Question 2 - Side inequality - Chapter 7 Class 9 Triangles

Last updated at Dec. 13, 2024 by

Ex 7.4, 2 - Sides AB & AC of ABC are extended to points P, Q - Ex 7.4

Ex 7.4, 2 - Chapter 7 Class 9 Triangles - Part 2

Ex 7.4, 2 - Chapter 7 Class 9 Triangles - Part 3

Ex 7.4, 2 - Chapter 7 Class 9 Triangles - Part 4

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Transcript

Ex7.4, 2 (Method 1) In the given figure sides AB and AC of ABC are extended to points P and Q respectively. Also, PBC < QCB. Show that AC > AB. Similarly, Now, PBC < QCB A + ACB < A + ABC ACB < ABC AB < AC Thus, AC > AB Hence proved. Ex7.4, 2 (Method 2) In the given figure sides AB and AC of ABC are extended to points P and Q respectively. Also, PBC < QCB. Show that AC > AB. Similarly, Now, PBC < QCB PBC > QCB 180 PBC > 180 QCB

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo