Ex 9.3

Ex 9.3, 1
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Ex 9.3, 2 Important Deleted for CBSE Board 2022 Exams You are here

Ex 9.3, 3 Deleted for CBSE Board 2022 Exams

Ex 9.3, 4 Deleted for CBSE Board 2022 Exams

Ex 9.3, 5 Important Deleted for CBSE Board 2022 Exams

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Ex 9.3, 8 Deleted for CBSE Board 2022 Exams

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Ex 9.3, 10 Deleted for CBSE Board 2022 Exams

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Ex 9.3, 12 Important Deleted for CBSE Board 2022 Exams

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Ex 9.3, 16 Important Deleted for CBSE Board 2022 Exams

Chapter 9 Class 9 - Areas of Parallelograms and Triangles (Deleted)

Serial order wise

Last updated at March 2, 2017 by Teachoo

Ex 9.3, 2 In a triangle ABC , E is the mid-point of median AD show that ar(BED) = 1/4 ar (ABC). Given: Δ ABC, with AD as median i.e. BD = CD & E is the mid-point of AD, i.e., AE = DE To prove: ar (BED) = 1/4 ar (ABC). Proof : AD is a median of Δ ABC & median divides a triangle into two triangles of equal area ∴ ar (ABD) = ar (ACD) ⇒ ar (ABD) = 1/2 ar (ABC) In Δ ABD, BE is the median median divides a triangle into two triangles of equal area ∴ ar (BED) = ar (BEA) ⇒ ar (BED) = 1/2 ar (ABD) ⇒ ar (BED) = 1/2× 1/2 ar (ABC) ⇒ ar (BED) = 1/4 ar (ABC) Hence proved