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Measuring Space: Perimeter and Area

NCERT Class 9 · Mathematics Based on NCERT Class 9 Mathematics textbook · Free CBSE study kit

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Chapter Notes

Overview
This chapter explores the concepts of perimeter and area, starting with the practical problem of calculating staggers on an athletics track. It builds from the perimeter of simple shapes like squares and rectangles to the circumference of a circle, introducing the constant π and its historical development. The chapter then covers area formulas for rectangles, parallelograms, triangles (including Heron's formula), and circles, and concludes with Brahmagupta's formula for cyclic quadrilaterals and Baudhāyana's method of squaring a rectangle. The one big idea is that perimeter is the total distance around a shape, while area is the space it encloses, and both can be calculated using formulas that often involve π or are derived from simpler shapes.
Key Concepts (15)
Perimeter
The total length around the border of a shape.
e.g. The distance an athlete runs around a 400 m track is the perimeter of the track.
Circumference
The perimeter of a circle.
e.g. The distance around a circular tyre is its circumference.
π (pi)
The ratio of the circumference of a circle to its diameter, a constant approximately equal to 3.14159...
e.g. If a circle has a diameter of 7 cm, its circumference is π × 7 ≈ 22 cm.
Arc
A part of the circumference of a circle.
e.g. The curved part of a sector of a circle is an arc.
Sector
A region bounded by two radii and the arc between them.
e.g. A slice of pizza is a sector of a circle.
Area
The amount of space occupied by a two-dimensional region in a plane, measured in square units.
e.g. The space inside a rectangular garden is its area.
Semi-perimeter (s)
Half the perimeter of a shape, used in Heron's formula.
e.g. For a triangle with sides 3, 4, and 5 cm, s = (3+4+5)/2 = 6 cm.
Median of a triangle
A line segment joining a vertex of a triangle to the midpoint of the opposite side.
e.g. In triangle ABC, if D is the midpoint of BC, then AD is a median.
Cyclic quadrilateral
A quadrilateral whose all four vertices lie on a single circle.
e.g. A rectangle is always a cyclic quadrilateral.
Incircle
The circle that fits tightly inside a triangle, touching all three sides.
e.g. The incircle of a triangle touches each side at exactly one point.
Circumcircle
The circle that passes through all three vertices of a triangle.
e.g. The circumcircle of a right-angled triangle has its centre at the midpoint of the hypotenuse.
Stagger
The distance between the starting points of adjacent lanes on a curved track to compensate for the longer distance run on outer lanes.
e.g. In a 400 m race, runners in outer lanes start ahead of those in inner lanes by a calculated stagger.
Irrational number
A number that cannot be written as a ratio of two integers, e.g., π and √2.
e.g. π is irrational because its decimal expansion goes on forever without a repeating pattern.
Special case
A result obtained from a general result by applying an extra condition.
e.g. A square is a special case of a rectangle, so the area formula for a square (a²) is a special case of the rectangle formula (ab) with b = a.
Generalisation
The process of extending a result to cover a broader set of cases.
e.g. Brahmagupta's formula for cyclic quadrilaterals is a generalisation of Heron's formula for triangles.
Formulas & Equations (13)
Perimeter of a square = 4a
Reads as: Perimeter of a square equals 4 times the side length
Variables: a = side length of the square (units)
Use when: Finding the total distance around a square
⚠ Using 'a' as the area instead of side length
Perimeter of an equilateral triangle = 3a
Reads as: Perimeter of an equilateral triangle equals 3 times the side length
Variables: a = side length of the triangle (units)
Use when: Finding the total distance around an equilateral triangle
⚠ Forgetting that all sides are equal
Perimeter of a rectangle = 2(a + b)
Reads as: Perimeter of a rectangle equals 2 times the sum of length and width
Variables: a = length, b = width (units)
Use when: Finding the total distance around a rectangle
⚠ Writing 2a + b instead of 2(a + b)
Circumference of a circle = πd = 2πr
Reads as: Circumference of a circle equals pi times diameter, or 2 pi times radius
Variables: d = diameter, r = radius (units), π ≈ 3.1416 or 22/7
Use when: Finding the perimeter of a circle
⚠ Using radius instead of diameter in πd, or forgetting the factor 2 in 2πr
Length of an arc = 2πr × (θ/360°)
Reads as: Length of an arc equals 2 pi r times theta over 360 degrees
