CBSE Class 10 Maths Triangles
Sides opposite to the equal angles in a triangle are
In triangle KMN, PQ intersects KM and KN at P and Q, respectively. If KP = 10 cm, PM = 5 cm, KQ = 14 cm and QN = 7 cm, then
In triangle ABC, PQ intersects AB and AC at P and Q, respectively. If AP = 6 cm, PB = 4 cm, AQ = 8 cm and QC = 4 cm, then
- In quadrilateral ABCD; P,Q,R and S are points of trisection of sides AB, BC, CA and AD. Prove PQRS is a Parallelogram. In the solution of this question shown in the video(exam decoded of converse of BPT section), its shown that : BP = 2AP, BQ = 2QC, DR = 2RC and DS = 2SA. explain this step in detail that how is this possible
- converse of bpt
- State the converse of Pythagoras theorem
- Converse of Thales theorm
- Converse pythagoras therom
- given a triangle ABC and a trapezium BCED AD=1,AB=3,AE=2,AC=x,EC=?
- In a trapezium ABCD, E and F are points on AB and AD respectively. Also G and H are points on CB and CD respectively such that AE=3, AB=9, AF=2, AD=6, CG=1, CB=4, CH=3, CD=4. Check if EF || GH.
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