Question 5 - 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
You are here
You are here
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
-
End-of-Chapter Exercises
Transcript
Question 5 The sides of a triangle are in the ratio 2:3:4, and its perimeter is 45 cm. Find its area. Since sides are in ratio 2: 3: 4 We can assume sides are 2x, 3x, 4x Let a = 2x, b = 3x, c = 4x Given, Perimeter = a + b + c 45 = 2x + 3x + 4x 45 = 9x 9x = 45 x = 45/9 x = 5 Thus, our sides are a = 2x = 2 × 5 = 10 cm b = 3x = 3 × 5 = 15 cm c = 4x = 4 × 5 = 20 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 = 45/2 = 22.5 cm Now , Area of Triangle = √(𝑠 (𝑠−𝑎)(𝑠−𝑏)(𝑠−𝑐)) = √(𝟐𝟐.𝟓(𝟐𝟐.𝟓−𝟏𝟎) × (𝟐𝟐.𝟓−𝟏𝟓) × (𝟐𝟐.𝟓−𝟐𝟎)) = √(22.5 × 12.5 × 7.5 × 2.5) = √(225/10 ×125/10 ×75/10 ×25/10) = √((15^2 × 5 × 25 × 3 × 25 × 25)/10^4 ) =√((15^2 × 5 × 5^2× 3 × 25^2)/10^4 ) = (15 × 5 × 25)/10^2 √15 = 1875/100 √15 = 18.75√𝟏𝟓 cm2