Variables: r = radius (units), θ = angle subtended at the centre (degrees)
Use when: Finding the length of a part of a circle's circumference
⚠ Forgetting to convert the angle to a fraction of 360°
Area of a rectangle = a × b
Reads as: Area of a rectangle equals length times width
Variables: a = length, b = width (units)
Use when: Finding the space inside a rectangle
⚠ Confusing area with perimeter
Area of a square = a²
Reads as: Area of a square equals side squared
Variables: a = side length (units)
Use when: Finding the space inside a square
⚠ Writing 4a instead of a²
Area of a parallelogram = base × height = bh
Reads as: Area of a parallelogram equals base times height
Variables: b = base, h = height (units)
Use when: Finding the space inside a parallelogram
⚠ Using the slant side instead of the perpendicular height
Area of a triangle = (1/2) × base × height = (1/2)bh
Reads as: Area of a triangle equals half times base times height
Variables: b = base, h = height (units)
Use when: Finding the space inside a triangle
⚠ Forgetting the factor 1/2
Heron's formula: Area = √[s(s-a)(s-b)(s-c)], where s = (a+b+c)/2
Reads as: Area equals the square root of s times s minus a times s minus b times s minus c, where s is the semi-perimeter
Variables: a, b, c = side lengths of triangle (units), s = semi-perimeter (units)
Use when: Finding the area of a triangle when only side lengths are known
⚠ Forgetting to take the square root of the product
Area of a triangle using circumradius: Area = abc/(4R)
Reads as: Area equals product of sides divided by 4 times the circumradius
Variables: a, b, c = side lengths (units), R = circumradius (units)
Use when: Finding area when sides and circumradius are known
⚠ Confusing R with the inradius
Area of a triangle using inradius: Area = r × (a+b+c)/2 = r × s
Reads as: Area equals inradius times semi-perimeter
Variables: r = inradius (units), s = semi-perimeter (units)
Use when: Finding area when inradius and side lengths are known
⚠ Forgetting that s is half the perimeter
Brahmagupta's formula for cyclic quadrilateral: Area = √[(s-a)(s-b)(s-c)(s-d)], where s = (a+b+c+d)/2
Reads as: Area equals the square root of s minus a times s minus b times s minus c times s minus d
Variables: a, b, c, d = side lengths (units), s = semi-perimeter (units)
Use when: Finding the area of a cyclic quadrilateral when all four sides are known
⚠ Applying it to a non-cyclic quadrilateral
How It Works
Estimating the C/D ratio (π) by home measurement
  1. Take a cotton reel with thin thread wrapped around it.
  2. Measure the diameter D of the reel accurately.
  3. Unwrap and then tightly wrap the thread around the reel 20 times.
  4. Unwrap the thread again and measure its total length L.
  5. Calculate the ratio L/(20D). This gives an estimate of π.
💡 The thread must be very thin for accuracy; the ratio should be between 3 and 4, ideally between 3.1 and 3.2.
Archimedes' method to trap π between perimeters of polygons
  1. Inscribe a regular polygon (e.g., hexagon) inside a circle of radius 1.
  2. Circumscribe a regular polygon of the same number of sides around the same circle.
  3. Calculate the perimeters of both polygons.
  4. The circle's circumference lies between these two perimeters.
  5. Increase the number of sides (e.g., to 96) to get tighter bounds for π.
💡 Archimedes used 96-sided polygons to find that π is between 3 10/71 and 3 1/7.
Deriving the area of a triangle from a parallelogram
  1. Take two congruent copies of the triangle.
  2. Fit them together along a common side to form a parallelogram.
  3. The area of the parallelogram is base × height.
  4. Since the parallelogram is made of two equal triangles, the area of one triangle is half of that: (1/2) × base × height.
💡 The two triangles must be congruent; the base and height are perpendicular.
Using a median to divide a triangle into two equal areas
  1. Draw a median from a vertex to the midpoint of the opposite side.
  2. The two smaller triangles have equal bases (since the median bisects the opposite side).
  3. They share the same height (the perpendicular distance from the vertex to the base line).
  4. Therefore, their areas are equal.
💡 The two triangles are not necessarily congruent, but they always have equal area.
Baudhāyana's method to square a rectangle
  1. Given rectangle ABCD with AD = a, AB = b, and a > b.
  2. Locate E on AD such that AE = AB = b.
  3. Find the midpoint F of ED.
  4. Draw square AFGH with side AF, extending AB to H.
  5. Draw an arc with centre H and radius AG, cutting side BC at K.
  6. Draw a line through K parallel to AH, cutting GH at P.
