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Chapter 9 Class 9 Areas of parallelograms and Triangles
Ex 9.2, 5 Important Deleted for CBSE Board 2023 Exams
Example 4 Important Deleted for CBSE Board 2023 Exams
Ex 9.3, 2 Important Deleted for CBSE Board 2023 Exams
Ex 9.3, 7 Important Deleted for CBSE Board 2023 Exams
Ex 9.3, 11 Important Deleted for CBSE Board 2023 Exams
Misc 3 Important Deleted for CBSE Board 2023 Exams
Chapter 9 Class 9 Areas of parallelograms and Triangles
Last updated at May 29, 2018 by Teachoo
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