Question 12 - 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 12 By dividing a trapezium into two triangles show that its area is, half the sum of the parallel sides multiplied by the height (the same formula as the one given above). Let’s draw the figure We divide trapezium ABCD into two triangles - ∆ ABD & ∆ BCD Let’s find area of both Area ∆ ABD ∆ ABD has Base = AB = b Height = h So, Area of ∆ ABD = 1/2 × Base × Height = 𝟏/𝟐 × b × h Area ∆ BCD ∆ BCD has Base = CD = a Height = h So, Area of ∆ BCD = 1/2 × Base × Height = 𝟏/𝟐 × a × h Now, Area of Trapezium = Area of Triangle ABD + Area of Triangle BCD = 𝟏/𝟐 × b × h + 𝟏/𝟐 × b × h = 1/2 × h × (b + a) = 𝟏/𝟐 × h × (a + b) Thus, area of a trapezium is half the sum of the parallel sides × height Hence proved
