10 mathematics ncert ch06 triangles ex 6 6 ans cxs

NCERT Solutions Class – X Mathematics Chapter-06 Triangles (Exercise 6.6)

Answers 1.

Given: PQR is a triangle and PS is the internal bisector of ∠ QPR meeting QR at S. ∠ QPS = ∠ SPR ∴ QS PQ To prove: = SR PR Construction: Draw RT SP to cut QP produced at T. Proof: Since PS

TR and PR cuts them, hence,

∠ SPR = ∠ PRT ∠ QPS = ∠ PTR ∠ QPS = ∠ SPR ∠ PRT = ∠ PTR PT = PR

And But ∴ ⇒

Now, in ∆ QRT, RT SP QS PQ = SR PT QS PQ ⇒ = SR PR Since, AB ⊥ BC and DM ⊥ BC ⇒ AB DM ∴

2.

Similarly, ⇒ CB ∴ ∴ (i)

……….(i) [Alternate ∠ s] ……….(ii) [Corresponding ∠ s] [Given] [From eq. (i) & (ii)] ……….(iii) [Sides opposite to equal angles are equal] [By construction] [Thales theorem] [From eq. (iii)]

BC ⊥ AB and DN ⊥ AB DN

quadrilateral BMDN is a rectangle. BM = ND In ∆ BMD, ∠ 1 + ∠ BMD + ∠ 2 = 180° ⇒ ∠ 1 + 90 + ∠ 2 = 180° ⇒ ∠ 1 + ∠ 2 = 90 Similarly in ∆ DMC, ∠ 3 + ∠ 4 = 90 Since BD ⊥ AC, ∠ 2 + ∠ 3 = 90 ∴ Now, ∠ 1 + ∠ 2 = 90 and ∠ 2 + ∠ 3 = 90 ⇒ ∠1+ ∠2= ∠2+ ∠3 ⇒ ∠1= ∠3 Also, ∠ 3 + ∠ 4 = 90 and ∠ 2 + ∠ 3 = 90 ⇒ ∠3+ ∠4= ∠2+ ∠3 ⇒ ∠4= ∠2 Thus, in ∆ BMD and ∆ DMC, Material downloaded from http://myCBSEguide.com and http://onlineteachers.co.in Portal for CBSE Notes, Test Papers, Sample Papers, Tips and Tricks

NCERT Solutions

3.

4.

5.

∠ 1 = ∠ 3 and ∠ 4 = ∠ 2 ∴ ∆ BMD ∼ ∆ DMC BM MD DN DM ⇒ ⇒ [BM = ND] = = DM MC DM MC ⇒ DM2 = DN.MC (ii) Processing as in (i), we can prove that ∆ BND ∼ ∆ DNA BN ND DM DN ⇒ ⇒ [BN = DM] = = DN NA DN AN ⇒ DN2 = DM.AN Given: ABC is a triangle in which ∠ ABC > 90 and AD ⊥ CB produced. To prove: AC2 = AB2 + BC2 + 2BC.BD Proof: Since ∆ ADB is a right triangle, right angled at D, therefore, by Pythagoras theorem, ………(i) AB2 = AD2 + DB2 Again, ∆ ADB is a right triangle, right angled at D, therefore, by Pythagoras theorem, AC2 = AD2 + DC2 ⇒ AC2 = AD2 + (DB + BC)2 ⇒ AC2 = AD2 + DB2 + BC2 + 2DB.BC ⇒ AC2 = (AD2 + DB2)+ BC2 + 2DB.BC ⇒ AC2 = AB2 + BC2 + 2DB.BC [Using eq. (i)] Given: ABC is a triangle in which ∠ ABC < 90 and AD ⊥ BC produced. To prove: AC2 = AB2 + BC2 – 2BC.BD Proof: Since ∆ ADB is a right triangle, right angled at D, therefore, by Pythagoras theorem, AB2 = AD2 + BD2 ………(i) Again, ∆ ADB is a right triangle, right angled at D, therefore, by Pythagoras theorem, AC2 = AD2 + DC2 ⇒ AC2 = AD2 + (BC – BD)2 ⇒ AC2 = AD2 + BC2 + BD2 – 2BC.BD ⇒ AC2 = (AD2 + DB2)+ BC2 – 2DB.BC ⇒ AC2 = AB2 + BC2 – 2DB.BC [Using eq. (i)] Since ∠ AMD = 90 , therefore ∠ ADM < 90 and ∠ ADC > 90 Thus, ∠ ADC is acute angle and ∠ ADC is obtuse angle. (i) In ∆ ADC, ∠ ADC is an obtuse angle. AC2 = AD2 + DC2 + 2DC.DM ∴ 2

