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Ex 6.3, 2
Last updated at June 3, 2026 by Teachoo
Class 9
Chapter 6 Class 9 - Measuring Space: Perimeter and Area (Ganita Manjar
- Definition (and Quesitons)
- Perimeter of Different Shapes
- Perimeter of a Circle - C/D Ratio
- Pi is Irrational
- Length of an Arc of a Circle
- Problems, Puzzles, and Paradoxes on Perimeter
- Exercise Set 6.1
- Area of Square & Rectangle
- Area of a Parallelogram
- Area of a Triangle
- Heron's Formula
- Circumcircle and Incircle of a Triangle
- Brahmaguptaβs Formula for the Area of a Cyclic 4-gon
- Squaring a Rectangle
- Exercise Set 6.2
- Area of a circle
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Exercise Set 6.3
- End-of-Chapter Exercises
Transcript
Ex 6.3, 2 Find the area of a quadrant of a circle whose circumference is 44 cm. Quadrant is ΒΌth of a circle So, we can write Area of quadrant = 1/4Γ area of circle = π/π π π^π We need to find r It is given that Circumference = 2Οr 44 = 2Οr 44 = 2 Γ 22/7 Γ r 44 = 44/7 Γ r 44 Γ 7/44 = r 7 = r r = 7 cm Now, Area of quadrant = 1/4 ππ^2 = π/π Γππ/π Γ π^π = 1/4 Γ 22 Γ 7 = 1/2 Γ 11 Γ 7 = 77/2 = 38.5 cm2
