MathematicsClass 9Measuring Space

Measuring Space | Class 9 Mathematics Notes

By ConceptScroll Team · Published on 17 July 2026 · 4 min read

Measuring Space | Class 9 Mathematics Notes

Measuring Space – this guide gives you a concise, exam-ready overview of Measuring Space from Class 9 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.

6.2 PERIMETER OF A CIRCLE—THE C/D RATIO

This section explores the perimeter of a circle, called the circumference, and the constant ratio of circumference to diameter, known as π (pi). Ancient civilizations observed that this ratio remains constant regardless of the size of the circle. The section describes a simple home experiment to estimate π by measuring the diameter of a cotton reel and the length of thread wrapped around it multiple times, calculating the ratio L/(20D). Historical approximations of π are discussed, including the Mesopotamian value 3.125, Archimedes' method of inscribed and circumscribed polygons bounding π between 3 and approximately 3.46, and later refinements by Ptolemy, Liu Hui, Zu Chongzhi, Aryabhata, and Brahmagupta. The section highlights the infinite series formula for π discovered by Mādhava, marking the beginning of calculus. It concludes by noting that π is irrational, meaning it cannot be expressed exactly as a fraction, and its decimal expansion is infinite and non-repeating.

📊 Diagram: Fig. 6.6 The Mesopotamian Hexagon-to-Circle comparison; Fig. 6.7: Archimedes’ method utilising inscribed and circumscribed polygons.

🧪 Activity: Home Measurement: Measure diameter and thread length to estimate π; reflect on historical approximations.

🔗 Connection: Prepares for the next section on calculating lengths of arcs of circles using π.

Frequently asked questions

Two circles of equal radius are located such that each circle passes through the centre of the other circle (Fig. 6.12). Given that the radius of each circle is r units, find the perimeter of the shape formed by the two circles in terms of r units. (Ignore the dotted portions that lie within the circles.)

Let the two circles have centers A and B, each with radius r units. Each circle passes through the center of the other, so AB = r. The circles intersect at points C and D. Triangle ABC is equilateral with sides r, so angle CAB = 60°. The arcs forming the perimeter are each 120° arcs (since the angle subtended by the chord CD at the center is 120°). Each red arc is therefore 1/3 of the circumference of a circle with radius r.

The perimeter consists of two such arcs, so total perimeter = 2 × (1/3

In Fig. 6.13, we see points P and Q and two paths connecting them. The first path is made up of the semicircle a. The other path is made up of three semicircles (b, c and d). Which path is longer? Choose one: (i) Path a is longer. (ii) Path b + c + d is longer. (iii) The two paths have equal length. (Try to answer this before reading on.)

The two paths have equal length.

Explanation: Let the radii of semicircles a, b, c, d be a', b', c', d' respectively. Length of semicircle a = πa'. Length of semicircles b, c, d are πb', πc', πd' respectively. Total length of second path = π(b' + c' + d'). Since the length of PQ = 2a' = 2b' + 2c' + 2d', it follows that a' = b' + c' + d'. Hence, lengths of both paths are equal.

1. The perimeter of a circle is 44 cm. What is its radius?

Given perimeter (circumference) C = 44 cm. We know circumference C = 2πr. Using π = 22/7, 44 = 2 × (22/7) × r => 44 = (44/7) × r => r = 44 × (7/44) = 7 cm. So, the radius of the circle is 7 cm.

2. Calculate, correct to 3 significant figures, the circumference of a circle with: (i) radius 7 cm (ii) radius 10 cm (iii) radius 12 cm.

Formula: Circumference C = 2πr. (i) r = 7 cm C = 2 × (22/7) × 7 = 44 cm (ii) r = 10 cm C = 2 × (22/7) × 10 = (44/7) × 10 = 440/7 ≈ 62.857 cm Rounded to 3 significant figures: 62.9 cm (iii) r = 12 cm C = 2 × (22/7) × 12 = (44/7) × 12 = 528/7 ≈ 75.429 cm Rounded to 3 significant figures: 75.4 cm

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