NCERT Solutions Class 9th Maths Chapter – 7 Triangles
| Textbook | NCERT |
| Class | 9th |
| Subject | Mathematics |
| Chapter | 7th |
| Chapter Name | Triangles |
| Category | Class 9th Math Solutions |
| Medium | English |
| Source | Last Doubt |
NCERT Solutions Class 9th Maths Chapter – 7 Triangles
Chapter – 7
Triangles
Exercise 7.1
Question 1. In quadrilateral ACBD, AC = AD and AB bisects ∠ A (see figure). Show that ∆ABC ≅ ∆ABD. What can you say about BC and BD? Solution In quadrilateral ACBD, we have AC = AD and AB being the bisector of ∠A. Now, In ∆ABC and ∆ABD, AC = AD (Given) ∠ CAB = ∠ DAB ( AB bisects ∠ CAB) and AB = AB (Common) ∴ ∆ ABC ≅ ∆ABD (By SAS congruence axiom) ∴ BC = BD (By CPCT) |
Question 2. ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see figure). Prove that (i) ∆ABD ≅ ∆BAC (ii) BD = AC (iii) ∠ABD = ∠ BAC Solution In quadrilateral ACBD, we have AD = BC and ∠ DAB = ∠ CBA (i) In ∆ ABC and ∆ BAC, (ii) Since ∆ABD ≅ ∆BAC (iii) Since ∆ABD ≅ ∆BAC |
Question 3. AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB. Solution In ∆BOC and ∆AOD, we have ∠BOC = ∠AOD BC = AD [Given] ∠BOC = ∠AOD [Vertically opposite angles] ∴ ∆OBC ≅ ∆OAD [By AAS congruency] ⇒ OB = OA [By C.P.C.T.] i.e., O is the mid-point of AB. Thus, CD bisects AB. |
Question 4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that ∆ABC = ∆CDA. Solution ∵ p || q and AC is a transversal, ∴ ∠BAC = ∠DCA …(1) [Alternate interior angles] Also l || m and AC is a transversal, ∴ ∠BCA = ∠DAC …(2) [Alternate interior angles] Now, in ∆ABC and ∆CDA, we have ∠BAC = ∠DCA [From (1)] CA = AC [Common] ∠BCA = ∠DAC [From (2)] ∴ ∆ABC ≅ ∆CDA [By ASA congruency] |
| Question 5. Line l is the bisector of an ∠ A and ∠ B is any point on l. BP and BQ are perpendiculars from B to the arms of LA (see figure). Show that (i) ∆APB ≅ ∆AQB (ii) BP = BQ or B is equidistant from the arms ot ∠A.  Solution We have, l is the bisector of ∠QAP. ∴ ∠QAB = ∠PAB ∠Q = ∠P [Each 90°] ∠ABQ = ∠ABP [By angle sum property of A] Now, in ∆APB and ∆AQB, we have ∠ABP = ∠ABQ [Proved above] AB = BA [Common] ∠PAB = ∠QAB [Given] ∴ ∆APB ≅ ∆AQB [By ASA congruency] Since ∆APB ≅ ∆AQB ⇒ BP = BQ [By C.P.C.T.] i. e., [Perpendicular distance of B from AP] = [Perpendicular distance of B from AQ] Thus, the point B is equidistant from the arms of ∠A. |
Question 6. In figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE. Solution We have, ∠BAD = ∠EAC Adding ∠DAC on both sides, we have ∠BAD + ∠DAC = ∠EAC + ∠DAC ⇒ ∠BAC = ∠DAE Now, in ∆ABC and ∆ADE. we have ∠BAC = ∠DAE [Proved above] AB = AD [Given] AC = AE [Given] ∴ ∆ABC ≅ ∆ADE [By SAS congruency] ⇒ BC = DE [By C.P.C.T.] |
| Question 7. AS is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB. (see figure). Show that (i) ∆DAP ≅ ∆EBP (ii) AD = BE  Solution We have, P is the mid-point of AB. ∴ AP = BP ∠EPA = ∠DPB [Given] Adding ∠EPD on both sides, we get ∠EPA + ∠EPD = ∠DPB + ∠EPD ⇒ ∠APD = ∠BPE (i) Now, in ∆DAP and ∆EBP, we have (ii) Since, ∆ DAP ≅ ∆ EBP |
| Question 8. In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see figure). Show that (i) ∆AMC ≅ ∆BMD (ii) ∠DBC is a right angle (iii) ∆DBC ≅ ∆ACB (iv) CM = 1/2 AB
Solution – Since M is the mid – point of AB. (i) In ∆AMC and ∆BMD, we have (ii) Since ∆AMC ≅ ∆BMD (iii) Again, ∆AMC ≅ ∆BMD [Proved above] (iv) As ∆DBC ≅ ∆ACB |
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