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Ex 9.3

Ex 9.3, 1
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Ex 9.3, 2 Important Deleted for CBSE Board 2023 Exams

Ex 9.3, 3 Deleted for CBSE Board 2023 Exams

Ex 9.3, 4 Deleted for CBSE Board 2023 Exams

Ex 9.3, 5 Important Deleted for CBSE Board 2023 Exams

Ex 9.3, 6 Important Deleted for CBSE Board 2023 Exams

Ex 9.3, 7 Important Deleted for CBSE Board 2023 Exams

Ex 9.3, 8 Deleted for CBSE Board 2023 Exams You are here

Ex 9.3, 9 Important Deleted for CBSE Board 2023 Exams

Ex 9.3, 10 Deleted for CBSE Board 2023 Exams

Ex 9.3, 11 Important Deleted for CBSE Board 2023 Exams

Ex 9.3, 12 Important Deleted for CBSE Board 2023 Exams

Ex 9.3, 13 Deleted for CBSE Board 2023 Exams

Ex 9.3, 14 Deleted for CBSE Board 2023 Exams

Ex 9.3, 15 Deleted for CBSE Board 2023 Exams

Ex 9.3, 16 Important Deleted for CBSE Board 2023 Exams

Chapter 9 Class 9 - Areas of Parallelograms and Triangles [Deleted]

Serial order wise

Last updated at May 29, 2018 by Teachoo

Ex 9.3, 8 XY is a line parallel to side BC of a triangle ABC. If BE AC and CF AB meet XY at E and F respectively, show that ar (ABE) = ar (ACF) Given: ABC Where XY BC & BE AC & CF AB To prove: ar (ABE) = ar (ACF) Proof : Let XY intersect AB & BC at M and N respectively Given that XY BC EN BC Also, it is given that BE AC BE CN In BCNE, both pair of opposite sides are parallel. BCNE is a parallelogram. Parallelogram BCNE and AEB lie on the same base BE and are between the same parallels BE and AC. ar ( ABE) = 1/2 ar (BCNE) Similarly BCFM is a parallelogram. and ar ( ACF) = 1/2 ar (BCFM) Parallelograms BCNE and BCFM are on the same base BC and between the same parallels BC and EF. ar (BCNE) = ar (BCFM) From (1) , (2) & (3) ar (ABE) = ar (ACF)