NCERT Solutions Maths Class 10 Exercise 6.6

# NCERT Solutions Maths Class 10 Exercise 6.6

## Maths Class 10 Exercise 6.6

1. In figure, PS is the bisector of QPR of PQR. Prove that QS/SR = PQ/PR.

Solution: Given: PS is the internal bisector of QPR of PQR meeting QR at S. Therefore, QPS = SPR

To proveQS/SR = PQ/PR

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

Proof: Since PS TR and PR is a transversal, then

SPR = PRT        ………. (i)                    [Alternate angles]

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

But QPS = SPR            [Given]

Therefore, PRT = PTR          [From equations (i) and (ii)]

PT = PR          ………. (iii)         [Sides opposite to equal angles are equal]

Now, in QRT, RT SP              [By construction]

Therefore, QS/SR = PQ/PT       [By Basic Proportionality theorem]

QS/SR = PQ/PR                      [From equation (iii)]                                Hence proved.

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

(i) DM2 = DN . MC

(ii) DN2 = DM . AN

Solution: Since, AB BC and DM BC

AB DM

Similarly, BC AB and DN AB

CB DN

Therefore, quadrilateral BMDN is a rectangle.

So, BM = ND

(i) In BMD, MBD + BMD + BDM = 180°

MBD + 90° + BDM = 180°

MBD + BDM = 90°        ……….. (i)

Similarly, in DMC, MDC + MCD = 90°        ………… (ii)

Since BD AC,

Therefore, MDB + MDC = 90°           …………. (iii)

From equations (i) and (iii), we get

MBD + BDM = MDB + MDC

MBD = MDC

From equations (ii) and (iii), we get

MDC + MCD = MDB + MDC

MCD =  MDB

Thus, in BMD and DMC,

MBD = MDC and MCD =  MDB

Therefore, ∆BMD ~ ∆DMC

BM/DM = MD/MC

DN/DM = DM/MC       [BM = DN]

DM2 = DN . MC

(ii) As proved in (i), we can prove that BND ~ ∆DNA.

⇒ BN/DN = DN/AN

DM/DN = DN/AN        [BN = DM]

DN2 = DM . AN

3. In figure, ABC is a triangle in which ABC > 90° and AD CB produced. Prove that:

AC2 = AB2 + BC2 + 2BC . BD

Solution: Given: ABC is a triangle in which ABC > 90° and AD CB produced.

To prove: AC2 = AB2 + BC2 + 2BC . BD

Proof: SinceADB is a right triangle, right angled at D,

Therefore, AB2 = AD2 + DB2             ……… (i)             [By Pythagoras theorem]

Again,ADC is a right triangle, right angled at D.

Therefore, AC2 = AD2 + DC2                                  [by Pythagoras theorem]

AC2 = AD2 + (DB + BC)2                                        [Since DC = DB + BC]

AC2 = AD2 + DB2 + BC2 + 2DB . BC

AC2 = (AD2 + DB2) + BC2 + 2DB . BC

AC2 = AB2 + BC2+ 2DB.BC                     [Using equation (i)]            Hence proved.

4. In figure, ABC is a triangle in which ABC < 90° and AD BC. Prove that:

AC2 = AB2 + BC2 – 2BC . BD

Solution: Given: ABC is a triangle in which ABC < 90° and AD BC.

To prove: AC2 = AB2 + BC2 – 2BC . BD

Proof: SinceADB is a right triangle, right angled at D.

Therefore, AB2 = AD2 + BD2                 ……… (i)                        [By Pythagoras theorem]

Again, ADC is a right triangle, right angled at D.

Therefore, AC2 = AD2 + DC2                                  [By Pythagoras theorem]

AC2 = AD2 + (BC – BD)2                                       [Since DC = BC – BD]

AC2 = AD2 + BC2 + BD2 – 2BC . BD

AC2 = (AD2 + BD2) + BC2 – 2 BC . BD

AC2 = AB2 + BC2 – 2BC . BD         [Using equation (i)]          Hence proved.

