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Ex 6.2, 3
Last updated at May 31, 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
- Brahmagupta’s Formula for the Area of a Cyclic 4-gon
- Squaring a Rectangle
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Exercise Set 6.2
Transcript
Ex 6.2, 3 Find the area of a triangle, given that its sides are 8 cm and 11 cm long, and its perimeter is 32 cm. Let sides be a, b, c ∴ a = 8 cm, b = 11 cm, c = ? Now, Perimeter = a + b + c 32 = 8 + 11 + c 32 = 19 + c 32 – 19 = c 13 = c c = 13 cm We find Area using Herons formula Area of Triangle = √(𝑠 (𝑠−𝑎)(𝑠−𝑏)(𝑠−𝑐)) Here, s = (𝒂 + 𝒃 + 𝒄)/𝟐 = 𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟/2 = 32/2 = 16 cm Now , Area of Triangle = √(𝑠 (𝑠−𝑎)(𝑠−𝑏)(𝑠−𝑐)) = √(𝟏𝟔 (𝟏𝟔−𝟖) × (𝟏𝟔−𝟏𝟏) × (𝟏𝟔−𝟏𝟑)) = √(16 × 8 × 5 × 3) = √(4^2 × 4 × 2 × 5 × 3) = √(4^2 × 2^2 × 2 × 5 × 3) = √(4^2 ) × √(2^2 ) ×√(2 × 5 × 3) = 4 × 2 × √30 = 8√𝟑𝟎 cm2
