| --- |
| source: ncert_exercises |
| topic: geometry |
| class_level: class_9 |
| chapter: ch7 |
| exercise: ex7.1 |
| difficulty: intermediate |
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| Class 9 | Ch7: Triangles | Exercise 7.1 |
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| Congruence Rules: |
| SAS: Two sides and included angle equal → triangles congruent. |
| ASA: Two angles and included side equal → triangles congruent. |
| AAS: Two angles and non-included side equal → triangles congruent. |
| SSS: All three sides equal → triangles congruent. |
| RHS: Right angle, hypotenuse, one side equal → right triangles congruent. |
| CPCT: Corresponding Parts of Congruent Triangles are equal. |
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| Q1. In quadrilateral ACBD, AC=AD and AB bisects ∠A. Show △ABC≅△ABD. What about BC and BD? |
| Answer: In △ABC and △ABD: AC=AD (given), ∠CAB=∠DAB (AB bisects ∠A), AB=AB (common). ∴ △ABC≅△ABD by SAS. ∴ BC=BD by CPCT. ✅ |
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| Q2. ABCD is a quadrilateral with AD=BC and ∠DAB=∠CBA. Prove (i) △ABD≅△BAC (ii) BD=AC (iii) ∠ABD=∠BAC. |
| Answer: In △ABD and △BAC: AD=BC (given), ∠DAB=∠CBA (given), AB=AB (common). ∴ △ABD≅△BAC by SAS. (i) Proved. (ii) BD=AC by CPCT. (iii) ∠ABD=∠BAC by CPCT. ✅ |
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| Q3. AD and BC are equal perpendiculars to line segment AB. Show that CD bisects AB. |
| Answer: In △AOD and △BOC: ∠AOD=∠BOC (vertically opposite), ∠DAO=∠CBO=90°, AD=BC (given). ∴ △AOD≅△BOC by AAS. ∴ AO=BO by CPCT. So CD bisects AB at O. ✅ |
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| Q4. l and m are two parallel lines intersected by another pair of parallel lines p and q. Show △ABC≅△CDA. |
| Answer: AC is transversal to l∥m: ∠BAC=∠DCA (alternate interior). AC is transversal to p∥q: ∠BCA=∠DAC (alternate interior). AC=CA (common). ∴ △ABC≅△CDA by ASA. ✅ |
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| Q5. Line l is bisector of ∠A and B is any point on l. BP and BQ are perpendiculars from B to arms of ∠A. Show (i) △APB≅△AQB (ii) BP=BQ (B equidistant from arms). |
| Answer: In △APB and △AQB: ∠APB=∠AQB=90°, AB=AB (common), ∠PAB=∠QAB (l bisects ∠A). ∴ △APB≅△AQB by AAS. ∴ BP=BQ by CPCT. ✅ |
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| Q6. In the figure, AC=AE, AB=AD and ∠BAD=∠EAC. Show BC=DE. |
| Answer: ∠BAD=∠EAC (given). Add ∠DAC both sides: ∠BAC=∠DAE. In △ABC and △ADE: AB=AD, AC=AE, ∠BAC=∠DAE. ∴ △ABC≅△ADE by SAS. ∴ BC=DE by CPCT. ✅ |
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| Q7. AB is a line segment and P is its midpoint. D and E are on same side of AB such that ∠BAD=∠ABE and ∠EPA=∠DPB. Show (i) △DAP≅△EBP (ii) AD=BE. |
| Answer: ∠EPA=∠DPB (given). Add ∠EPD: ∠DPE+∠EPA=∠DPE+∠DPB → ∠DPA=∠EPB. In △DAP and △EBP: ∠DAP=∠EBP (given), AP=BP (P midpoint), ∠DPA=∠EPB. ∴ △DAP≅△EBP by ASA. ∴ AD=BE by CPCT. ✅ |
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| Q8. In right triangle ABC, right angle at C, M is midpoint of hypotenuse AB. C is joined to M and produced to point D such that DM=CM. Point D is joined to B. Show (i) △AMC≅△BMD (ii) ∠DBC=90° (iii) △DBC≅△ACB (iv) CM=½AB. |
| Answer: (i) AM=BM (M midpoint), CM=DM (given), ∠AMC=∠BMD (vertically opposite). ∴ △AMC≅△BMD by SAS. (ii) ∠ACM=∠BDM by CPCT → AC∥DB → ∠DBC+∠ACB=180° → ∠DBC=90°. (iii) DB=AC (CPCT), BC=BC (common), ∠DBC=∠ACB=90°. ∴ △DBC≅△ACB by SAS. (iv) DC=AB (CPCT) → 2CM=AB → CM=½AB. ✅ |
