Summary
Chapter 6 of NCERT Class 9 Maths, "Measuring Space: Perimeter and Area", teaches how to measure the perimeter and area of shapes, covering the circle's circumference (2*pi*r), arc length, sectors, the area of triangles, Heron's formula, and the area of a circle (pi*r^2).
This chapter builds from perimeter to area. It defines perimeter as the total length around a shape's border, then explores the circle's circumference through the constant C/D ratio called pi, tracing its history from ancient approximations like 22/7 and 355/113 to Madhava's infinite series. It shows pi is irrational, derives arc length and sector area using central angles, and develops area formulas for rectangles, parallelograms, and triangles. It introduces Heron's formula, Brahmagupta's formula for cyclic quadrilaterals, squaring a rectangle, and proves the area of a circle is pi*r^2.
Key points & formulas
- 01Perimeter of a circle (circumference) is C = 2*pi*r, with diameter d giving pi*d
- 02pi is the irrational circumference-to-diameter (C/D) ratio, about 22/7 or 3.14
- 03Arc length subtending angle t degrees is 2*pi*r * (t/360)
- 04Area of a triangle equals half base times height (1/2 * b * h)
- 05Heron's formula: area = sqrt(s(s-a)(s-b)(s-c)), where s = (a+b+c)/2
- 06Area of a circle is A = pi*r^2; area of a sector is pi*r^2 * (t/360)
- 07Brahmagupta's formula gives the area of a cyclic quadrilateral and generalises Heron's formula
Frequently asked questions
01What is Heron's formula in Class 9 Maths Chapter 6?
Heron's formula finds a triangle's area from its three side lengths a, b, and c. First compute the semi-perimeter s = (a + b + c)/2, then area = sqrt(s(s-a)(s-b)(s-c)). For a 3-4-5 triangle, s = 6 and the area works out to 6 square units.
02What is the value of pi used in this chapter?
pi is the constant ratio of a circle's circumference to its diameter and is approximately 22/7 or 3.14. The chapter notes pi is irrational, so it is close but not equal to 22/7; a far better approximation is 355/113 (about 3.1415929).
03How do you find the area of a circle and a sector?
The area of a circle of radius r is A = pi*r^2, a result Archimedes proved around 250 BCE. The area of a sector that subtends an angle t degrees at the centre is pi*r^2 * (t/360), which is that fraction of the whole circle's area.
04What is Brahmagupta's formula and how does it relate to Heron's formula?
Brahmagupta (628 CE) found the area of a cyclic quadrilateral with sides a, b, c, d as sqrt((s-a)(s-b)(s-c)(s-d)), where s = (a+b+c+d)/2. Setting the fourth side d = 0 reduces the quadrilateral to a triangle, so Brahmagupta's formula generalises Heron's formula.
More chapters in Ganita Manjari
This is the complete Ganita Manjari Chapter 6 as published by NCERT — every diagram, solved example, and exercise included, free. Browse all NCERT Class 9 textbooks.
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