Question 2 - End-of-Chapter Exercises - Chapter 6 Class 9 - Measuring Space: Perimeter and Area (Ganita Manjar
Last updated at June 3, 2026 by Teachoo
End-of-Chapter Exercises
Question 1
Important
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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
- Exercise Set 6.2
- Area of a circle
- Exercise Set 6.3
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End-of-Chapter Exercises
Transcript
Question 2 An isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area of the triangle. Let sides be a, b, c ∴ a = 15 cm, b = ?, c = 15 cm Now, Perimeter = a + b + c 40 = 15 + b + 15 40 = 30 + b 40 – 30 = b 10 = b b = 10 cm 15 cm 15 cm b We find Area using Herons formula Area of Triangle = √(𝑠 (𝑠−𝑎)(𝑠−𝑏)(𝑠−𝑐)) Here, s = (𝒂 + 𝒃 + 𝒄)/𝟐 = 𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟/2 = 40/2 = 20 cm Now , Area of Triangle = √(𝑠 (𝑠−𝑎)(𝑠−𝑏)(𝑠−𝑐)) = √(𝟐𝟎(𝟐𝟎−𝟏𝟓) × (𝟐𝟎−𝟏𝟎) × (𝟐𝟎−𝟏𝟓)) 15 cm 15 cm 10 cm = √(20 × 5 × 10 × 5) = √(2 × 10 × 5 × 10 × 5) = √(2 × 10^2 × 5^2 ) = √2 × √(10^2 ) × √(5^2 ) = √2 × 10 × 5 = 50√𝟐 cm2
