| --- |
| source: ncert_exercises |
| topic: geometry |
| class_level: class_10 |
| chapter: ch6 |
| exercise: ex6.2 |
| difficulty: intermediate |
| --- |
| |
| Class 10 | Ch6: Triangles | Exercise 6.2 |
|
|
| Basic Proportionality Theorem (BPT / Thales Theorem): |
| If a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally. |
| If DE∥BC in △ABC, then AD/DB = AE/EC. |
| Converse: If AD/DB = AE/EC, then DE∥BC. |
|
|
| Q1. In △ABC, DE∥BC. Find EC or DB using BPT: |
| (i) AD=1.5cm, DB=3cm, AE=1cm. Find EC. |
| (ii) AD=4cm, AE=8cm, EC=3cm. Find AB. |
| Answer: |
| (i) AD/DB=AE/EC → 1.5/3=1/EC → EC=2cm. ✅ |
| (ii) AD/DB=AE/EC → 4/DB=8/3 → DB=3/2=1.5cm. AB=AD+DB=5.5cm. ✅ |
|
|
| Q2. E and F are points on PQ and PR of △PQR. State whether EF∥QR: |
| (i) PE=3.9cm, EQ=3cm, PF=3.6cm, FR=2.4cm |
| (ii) PE=4cm, QE=4.5cm, PF=8cm, RF=9cm |
| (iii) PQ=1.28cm, PR=2.56cm, PE=0.18cm, PF=0.36cm |
| Answer: |
| (i) PE/EQ=3.9/3=1.3. PF/FR=3.6/2.4=1.5. Not equal → EF NOT∥QR. ✅ |
| (ii) PE/EQ=4/4.5=8/9. PF/RF=8/9. Equal → EF∥QR. ✅ |
| (iii) PE/EQ=0.18/(1.28-0.18)=0.18/1.1. PF/RF=0.36/(2.56-0.36)=0.36/2.2=0.18/1.1. Equal → EF∥QR. ✅ |
|
|
| Q3. In the figure, if LM∥CB and LN∥CD, prove AM/AB=AN/AD. |
| Answer: In △ABC: LM∥CB → AL/LB=AM/MC... Use BPT: AM/MB=AL/LB. In △ACD: LN∥CD → AN/ND=AL/LC. Since AL/LB=AL/LC (L is same)... Actually: BPT in △ABC: AM/AB=AL/AC (if LM∥BC). BPT in △ACD: AN/AD=AL/AC. ∴ AM/AB=AN/AD. ✅ |
|
|
| Q4. In the figure, DE∥AC and DF∥AE. Prove BF/FE=BE/EC. |
| Answer: In △ABC: DE∥AC (BPT) → BD/DA=BE/EC ...(i). In △ABE: DF∥AE (BPT) → BD/DA=BF/FE ...(ii). From (i) and (ii): BF/FE=BE/EC. ✅ |
|
|
| Q5. In the figure, DE∥OQ and DF∥OR. Show EF∥QR. |
| Answer: In △PQO: DE∥OQ (BPT) → PE/EQ=PD/DO ...(i). In △POR: DF∥OR (BPT) → PF/FR=PD/DO ...(ii). PE/EQ=PF/FR → EF∥QR (converse BPT in △PQR). ✅ |
|
|
| Q6. In the figure, A,B,C are points on OP,OQ,OR such that AB∥PQ and AC∥PR. Show BC∥QR. |
| Answer: AB∥PQ in △OPQ: OA/AP=OB/BQ ...(i). AC∥PR in △OPR: OA/AP=OC/CR ...(ii). OB/BQ=OC/CR → BC∥QR (converse BPT in △OQR). ✅ |
|
|
| Q7. Using BPT, prove that line drawn through midpoint of one side of triangle parallel to another side bisects third side. |
| Answer: Let M be midpoint of AB. Draw MN∥BC. In △ABC: MN∥BC (BPT) → AM/MB=AN/NC. AM=MB (M midpoint) → AN=NC. N is midpoint of AC. ✅ |
|
|
| Q8. Using the converse of BPT, prove that the line joining the midpoints of any two sides of a triangle is parallel to the third side. |
| Answer: Let M,N be midpoints of AB,AC. AM=MB and AN=NC → AM/MB=1=AN/NC. By converse BPT: MN∥BC. ✅ |
|
|
| Q9. ABCD is a trapezium with AB∥DC. E and F are midpoints of AD and BC. EF∥AB and EF=1/2(AB+DC). |
| Answer: Draw diagonal AC. In △ABC: E is mid of... Standard midpoint theorem on trapezium. EG∥AB in △ABD gives G is mid of BD. GF∥DC in △BDC gives F is mid of BC (confirmed). EF=EG+GF=AB/2+DC/2=½(AB+DC). ✅ |
|
|
| Q10. Prove that the diagonals of a trapezium divide each other proportionally. |
| Answer: ABCD trapezium, AB∥DC. Diagonals AC,BD meet at O. In △AOB and △COD: ∠AOB=∠COD (vertical), ∠OAB=∠OCD (alternate, AB∥DC). △AOB∼△COD (AA). ∴ AO/CO=BO/DO. ✅ |
