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Question 2 - Side inequality - Chapter 7 Class 9 Triangles
Last updated at August 5, 2026 by Teachoo
Chapter 7 Class 9 Triangles
Concept wise
- Congruency - Introduction
- SAS
- ASA/AAS
- Opposite Angles of equal sides
- SSS
- RHS
-
Side inequality
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
