HBSE 10th Class Maths Solutions Chapter 6 Triangles Ex 6.6

Haryana State Board HBSE 10th Class Maths Solutions Chapter 6 Triangles Ex 6.6 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 6 Triangles Ex 6.6

Question 1.
In the given figure, PS is the bisector of ∠QPR of ∆PQR. Prove that \(\frac{\mathrm{QS}}{\mathrm{SR}}=\frac{\mathrm{PQ}}{\mathrm{PR}}\).

Haryana Board 10th Class Maths Solutions Chapter 6 Triangles Ex 6.6 1

Solution :
Given : A ∆PQR in which PS is the bisector of ∠QPR
To Prove: \(\frac{\mathrm{QS}}{\mathrm{SR}}=\frac{\mathrm{PQ}}{\mathrm{PR}}\)
Construction : Draw RT || PS to meet QP produced at T.
Proof : Since RT || SP and PR is the trans-versal :
∠2 = ∠3 ……………..(1) [alternate interior ∠s]
∠1 = ∠4 ……………..(2) [Corresponding ∠s]
∠1 = ∠2 ……………..(3) [PS is the bisector of ∠QPR]
From (1), (2) and (3) we get

Haryana Board 10th Class Maths Solutions Chapter 6 Triangles Ex 6.6 2

∠3 = ∠4
∴ PR = PT
Now in ∆QRT, PS || TR
⇒ \(\frac{\mathrm{QS}}{\mathrm{SR}}=\frac{\mathrm{PQ}}{\mathrm{PT}}\) [By BPT]
⇒ \(\frac{\mathrm{QS}}{\mathrm{SR}}=\frac{\mathrm{PQ}}{\mathrm{PR}}\) [∵ PT = PR]
Hence Proved.

Haryana Board Solutions for 10th Class Maths Chapter 6 Triangles Ex 6.6 width=

Question 2.
In the given figure D is a point on hypotenuse AC of ∆ABC, such that BD ⊥ AC, DM ⊥ BC and DN ⊥ AB. Prove that:
(i) DM2 = DN.MC
(ii) DN2 = DM.AN

Haryana Board 10th Class Maths Solutions Chapter 6 Triangles Ex 6.6 3

Solution :
Given : D is a point on AC such that BD ⊥ AC, DM ⊥ BC and DN ⊥ AB.
To Prove :
(i) DM2 = DN.MC
(ii) DN2 = DM.AN
Proof: ∠ABC = 90° [∵ AC is hypotenuse of right ∆ABC]
∠DMB = 90° [∵ DM ⊥ BC]
∠DNB = 90° [∵ DN ⊥ AB]
∠MDN = 90° [Fourth ