Check sibling questions

Ex 9.2, 2 - If E, F, G and H are mid-points of sides - Ex 9.2

Ex 9.2, 2 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 2
Ex 9.2, 2 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 3

This video is only available for Teachoo black users

Solve all your doubts with Teachoo Black (new monthly pack available now!)


Transcript

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

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.