  7. Draw square HPQS with side HP. This square has the same area as the original rectangle.
💡 This method is a geometric translation of the algebraic identity: ab = [(a+b)/2]² - [(a-b)/2]².
Calculating stagger on an athletics track
  1. Identify the radius of the inner lane's semicircle (e.g., 36.5 m) and lane width (e.g., 1.22 m).
  2. For a runner in a given lane, add 0.3 m to the inner radius to account for running path.
  3. Calculate the circumference of the full circle formed by the two semicircles: 2πr.
  4. The difference in circumference between two adjacent lanes gives the stagger distance.
  5. On straight sections, all runners cover the same distance, so stagger only compensates for curved portions.
💡 The stagger ensures that all runners cover exactly the same total distance (e.g., 400 m) despite different lane radii.
Exam Traps — Don't Make These Mistakes
⚠ Confusing π with 22/7
✗ π = 22/7
→ π is an irrational number approximately equal to 22/7, but they are not exactly equal. The NCERT explicitly states π ≠ 22/7.
✓ π ≈ 22/7 (or π is approximately 22/7, but not equal to it)
⚠ Using slant height instead of perpendicular height in area of parallelogram
✗ Area = base × slant side
→ The area formula requires the perpendicular height (distance between parallel sides), not the length of the slant side.
✓ Area = base × perpendicular height (h)
⚠ Forgetting the factor 1/2 in triangle area
✗ Area of triangle = base × height
→ A triangle is half of a parallelogram, so the area is half the product of base and height.
✓ Area of triangle = (1/2) × base × height
⚠ Applying Heron's formula without calculating semi-perimeter correctly
✗ Area = √[(a+b+c)(a+b+c-a)(a+b+c-b)(a+b+c-c)]
→ The semi-perimeter s must be calculated first as (a+b+c)/2, then used in the formula √[s(s-a)(s-b)(s-c)].
✓ First find s = (a+b+c)/2, then Area = √[s(s-a)(s-b)(s-c)]
⚠ Using Brahmagupta's formula for any quadrilateral
✗ Area of any quadrilateral = √[(s-a)(s-b)(s-c)(s-d)]
→ Brahmagupta's formula applies only to cyclic quadrilaterals (those whose vertices lie on a circle). For non-cyclic quadrilaterals, the formula gives an incorrect result.
✓ Area of a cyclic quadrilateral = √[(s-a)(s-b)(s-c)(s-d)], where s = (a+b+c+d)/2
⚠ Confusing arc length with area of sector
✗ Length of arc = (θ/360) × πr²
→ The formula (θ/360) × πr² gives the area of a sector, not the arc length. Arc length is (θ/360) × 2πr.
✓ Length of arc = (θ/360) × 2πr
Compare & Contrast
Definition and measurement
Perimeter
Area
Perimeter is the total distance around a shape (measured in units like cm, m), while area is the space enclosed by the shape (measured in square units like cm², m²).
Applicability
Heron's formula
Brahmagupta's formula
Heron's formula gives the area of a triangle using its three sides, while Brahmagupta's formula gives the area of a cyclic quadrilateral using its four sides. Brahmagupta's formula generalises Heron's formula (when one side of the quadrilateral is zero, it reduces to Heron's formula).
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 SA-I (2 marks), 1 SA-II (3 marks), 1 LA (5 marks) — total approximately 12 marks
Most asked: Circumference and area of a circle · Area of a triangle using Heron's formula · Area of a parallelogram and triangle (half base times height) · Length of an arc and area of a sector
Don't skip: Application of Heron's formula to find area of a triangle given three sides; Word problems involving circumference and area of a circle (e.g., tyre revolutions, track staggers); Finding area of combined shapes (e.g., rectangle with semicircular ends)
One-Line Revision (read aloud before the exam)
  1. Perimeter is the total distance around a shape; area is the space it encloses.
  2. Circumference of a circle = πd = 2πr.
  3. π is an irrational number approximately equal to 3.1416 or 22/7, but π ≠ 22/7.
  4. Length of an arc = 2πr × (θ/360°).
  5. Area of a rectangle = length × width.
  6. Area of a parallelogram = base × perpendicular height.
  7. Area of a triangle = (1/2) × base × height.
  8. A median of a triangle divides it into two triangles of equal area.
  9. Heron's formula: Area = √[s(s-a)(s-b)(s-c)], where s = (a+b+c)/2.
  10. Brahmagupta's formula for cyclic quadrilaterals: Area = √[(s-a)(s-b)(s-c)(s-d)], where s = (a+b+c+d)/2.
  11. Brahmagupta's formula generalises Heron's formula (when d = 0).
  12. Stagger on a track compensates for the longer distance on outer lanes due to larger radius of curved portions.
  13. A square is a special case of a rectangle; an isosceles right triangle is a special case of a right triangle.
  14. The ratio of circumference to diameter (C/D) is constant for all circles and is called π.