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

(i) AC2 = AD2 + BC . DM + (BC/2)2

(ii) AB2 = AD2 BC . DM + (BC/2)2

(iii) AC2 + AB2 = 2AD2 + ½ BC2

Solution: Since AMD = 90°, therefore, ADM < 90° and ADC > 90°

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

Therefore, AC2 = AD2 + DC2 + 2DC . DM

AC2 = AD2 + (BC/2)2 + 2BC/2 . DM              [Since DC = BC/2]

AC2 = AD2 + (BC/2)2 + BC . DM

AC2 = AD2 + BC . DM + (BC/2)2                 ………….. (i)

(ii) In ABD, ADM is an acute angle.

Therefore, AB2 = AD2 + BD2 – 2BD . DM

AB2 = AD2 + (BC/2)2 – 2BC/2 . DM                [Since BD = BC/2]

AB2 = AD2 + (BC/2)2 – BC . DM

AB2 = AD2 – BC . DM + (BC/2)2                 ………….. (ii)

(iii) Adding equations (i) and (ii), we get

AC2 + AB2 = 2AD2 + ½ BC2

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

Solution: Given: A parallelogram ABCD whose diagonals AC and BD intersect at point O.

To prove: AB2 + BC2 + AD2 + CD2 = AC2 + BD2

If AD is a median of ABC, then

AC2 + AB2 = 2AD2 + ½ BC2         [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.

Therefore, AB2 + BC2 = 2BO2 + ½ AC2         ………. (i)

And AD2 + CD2 = 2DO2 + ½ AC2                    ………. (ii)

Adding equations (i) and (ii), we get

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

AB2 + BC2 + AD2 + CD2 = 2(¼ BD2 + ¼ BD2) + AC2                        [DO = ½ BD]

AB2 + BC2 + AD2 + CD2 = AC2 + BD2                         Hence proved.

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

(i) APC ~ ∆DPB

(ii) AP . PB = CP . DP

Solution: (i) In the triangles APC and DPB,

APC = DPB          [Vertically opposite angles]

CAP = BDP          [Angles in the same segment of a circle are equal]

Therefore, ∆APC ~ ∆DPB                   [By AAA criterion of similarity]

(ii) Since APC ~ ∆DPB

Therefore, AP/DP = CP/PB

AP . PB = CP . DP

8. In 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

Solution: (i) In the PAC and PDB, we have

APC = DPB          [Common]

BDP = 180°– BAC

= 180°– (180° PAC) = PAC      [Since BAC = 180°– PAC]

PAC = BDP

Therefore, ∆APC ~ ∆PDB                          [By AAA criterion of similarity]

(ii) Since APC ~ ∆PDB

Therefore, PA/PD = PC/PB

PA . PB = PC . PD

9. In figure, D is a point on side BC of ABC such that BD/CD = AB/AC. Prove that AD is the bisector of BAC.

Solution: Given: ABC is a triangle and D is a point on BC such that BD/CD = AB/AC.

To prove: AD is the internal bisector of BAC.

Construction: Produce BA to E such that AE = AC. Join CE.

Proof: In AEC, we have

AE = AC

Therefore, AEC = ACE   ………. (i) [Angles opposite to equal side of a triangle are equal]

Now, BD/CD = AB/AC        [Given]

BD/CD = AB/AE             [Since, AC = AE, by construction]

Therefopre, DA CE         [By converse of Basic Proportionality theorem]

Now, since CA is a transversal.

Therefore, BAD = AEC          ………. (ii)               [Corresponding angles]

And DAC = ACE           ………. (iii)                         [Alternate angles]

Also AEC = ACE                [From equation (i)]

Hence, BAD = DAC          [From equations (ii) and (iii)]

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 figure)? 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?

Solution: (i) To find the length of AC, we have

AC2 = AB2 + BC2                            [By Pythagoras theorem]

= (2.4)2 + (1.8)2

AC2 = 5.76 + 3.24 = 9.00

AC = 3 m

Therefore, the length of the string she has out = 3 m

(ii) The length of the string pulled at the rate of 5 cm/sec in 12 seconds

= (5 × 12) cm = 60 cm = 0.60 m

Therefore,  the remaining string left out = 3 – 0.6 = 2.4 m To find the length of PB, we have

PB2 = PC2 – BC2               [By Pythagoras theorem]

= (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.)