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Measuring Space: Perimeter and Area Class 9 Notes

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Chapter mind map

The whole chapter at a glance: the big idea, then each branch and what sits under it.

Measuring Space: Perimeter and Area

Exploring the geometry of boundaries and enclosed regions through historical methods and mathematical formulas.

  1. The Constant Pi (π)

    The fixed ratio of a circle's circumference to its diameter, an irrational number with a rich history of approximation.

    • Historical Approximations — Mesopotamians used 3.125; Archimedes trapped π between 3 10/71 and 3 1/7; Zu Chongzhi found 355/113.
    • Indian Contributions — Āryabhaṭa gave 3.1416; Brahmagupta suggested √10; Mādhava discovered the first exact infinite series formula.
  2. Triangles and Quadrilaterals

    Calculating area using base, height, or side lengths for rectilinear shapes.

    • Heron's Formula — Area = √[s(s-a)(s-b)(s-c)] where s is the semi-perimeter; ideal for scalene triangles where height is unknown.
    • Brahmagupta's Generalization — For cyclic quadrilaterals: Area = √[(s-a)(s-b)(s-c)(s-d)]. Reduces to Heron's formula if one side d = 0.
    • Parallelogram to Triangle — A triangle is half a parallelogram; joining two congruent triangles forms a parallelogram with area bh.
  3. Circular Measurements

    Formulas for the full circle and its fractional parts based on radius and central angle.

    • Arc Length and Track Staggers — Length = 2πr × (θ/360). Used in athletics to ensure runners in outer lanes travel the same 400m distance.
    • Sector and Segment Area — Sector area is πr² × (θ/360). Segment area is found by subtracting the triangle area from the sector area.
    • Archimedes' Circle Area — Area = πr². Visualized as a right triangle with base 2πr and height r, or rearranged wedges forming a parallelogram.
  4. Geometric Constructions

    Ancient methods for transforming shapes while preserving area.

    • Baudhāyana's Squaring — Constructing a square equal to a rectangle's area using the identity ab = [(a+b)/2]² - [(a-b)/2]².
    • Śhulbasūtra Geometry — 800 BCE Indian texts using geometric constructions to solve algebraic problems before modern notation.

Chapter notes

This chapter explores the measurement of boundaries and internal spaces of geometric shapes, covering the history of pi, arc lengths, Heron's formula for triangles, and Brahmagupta's formula for cyclic quadrilaterals.

Perimeter of a Shape

Perimeter is the total length around the border of a shape. It can be visualized as the distance a tiny insect travels while walking once around the boundary of a figure until it returns to the start.

For regular polygons, the perimeter is calculated by multiplying the length of one side by the total number of sides. For example, a square with side 'a' has a perimeter of 4a, and an equilateral triangle with side 'a' has a perimeter of 3a. A rectangle with length 'a' and width 'b' has a perimeter of 2(a + b).

The square is a special case of a rectangle where a = b, resulting in 2(a + a) = 4a. In all squares, the ratio of perimeter to side remains fixed at 4:1, regardless of size. Similarly, for equilateral triangles, this ratio is always 3:1.

When dealing with circles, the perimeter is specifically called the circumference. Just as squares have a fixed ratio of perimeter to side, circles have a fixed ratio of circumference (C) to diameter (D), which mathematicians call π (pi).

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If the side of an equilateral triangle is doubled, what happens to its perimeter?

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NCERT reference: chapter PDF pages 2, 3.

The C/D Ratio and the History of Pi

The ratio of a circle's circumference (C) to its diameter (D) is a constant denoted by the Greek letter π (pi).

Historically, civilizations estimated π with increasing accuracy. Mesopotamians (c. 1900 BCE) used 3.125, while Archimedes (250 BCE) used polygons to 'trap' π between 3 10/71 and 3 1/7. In China, Zu Chongzhi (480 CE) discovered the 'Close Ratio' 355/113, which remained the most accurate for 800 years.

In India, Āryabhaṭa (499 CE) provided the value 3.1416, describing it as 'asanna' (approximate). Brahmagupta (628 CE) suggested √10 ≈ 3.1622 for its mathematical elegance. Later, Mādhava of Sangamagrāma discovered the first exact formula for π using an infinite series: π/4 = 1 - 1/3 + 1/5 - 1/7 + ... (or π = 4(1 - 1/3 + 1/5 - 1/7 + ...)).

π is an irrational number, meaning it cannot be expressed as a simple fraction a/b where a and b are integers. Its decimal expansion goes on forever without a repeating pattern. Common approximations include 22/7 and 3.14.