MCQs — 10 Questions with Answers

Q1. The perimeter of a square with side 5 cm is:

  • A. 20 cm ✓
  • B. 25 cm
  • C. 15 cm
  • D. 10 cm

Answer: A — Perimeter of square = 4a = 4 × 5 = 20 cm.

Q2. The perimeter of a rectangle with length 8 m and width 5 m is:

  • A. 26 m ✓
  • B. 40 m
  • C. 13 m
  • D. 30 m

Answer: A — Perimeter = 2(a+b) = 2(8+5) = 2(13) = 26 m.

Q3. What is the C/D ratio of a circle?

  • A. Always between 3 and 4, regardless of circle size ✓
  • B. Different for each circle depending on its radius
  • C. Exactly 3.5 for all circles
  • D. Changes as the circumference changes

Answer: A — The C/D ratio is constant for all circles and equals π, which is approximately 3.14159, always between 3 and 4.

Q4. In a 400 m athletics track, why do athletes in outer lanes start ahead?

  • A. To give them an advantage over inner lane runners
  • B. Because outer lanes have larger circumferences and must cover more distance along the curve ✓
  • C. To make the race more difficult for strong runners
  • D. Because the finish line is different for each lane

Answer: B — Outer circular lanes have larger circumferences (C = πD), so runners must start farther ahead to run the same distance to the same finish line.

Q5. Which ancient mathematician first moved beyond the crude integer value of 3 for π?

  • A. Archimedes
  • B. Mesopotamian mathematicians ✓
  • C. Zu Chongzhi
  • D. Āryabhaṭa

Answer: B — Mesopotamian mathematicians (c. 1900 BCE) set π = 3 + 1/8 = 3.125 by comparing the hexagon inscribed in a circle.

Q6. Archimedes found that π lies between which two values?

  • A. 3 and 3.5
  • B. 3 10/71 and 3 1/7 ✓
  • C. 3.1 and 3.2
  • D. 22/7 and 355/113

Answer: B — Using inscribed and circumscribed 96-sided polygons, Archimedes proved that 3 10/71 < π < 3 1/7.

Q7. Which of the following is NOT a correct statement about Zu Chongzhi's work?

  • A. He discovered the ratio 355/113 for π
  • B. His approximation remained the most accurate for over 800 years
  • C. He proved that π is exactly 22/7 ✓
  • D. He used a 24,576-sided polygon approximation method

Answer: C — Zu Chongzhi discovered both 22/7 and the more accurate 355/113, but neither is exact; π is irrational and cannot be expressed as a simple fraction.

Q8. Ramesh observes that two circular tracks have different radii, yet the ratio of their circumferences to their diameters appears the same. Which of the following best explains this observation?

  • A. The C/D ratio depends on the radius of the circle
  • B. The C/D ratio is a universal constant called π that holds for all circles ✓
  • C. Ramesh made a measurement error because larger circles should have larger C/D ratios
  • D. The C/D ratio only applies to tracks, not to mathematical circles

Answer: B — The C/D ratio equals π for all circles regardless of size; this fundamental constant connects circumference and diameter universally.

Q9. Āryabhaṭa described the value of π as 'asanna' (approaching). What insight does this reveal?

  • A. π can be approximated by simple fractions like 22/7
  • B. π is slightly larger than 3.1416
  • C. π cannot be expressed exactly as a simple fraction and can only be approached as a limit ✓
  • D. π is an integer between 3 and 4

Answer: C — By calling π 'approaching,' Āryabhaṭa profoundly recognized that π is irrational and can only be reached as a limit, not expressed as an exact fraction.

Q10. How does Mādhava's infinite series formula (π/4 = 1 − 1/3 + 1/5 − 1/7 + …) represent a breakthrough compared to earlier approximations?

  • A. It provides a simpler fraction than 355/113
  • B. It proves π is rational and can be calculated exactly
  • C. It shifts from finite geometric approximations to an infinite algebraic series that defines π as a limit to be reached ✓
  • D. It eliminates the need to study circles in geometry

Answer: C — Mādhava's formula revolutionized mathematics by showing π is not just a number to approximate but a limit defined by an infinite series, bridging geometry and algebra.

Flashcards

Define perimeter of a shape.

Perimeter is the total length around the border of a shape, measured by imagining an insect walking around it until returning to start.

What is the formula for the perimeter of a square with side a?

Perimeter of square = 4a units.

What is the formula for the perimeter of a rectangle with length a and width b?

Perimeter of rectangle = 2(a + b) units.

Define circumference of a circle.

