Question 13 - 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 13 Show how we can use two identical copies of a trapezium to make a parallelogram. How will this give us the formula for the area of a trapezium? Let’s draw figure So, by joining identical trapeziums ABCD and MNOP we get Parallelogram APMD Since both trapeziums are identical, we can write Area of parallelogram = 2 × Area of trapezium And, Parallelogram APMD has Base = AP = b + a Height = h Putting it in formula Area of parallelogram = 2 × Area of trapezium Base × Height = 2 × Area of trapezium (b + a) × h = 2 × Area of trapezium 2 × Area of trapezium = (b + a) × h Area of trapezium = 𝟏/𝟐 × (b + a) × h Thus, area of a trapezium is half the sum of the parallel sides × height Hence proved
