Question 11 - End-of-Chapter Exercises - Chapter 6 Class 9 - Measuring Space: Perimeter and Area (Ganita Manjar

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Question 11 You know that the area of a parallelogram is base Γ— height. Using this and the figure, show that the area of a trapezium is half the sum of the parallel sides Γ— height, i.e., 1/2(π‘Ž+𝑏)β„Ž. We can divide the given trapezium into Parallelogram Triangle Let’s find Area of each individually Area of Parallelogram Labelling the figure Since ABDE is a parallelogram And, opposite sides of parallelogram are equal ∴ AB = ED = a And, BC = AC – AB = b – a Now, In Parallelogram ABDe Base = AB = a Height = h So, Area of Parallelogram ABDE = Base Γ— Height = a Γ— h Area of Triangle βˆ† BCD has Base = BC = b – a Height = h So, Area of βˆ† BCD = 1/2 Γ— Base Γ— Height = a Γ— h And, opposite sides of parallelogram are equal ∴ AB = ED = a And, BC = AC – AB = b – a So, Area of βˆ† BCD = 1/2 Γ— Base Γ— Height = 𝟏/𝟐 Γ— (b – a) Γ— h Now, Area of Trapezium = Area of Parallelogram ABDE + Area of Triangle BCD = a Γ— h + 𝟏/𝟐 Γ— (b – a) Γ— h = a Γ— h + 1/2 Γ— b Γ— h – 1/2 Γ— a Γ— h = a Γ— h – 1/2 Γ— a Γ— h + 1/2 Γ— b Γ— h = 1/2 Γ— a Γ— h + 1/2 Γ— b Γ— h = 𝟏/𝟐 Γ— h Γ— (a + b) Thus, area of a trapezium is half the sum of the parallel sides Γ— height Hence proved

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