Last updated at May 29, 2018 by Teachoo

Transcript

Example 4 In figure, ABCD is a quadrilateral and BE ∥ AC and also BE meets DC produced at E. Show that area of Δ ADE is equal to the area of the quadrilateral ABCD. Given: A quadrilateral ABCD where BE ∥ AC To prove: ar (ADE) = ar (ABCD) Proof : We prove ar(ABC) = ar (ACE) & then add ar(ADC) both sides Since, Δ BAC and Δ EAC lie on the same base AC and between the same parallels AC and BE. ∴ ar(BAC) = ar(EAC) Adding ar(ADC) both sides So, ar(BAC) + ar(ADC) = ar(EAC) + ar(ADC) ⇒ ar(ABCD) = ar(ADE)

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.