NCERT Mathematics · Class 10 · English · 2026-27
Triangles — Exercise 6.2
Independently written, step-by-step solutions. Prompts below are concise paraphrases; use the official textbook for the complete question wording.
Choose a question
- Question 1
- Question 2
- Question 3
- Question 4
- Question 5
- Question 6
- Question 7
- Question 8
- Question 9
- Question 10
Question 1
With DE parallel BC in the two diagrams, find EC and AD.
Step-by-step solution
Use the Basic Proportionality Theorem: AD/DB=AE/EC. (i) 1.5/3=1/EC, giving EC=2 cm. (ii) AD/7.2=1.8/5.4=1/3, giving AD=2.4 cm.
Resource ID: CB-10-M-CH6-EX6.2-Q1
Question 2
Decide if EF is parallel QR in each of the three side-data cases.
Step-by-step solution
By the converse of BPT, compare PE/EQ with PF/FR, or equivalently PE/PQ with PF/PR. (i) 3.9/3=1.3 but 3.6/2.4=1.5, so not parallel. (ii) 4/4.5=8/9 and 8/9 agree, so parallel. (iii) 0.18/1.28=0.36/2.56=9/64, so both points divide their whole sides in the same fraction.
Resource ID: CB-10-M-CH6-EX6.2-Q2
Question 3
In Fig. 6.18, LM∥CB and LN∥CD. Prove AM/AB=AN/AD.
Step-by-step solution
In triangle ABC, BPT gives AM/MB=AL/LC. Converting part-to-remainder ratios to part-to-whole ratios gives AM/AB=AL/AC. In triangle ADC, similarly AN/AD=AL/AC. Equate the common ratios.
Resource ID: CB-10-M-CH6-EX6.2-Q3
Question 4
In Fig. 6.19, DE∥AC and DF∥AE. Prove BF/FE=BE/EC.
Step-by-step solution
Apply BPT in triangle BAE, with DF∥AE: BF/FE=BD/DA. Apply BPT in triangle BAC, with DE∥AC: BE/EC=BD/DA. Therefore both ratios are equal.
Resource ID: CB-10-M-CH6-EX6.2-Q4
Question 5
In Fig. 6.20, DE∥OQ and DF∥OR. Show EF∥QR.
Step-by-step solution
In triangle POQ, PE/EQ=PD/DO by BPT. In triangle POR, PF/FR=PD/DO. Hence PE/EQ=PF/FR. Points E,F lie on PQ,PR, so the converse of BPT in triangle PQR gives the required parallelism.
Resource ID: CB-10-M-CH6-EX6.2-Q5
Question 6
Given A on OP, B on OQ, C on OR, with AB∥PQ and AC∥PR, show BC∥QR.
Step-by-step solution
BPT in triangle OPQ gives OA/AP=OB/BQ. In triangle OPR it gives OA/AP=OC/CR. Thus OB/BQ=OC/CR. By the converse of BPT in triangle OQR, the joining line BC is parallel to QR.
Resource ID: CB-10-M-CH6-EX6.2-Q6
Question 7
Use BPT to prove that a parallel line through one side’s midpoint bisects the other side.
Step-by-step solution
In triangle ABC let D be the midpoint of AB and DE∥BC meet AC at E. BPT gives AE/EC=AD/DB=1. Therefore AE=EC, so E is the midpoint of AC.
Resource ID: CB-10-M-CH6-EX6.2-Q7
Question 8
Use the converse of BPT to prove the midpoint joining theorem.
Step-by-step solution
Let D,E be the midpoints of AB,AC. Then AD/DB=1 and AE/EC=1. Equal division ratios allow the converse of BPT, so DE∥BC.
Resource ID: CB-10-M-CH6-EX6.2-Q8
Question 9
For trapezium ABCD with AB∥CD and diagonals meeting at O, prove AO/BO=CO/DO.
Step-by-step solution
Triangles AOB and COD have vertically opposite angles at O. Also ∠ABO=∠CDO because AB∥CD. Thus they are similar by AA, with A↔C,B↔D. So AO/CO=BO/DO. Cross-multiplying gives AO×DO=BO×CO; divide by BO×DO.
Resource ID: CB-10-M-CH6-EX6.2-Q9
Question 10
If intersecting diagonals satisfy AO/BO=CO/DO, show the quadrilateral is a trapezium.
Step-by-step solution
Rearrange the given ratio as AO/CO=BO/DO. Vertically opposite angles AOB and COD are equal, and their including side pairs are proportional. Thus triangle AOB is similar to COD by SAS. Corresponding angles ABO and CDO are equal, implying AB∥CD by the converse alternate-angle rule.
Resource ID: CB-10-M-CH6-EX6.2-Q10