Mathematician/CultureValue or Approximation of πMethod Used
Mesopotamia3.125Hexagon comparison
Archimedes3 10/71 < π < 3 1/796-sided polygons
Zu Chongzhi355/113 (≈ 3.1415929)24,576-sided polygons
Āryabhaṭa3.141662832/20000
Brahmagupta√10 (≈ 3.1622)Algebraic elegance
Mādhavaπ = 4(1 - 1/3 + 1/5 - 1/7 + ...)Infinite series

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Why is 22/7 used for π if π is irrational?

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NCERT reference: chapter PDF pages 3, 4, 5, 6, 7, 37.

Length of an Arc of a Circle

An arc is a portion of the circumference of a circle. Its length depends on the radius of the circle and the angle it subtends at the centre.

The circumference of a full circle is 2πr. A semicircle represents half a circle (180°), so its arc length is (180/360) × 2πr = πr. Similarly, a quarter circle (90°) has an arc length of (90/360) × 2πr = πr/2.

For any arc that subtends an angle θ at the centre, the length is calculated as a fraction of the total circumference. This is expressed by the formula: Length = 2πr × (θ/360).

In an athletics track, staggers are used because runners in outer lanes travel along arcs with larger radii. To ensure everyone runs exactly 400 m, the starting positions are shifted forward for outer lanes.

Calculating Arc Length

Length = 2 × (22/7) × 7 × (60/360) = 44 × (1/6) ≈ 7.33 cm

For a circle with radius 7 cm and a central angle of 60°, we substitute the values into the formula. 2 × 22/7 × 7 = 44. Then 44 × (1/6) = 44/6 ≈ 7.33 cm (to 3 significant figures).

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What is the perimeter of a semicircle of radius r?

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NCERT reference: chapter PDF pages 8, 9, 10, 11, 12.

Area of Parallelograms and Triangles

The area of a shape is the amount of space it occupies in a two-dimensional plane, measured in square units.

A parallelogram can be transformed into a rectangle with the same base (b) and height (h). Therefore, the area of a parallelogram is given by the formula: Area = base × height (bh).

A triangle can be viewed as half of a parallelogram. By joining two congruent triangles, we form a parallelogram with the same base and height. Thus, the area of a triangle is 1/2 × base × height.

A core theorem states: A median of a triangle divides it into two triangles with equal area. Even if the two resulting triangles are not congruent, their areas are identical because they share the same height and have equal bases.

From Triangle to Parallelogram

  1. 1

    Single Triangle

    Start with a triangle of base 'b' and height 'h'.

  2. 2

    Congruent Copy

    Create an identical copy of the triangle.

  3. 3

    Joining

    Rotate and join the copy along one side to form a parallelogram.

  4. 4

    Area Calculation

    The parallelogram area is bh; since it's made of two triangles, one triangle is (1/2)bh.

This sequence shows how the triangle area formula is derived from the parallelogram area.

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If a triangle and a parallelogram have the same base and height, what is the ratio of their areas?

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NCERT reference: chapter PDF pages 13, 14, 15, 16.

Heron's Formula for Triangles

Heron's formula allows us to calculate the area of a triangle when only the lengths of its three sides are known, without needing the height.

If a triangle has sides a, b, and c, we first calculate the semi-perimeter (s), which is half the perimeter: s = (a + b + c) / 2.

The area is then found using the formula: Area = √[s(s - a)(s - b)(s - c)]. This formula is particularly useful for scalene triangles where the height is difficult to measure.

Other area formulas involve the circumcircle (radius R) and incircle (radius r). Area = abc / 4R and Area = r(a + b + c) / 2 (which is also Area = rs).

Area of a 3-4-5 Triangle

s = (3+4+5)/2 = 6; Area = √[6(6-3)(6-4)(6-5)]

First, find s = 6. Then Area = √[6 × 3 × 2 × 1] = √36 = 6 sq. units. This matches the (1/2) × base × height method for a right triangle (1/2 × 3 × 4 = 6).

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Can Heron's formula be used for an equilateral triangle?

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NCERT reference: chapter PDF pages 17, 18, 19, 37.

The rest of this chapter

Keep reading Measuring Space: Perimeter and Area, free

  1. Locked: 1. Brahmagupta's Formula for Cyclic Quadrilaterals
  2. Locked: 2. Baudhāyana's Squaring of a Rectangle
  3. Locked: 3. Area of a Circle
  4. Locked: 4. Area of a Sector

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