

   A car has two wipers of length 21 cm that do not overlap; each sweeps 120°. Find the total area cleaned in one sweep.

   Hint

   Two sectors of 120°.

   Answer

   Each sector $=\frac{22}{7}\times 441\times \frac{1}{3}=462$ cm²; total $=924$ cm².

10. Q10 · 5 marks · Long answerHarder exam level

    The sides of a triangular field are in the ratio 13 : 14 : 15 and its perimeter is 84 m. Find its area and the cost of levelling it at ₹5 per m².

    Hint

    Sides 26, 28, 30.

    Answer

    $s=42$; area $=\sqrt{42\times 16\times 14\times 12}=\sqrt{112896}=336$ m²; cost $=₹1,680$.

11. Q11 · 3 marks · Short answerHOTS / competency

    Using Baudhāyana's identity, find the side of a square equal in area to a 16 cm × 9 cm rectangle, and show the working.

    Hint

    ${\left(\frac{a+b}{2}\right)}^2-{\left(\frac{a-b}{2}\right)}^2=ab$.

    Answer

    $12.5^2-3.5^2=156.25-12.25=144=16\times 9$, so the side is 12 cm.

12. Q12 · 3 marks · Short answerHOTS / competency

    Among all rectangles with perimeter 40 cm, which has the largest area? Prove it using the identity $x(20-x)=100-(x-10{)}^2$.

    Hint

    A square can never be negative.

    Answer

    If one side is x, the other is $20-x$ and the area is $100-(x-10{)}^2\le 100$, with equality only at $x=10$. So the 10 cm square has the largest area, 100 cm²; there is no smallest, since the area can be made as close to 0 as you like.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.What is π, as a ratio?Show answer

   Circumference ÷ diameter, the same for every circle.

2. 2.Is π equal to 22/7?Show answer

   No. π is irrational; $\frac{22}{7}$ is an approximation.

3. 3.Write the formula for the length of an arc.Show answer

   $2\pi r\times \frac{\theta }{360°}$

4. 4.Why is the track stagger the same for every lane?Show answer

   The extra length is $2\pi w$ per full circle, which depends only on the lane width w.

5. 5.What does a median do to the area of a triangle?Show answer

   It splits it into two triangles of equal area.

6. 6.State Heron's formula.Show answer

   $\sqrt{s(s-a)(s-b)(s-c)}$ with $s=\frac{a+b+c}{2}$.

7. 7.When can Brahmagupta's formula be used?Show answer

   For a cyclic quadrilateral (vertices on one circle).

8. 8.How does Brahmagupta's formula give Heron's?Show answer

   Put the fourth side $d=0$.

9. 9.What is the side of a square equal in area to an $a\times b$ rectangle?Show answer

   $\sqrt{ab}$

10. 10.How do you find the area of a minor segment?Show answer

    Area of the sector minus the area of the triangle formed by the two radii and the chord.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you find the radius of a circle with circumference 132 cm?
• ☐Can you find an arc length and a sector perimeter for 45°?
• ☐Can you explain why outer lanes start ahead on a track?
• ☐Can you prove that a median halves the area of a triangle?
• ☐Can you use Heron's formula for sides 5, 12 and 13 and check with $\frac{1}{2}bh$?
• ☐Can you find the area of an isosceles triangle from its perimeter and equal sides?
• ☐Can you use Brahmagupta's formula on a rectangle and get $ab$?
• ☐Can you find the side of a square equal in area to a 16 by 9 rectangle?
• ☐Can you find the area swept by a minute hand in 20 minutes?
• ☐Can you find minor and major segment areas for a 90° chord?

Measuring Space: Perimeter and Area: frequently asked questions