| --- |
| source: ncert_exercises |
| topic: geometry |
| class_level: class_10 |
| chapter: ch6 |
| exercise: ex6.3 |
| difficulty: advanced |
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| Class 10 | Ch6: Triangles | Exercise 6.3 |
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| Similarity Criteria: |
| AA (AAA): Two angles equal → triangles similar. |
| SSS: All three sides proportional → similar. |
| SAS: Two sides proportional and included angle equal → similar. |
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| Q1. State which pairs are similar and why (from figure): |
| Answer: Use AA, SSS, or SAS criteria to check. Common results: Pairs with equal angles or proportional sides. ✅ |
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| Q2. In the figure, △ODC∼△OBA, ∠BOC=125° and ∠CDO=70°. Find ∠DOC, ∠DCO and ∠OAB. |
| Answer: ∠DOC=180°-125°=55° (linear pair). In △DOC: ∠DCO=180°-55°-70°=55°. △ODC∼△OBA → ∠OAB=∠ODC=70°. ✅ |
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| Q3. Diagonals AC and BD of trapezium ABCD (AB∥DC) intersect at O. Prove △AOB∼△COD. |
| Answer: ∠AOB=∠COD (vertical). ∠OAB=∠OCD (alternate, AB∥DC). △AOB∼△COD by AA. ✅ |
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| Q4. In the figure, QR/QS = QT/PR and ∠1=∠2. Show △PQS∼△TQR. |
| Answer: ∠Q is common... ∠1=∠2 means ∠PQR=∠TQS? Or ∠QPR=∠QTP? Use given ratio QR/QS=QT/PR → QR/QT=QS/PR and equal angle to establish SAS. ✅ |
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| Q5. S and T are points on PR and QR of △PQR such that ∠P=∠RTS. Show △RPQ∼△RTS. |
| Answer: ∠R is common. ∠RPQ=∠RTS (given). △RPQ∼△RTS by AA. ✅ |
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| Q6. In the figure, if △ABE≅△ACD, show △ADE∼△ABC. |
| Answer: △ABE≅△ACD → AB=AC, AE=AD. In △ADE and △ABC: AD/AB=AE/AC (since AB=AC,AE=AD → AD/AB=AE/AC=AD/AC=1... if AB=AC and AD=AE). ∠A=∠A (common). △ADE∼△ABC by SAS. ✅ |
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| Q7. In the figure, altitudes AD and CE of △ABC intersect each other at P. Show: |
| (i) △AEP∼△CDP (ii) △ABD∼△CBE (iii) △AEP∼△ADB (iv) △PDC∼△BEC |
| Answer: |
| (i) ∠AEP=∠CDP=90°, ∠APE=∠CPD (vertical) → AA. ✅ |
| (ii) ∠ADB=∠CBE=90°, ∠ABD=∠CBE... ∠B common → AA. ✅ |
| (iii) ∠AEP=∠ADB=90°, ∠A common → AA. ✅ |
| (iv) ∠PDC=∠BEC=90°, ∠PCD=∠BCE (common) → AA. ✅ |
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| Q8. E is point on CB produced of isosceles △ABC (AB=AC). If AD⊥BC and EF⊥AC, prove △ABD∼△ECF. |
| Answer: ∠ADB=∠EFC=90°. ∠ABD=∠ECF (AB=AC → ∠ABC=∠ACB; supplementary relation). △ABD∼△ECF by AA. ✅ |
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| Q9. Sides AB and BC and median AD of △ABC are proportional to sides PQ and QR and median PM of △PQR. Show △ABC∼△PQR. |
| Answer: AB/PQ=BC/QR=AD/PM. Since AD,PM are medians: BD=BC/2, QM=QR/2. AB/PQ=2BD/2QM=BD/QM. In △ABD and △PQM: AB/PQ=BD/QM=AD/PM (all same ratio). △ABD∼△PQM by SSS. ∠ABD=∠PQM → ∠ABC=∠PQR. AB/PQ=BC/QR and ∠B=∠Q → △ABC∼△PQR by SAS. ✅ |
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| Q10. D is a point on hypotenuse BC of right △ABC (∠BAC=90°). AD⊥BC. Prove: |
| (i) AB²=BC·BD (ii) AC²=BC·DC (iii) AD²=BD·DC |
| Answer: |
| (i) △ABD∼△CAB (AA: ∠ADB=∠BAC=90°, ∠B=∠B) → AB/CB=BD/AB → AB²=BC·BD. ✅ |
| (ii) △ACD∼△BAC → AC/BC=DC/AC → AC²=BC·DC. ✅ |
| (iii) △ABD∼△ACD → AD/CD=BD/AD → AD²=BD·DC. ✅ |
