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Last updated at May 29, 2018 by Teachoo

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Ex 9.2, 2 If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD show that ar (EFGH) = 1/2 ar (ABCD) Given: A parallelogram ABCD where E, F, G, H are the mid-points of AB,BC,CD & AD respectively To prove: ar (EFGH) = 1/2 ar (ABCD) Proof: Join H & F Now, So, AD ∥ BC ⇒ DH ∥ CF Also, AD = BC 1/2 AD = 1/2 BC DH = CF In DHFC, DH = CF & DH ∥ CF ‖ Since one pair of opposite sides are equal and parallel ∴ DHFC is a parallelogram Similarly, HABF is a parallelogram Since DHFC is a parallelogram, DC ∥ HF As ΔHGF and parallelogram DHFC are on the same base HF & between the same parallel lines DC and HF, ∴ Area (ΔHGF) = 1/2 Area (DHFC) Similarly, Since HABF is a parallelogram, HF ∥ AB As ΔHEF and parallelogram HABF are on the same base HF and between the same parallel lines AB and HF, ∴ Area (ΔHEF) = 1/2 Area (HFAB) Adding (1) & (2) Area (∆HGF) + Area (∆HEF) = 1/2 Area (DHFC) + 1/2 Area (HFAB) Area (HEFG) = 1/2 [ Area (DHFC) + Area (HFAB)] ⇒ Area (EFGH) = 1/2 Area (ABCD) 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.