Last updated at May 29, 2018 by Teachoo

Transcript

Ex 9.2, 5 In the given figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that (i) ar (PQRS) = ar (ABRS) Since PQRS is a parallelogram PQ ∥ RS & ABRS is also a parallelogram So, AB ∥ RS Since PQ ∥ RS & AB ∥ RS We can say that PB ∥ RS Now, PQRS & ABRS are two parallelograms with the same base RS and between the same parallels PB & RS ∴ ar (PQRS) = ar (ABRS) Ex 9.2, 5 In the given figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that (ii) ar (AXS) = 1/2 ar (PQRS) Since ABRS is a parallelogram, AS ∥ BR Δ AXS and parallelogram ABRS lie on the same base AS and are between the same parallel lines AS and BR, ∴ Area (ΔAXS) = 1/2 Area (ABRS) ⇒ Area (ΔAXS) = 1/2 Area (PQRS) Hence proved

Chapter 9 Class 9 Areas of parallelograms and Triangles

Class 9

Important Questions for Exam - Class 9

- Chapter 1 Class 9 Number Systems
- Chapter 2 Class 9 Polynomials
- Chapter 3 Class 9 Coordinate Geometry
- Chapter 4 Class 9 Linear Equations in Two Variables
- Chapter 5 Class 9 Introduction to Euclid's Geometry
- Chapter 6 Class 9 Lines and Angles
- Chapter 7 Class 9 Triangles
- Chapter 8 Class 9 Quadrilaterals
- Chapter 9 Class 9 Areas of parallelograms and Triangles
- Chapter 10 Class 9 Circles
- Chapter 11 Class 9 Constructions
- Chapter 12 Class 9 Herons Formula
- Chapter 13 Class 9 Surface Areas and Volumes
- Chapter 14 Class 9 Statistics
- Chapter 15 Class 9 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.