NCERT Solutions class 10 Maths Exercise 6.6 Ch 6 Triangles

NCERT Solutions for Class 10 Maths Exercise 6.6 Chapter 6 Triangles – FREE PDF Download

NCERT Class 10 Maths Ch 6 is one of the most important ones in the NCERT syllabus. Duly following NCERT Solutions for Class 10 Maths Chapter
6 ensures you that there will be no hindrance when you opt for more advanced branches of Maths. This is where CoolGyan comes in. Our free Class 10 Triangles solutions will help you understand the chapter thoroughly.

NCERT Solutions for Class 10 Maths Chapter 6 – Triangles

1. In the given figure, PS is the bisector of  QPR of   PQR. Prove that

Ans. Given: PQR is a triangle and PS is the internal bisector of QPR 

meeting QR at S.

QPS = SPR

To prove: 

Construction: Draw RT  SP to cut QP produced at T.

Proof: Since PS  TR and PR cuts them, hence,

SPR = PRT ……….(i) [Alternate s]

And QPS = PTR ……….(ii)[Corresponding s]

But QPS = SPR [Given]

PRT = PTR[From eq. (i) & (ii)]

PT = PR……….(iii)

[Sides opposite to equal angles are equal]

Now, in QRT,

RT  SP[By construction]

[Thales theorem]

[From eq. (iii)]

2. In the given figure, D is a point on hypotenuse AC of ABC, BD AC, DM BC and DN AB. Prove that:

(i) = DN.MC

(ii) = DM.AN

Ans. Since AB BC and DM BC 

AB  DM

Similarly, BC AB and DN AB

CB  DN

quadrilateral BMDN is a rectangle.

BM = ND

(i) In BMD, 1 + BMD + 2 = 

1 +  + 2 = 

1 + 2 = 

Similarly, in DMC,3 + 4 = 

Since BD AC,

2 + 3 = 

Now, 1 + 2 =  and 2 + 3 = 

1 + 2 = 2 + 3

 1 = 3

Also, 3 + 4 =  and 2 + 3 = 

3 + 4 = 2 + 3

 4 = 2

Thus, in BMD and DMC,

1 = 3 and 4 = 2

BMD  DMC

 [BM = ND]

 = DN.MC

(ii) Processing as in (i), we can prove that

BND DNA

 [BN = DM]

 = DM.AN

3. In the given figure, ABC is a triangle in which ABC >  and AD CB produced. Prove that:

.BD

Ans. Given: ABC is a triangle in which ABC >  and AD CB produced. 

To prove:.BD

Proof: SinceADB is a right triangle, right angled at D, therefore, by Pythagoras theorem,

………(i)

Again, ADB is a right triangle, right angled at D, therefore, by Pythagoras theorem,

.BC

.BC

.BC

[Using eq. (i)]

4. In the given figure, ABC is a triangle in which ABC <  and AD BC produced. Prove that:

.BD

Ans. Given: ABC is a triangle in which ABC <  and AD BC produced. 

To prove:.BD

Proof: SinceADB is a right triangle, right angled at D, therefore, by Pythagoras theorem,

………(i)

Again, ADB is a right triangle, right angled at D, therefore, by Pythagoras theorem,

– 2BC.BD

– 2DB.BC

– 2DB.BC

[Using eq. (i)]

5. In the given figure, AD is a median of a triangle ABC and AM BC. Prove that:

(i) 

(ii) 

(iii) BC

Ans. Since AMD =  therefore ADM <  and ADC >  

Thus, ADC is the acute angle and ADC is an obtuse angle.

(i) In ADC, ADC is an obtuse angle.

  + 2..DM

  + BC.DM

 ……….(i)

(ii) In ABD, ADM is an acute angle.

  – 2..DM

 ……….(ii)

(iii) From eq. (i) and eq. (ii),

6. Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Ans. If AD is a median of ABC, then 

 [See Q.5 (iii)]

Since the diagonals of a parallelogram bisect each other, therefore, BO and DO are medians of triangles ABC and ADC respectively.

AB2 + BC2 = 2BO2 + AC2 ……….(i)

And AD2 + CD2 = 2DO2 + AC2 ……….(ii)

Adding eq. (i) and (ii),

AB2 + BC2 + AD2 + CD2 = 2 (BO2 + DO2) + AC2

AB2 + BC2 + AD2 + CD2 =  + AC2

AB2 + BC2 + AD2 + CD2 = AC2 + BD2

7. In the given figure, two chords AB and CD intersect each other at the point P. Prove that:

(i) APC DPB

(ii) AP.PB = CP.DP

Ans. (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 x PB = CP x DP

8. In the give figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that:

(i) PAC PDB

(ii) PA.PB = PC.PD

Ans. (i) In the triangles PAC and PDB, 

APC = DPB [Common]

CAP = BDP [BAC = PAC and PDB = CDB]

= BAC =  = PAC]

By AA-criterion of similarity,

APC DPB

(ii) Since APC DPB

PA.PB = PC.PD

9. In the given figure, D is appointing on side BC of ABC such that  Prove that AD is the bisector of BAC.

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

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]

Now, [Given]

[ AE = AC, by construction]

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. Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Fig. )? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

Ans. 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 =  = 1.59 (approx.)

Hence, the horizontal distance of the fly from Nazima after 12 seconds

= 1.59 + 1.2 = 2.79 m (approx.)