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Ex 6.2, 6
Last updated at May 31, 2026 by Teachoo
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
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Exercise Set 6.2
- Area of a circle
- Exercise Set 6.3
- End-of-Chapter Exercises
Transcript
Ex 6.2, 6 ABCD is a parallelogram. P and Q are any two points on side AB . What can you say about the ratio area (△PCD ): area (△QCD )? Let’s draw the diagram Here, ∆ PCD & ∆ QCD have same base and lie between same parallel lines We use the formula Area of triangle = 𝟏/𝟐 × Base × Height Drawing perpendicular from point P & point Q to CD So, PM ⊥ CD and QN ⊥ CD Since ABCD is a parallelogram Sides AB & CD are parallel And distance between parallel lines is equal Distance will be perpendicular distance Thus, perpendiculars from any point on AB to CD (and vice versa) will be equal ∴ PM = QN So, ∆ PCD & ∆ QCD have Same base CD Same height PM Since both triangles have same Base & Height Their Area would be equal We need to find ratio area (△PCD ): area (△QCD ) Since Area is same, ratio would be 1 : 1