⇒

AC2

=

AD2

BC  BC  + .DM  + 2. 2  2 

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NCERT Solutions 2

⇒

(ii)

 BC  AC2 = AD2 +   + BC.DM  2 

 BC  ⇒ = + BC.DM +    2  In ∆ ABD, ∠ ADM is an acute angle. AB2 = AD2 + BD2 – 2BD.DM

AC2

AD2

AB2

AD2

2

……….(i)

2

⇒

=

BC  BC  + .DM  – 2. 2  2  2

 BC  ⇒ = – BC.DM +  ……….(ii)   2  (iii) From eq. (i) and eq. (ii), 1 AB2 + AC2 = 2AD2 + BC2 2 If AD is a median of ∆ ABC, then 1 AB2 + AC2 = 2AD2 + BC2 [See Q.5 (iii)] 2 Since the diagonals of a parallelogram bisect each other, therefore, BO and DO are medians of triangles ABC and ADC respectively. 1 AB2 + BC2 = 2BO2 + AC2 ……….(i) ∴ 2 1 And AD2 + CD2 = 2DO2 + AC2 ……….(ii) 2 Adding eq. (i) and (ii), AB2 + BC2 + AD2 + CD2 = 2 (BO2 + DO2) + AC2 1 1 1    ⇒ AB2 + BC2 + AD2 + CD2 = 2  BD 2 + BD 2  + AC2 DO = BD   4 2 4    ⇒ AB2 + BC2 + AD2 + CD2 = AC2 + BD2 (i) In the triangles APC and DPB, ∠ APC = ∠ DPB [Vertically opposite angles] ∠ CAP = ∠ BDP [Angles in same segment of a circle are equal] ∴ By AA-criterion of similarity, ∆ APC ∼ ∆ DPB (ii) Since ∆ APC ∼ ∆ DPB AP CP ⇒ AP x PB = CP x DP ∴ = DP PB (i) In the triangles PAC and PDB, ∠ APC = ∠ DPB [Common] ∠ CAP = ∠ BDP [∵ ∠ BAC = 180° − ∠ PAC and ∠ PDB = ∠ CDB]

AB2

6.

7.

8.

AD2

= 180° − ∠ BAC = 180° − (180° − ∠PAC ) = ∠ PAC]

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NCERT Solutions ∴

By AA-criterion of similarity, ∆ APC ∼ ∆ DPB (ii) Since ∆ APC ∼ ∆ DPB AP CP ⇒ PA.PB = PC.PD ∴ = DP PB 9.

Given: ABC is a triangle and D is a point on BC such that

BD AB = CD AC

To prove: AD is the internal bisector of ∠ BAC. Construction: Produce BA to E such that AE = AC. Join CE. Proof: In ∆ AEC, since AE = AC ∠ AEC = ∠ ACE ……….(i) ∴ [Angles opposite to equal side of a triangle are equal] BD AB Now, [Given] = CD AC BD AB ⇒ [∵ AE = AC, by construction] = CD AE By converse of Basic Proportionality Theorem, ∴ DA CE Now, since CA is a transversal, ∠ BAD = ∠ AEC ……….(ii) [Corresponding ∠ s] ∴ And ∠ DAC = ∠ ACE ……….(iii) [Alternate ∠ s] Also ∠ AEC = ∠ ACE [From eq. (i)] Hence, ∠ BAD = ∠ DAC [From eq. (ii) and (iii)] Thus, AD bisects ∠ BAC internally. 10. I. To find: The length of AC. By Pythagoras theorem, AC2 = (2.4)2 + (1.8)2 ⇒ AC2 = 5.76 + 3.24 = 9.00 ⇒ AC = 3 m ∴ Length of string she has out= 3 m Length of the string pulled at the rate of 5 cm/sec in 12 seconds = (5 x 12) cm = 60 cm = 0.60 m ∴ Remaining string left out = 3 – 0.6 = 2.4 m II. To find: The length of PB PB2 = PC2 – BC2 = (2.4)2 – (1.8)2 = 5.76 – 3.24 = 2.52 ⇒ PB = 2.52 = 1.59 (approx.) Hence, the horizontal distance of the fly from Nazima after 12 seconds = 1.59 + 1.2 = 2.79 m (approx.)

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