Circumference is the perimeter of a circle, meaning the total distance around its boundary.

What is the C/D ratio and why is it important?

The C/D ratio is the ratio of circumference to diameter, which is constant for all circles and equals π (approximately 3.14159...).

Why do athletes in outer lanes start ahead in a relay race?

Outer lanes have larger circumferences, so athletes must start ahead to ensure all run the same distance to the same finish line.

What did Archimedes contribute to finding the value of π?

Archimedes trapped π between the perimeters of inscribed and circumscribed polygons, showing that 3 10/71 < π < 3 1/7.

What was Zu Chongzhi's most important discovery about π?

Zu Chongzhi discovered the ratio 355/113 as an approximation for π, which remained the most accurate for over 800 years.

How did Āryabhaṭa's approach to π differ from earlier mathematicians?

Āryabhaṭa described π as 'asanna' (approaching), suggesting it could not be expressed exactly as a simple fraction.

What was revolutionary about Mādhava's formula for π?

Mādhava discovered an infinite series formula (π/4 = 1 − 1/3 + 1/5 − 1/7 + …), proving π is a limit rather than just an approximation.

Important Board Questions

Define the perimeter of a shape and write the formula for the perimeter of a square with side a units. [2 marks]

Perimeter is the total length around a shape's border. For a square, add all four equal sides: a + a + a + a = 4a.

Explain why athletes in the outer lanes of a 400 m relay race start ahead of athletes in the inner lanes, even though the finish line is the same for all. Use the concept of circumference in your explanation. [3 marks]

Outer lanes form larger circles with greater circumference (C = πD). Since circumference increases with diameter, runners in outer lanes travel a longer curved distance and must start farther ahead to run the same total distance.

Trace the historical journey of calculating π from ancient Mesopotamia to Mādhava. Explain how each mathematician's approach improved upon the previous one, and why Mādhava's infinite series formula represented a revolutionary breakthrough in mathematics. [5 marks]

Progress: Mesopotamia used 3.125 (integer with fraction) → Archimedes trapped π using polygon perimeters → Zu Chongzhi improved accuracy to 355/113 → Āryabhaṭa recognized π as irrational ('approaching') → Mādhava shifted from finite approximations to an infinite series, proving π is a limit not a simple fraction; this unified geometry and advanced algebra.

Assertion-Reason Questions for Measuring Space: Perimeter and Area

Q1. Assertion (A):

The perimeter of all squares is always 4 times their side length.

Reason (R):

A square has four equal sides, so the perimeter formula is P = 4s.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true, and the reason correctly explains why the assertion holds — the perimeter is the sum of all four equal sides.

Q2. Assertion (A):

The C/D ratio (circumference to diameter) remains constant for all circles regardless of their size.

Reason (R):

Larger circles have larger circumferences but proportionally larger diameters, maintaining the same ratio.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — The assertion is true, and the reason is also true, but it doesn't fully explain why — the real reason is that all circles are geometrically similar. The reason provides an observation rather than the fundamental geometric principle.

Q3. Assertion (A):

Brahmagupta's choice to use √10 as an approximation for π was based on its superior accuracy compared to other known values.

Reason (R):

The text states he chose √10 because of its mathematical elegance and ease of manipulation in equations.

Answer: D — A is false but R is true — The assertion is false — Brahmagupta did not choose √10 for accuracy (it was less accurate than 355/113). The reason is correct and actually contradicts the assertion, showing that practical mathematical advantage, not accuracy, guided his choice.

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Case-Based Questions (CBQ) for Measuring Space: Perimeter and Area

Case Study:

A school is designing a new running track with two circular lanes. The inner lane has a diameter of 56 m, and each lane is 1.5 m wide. Athletes use the inner lane for the 100 m dash and the outer lane for the 200 m dash. The school organisers need to determine the stagger (difference in starting positions) to ensure both athletes run the same distance around their respective circular paths. Use π ≈ 22/7 for calculations.

Q1. What is the circumference of the inner lane?

  • A. 176 m ✓
  • B. 112 m
  • C. 88 m
  • D. 264 m

Answer: A

Q2. What is the diameter of the outer lane?

  • A. 56 m
  • B. 57.5 m
  • C. 59 m ✓
  • D. 60 m

Answer: C

Q3. What is the circumference of the outer lane?

  • A. 176 m
  • B. 185.6 m
  • C. 198 m ✓
  • D. 209.4 m

Answer: C

Q4. By how many meters should the outer lane athlete start ahead to equalize their running distance?

  • A. 22 m ✓
  • B. 30 m
  • C. 32 m
  • D. 55 m

Answer: A

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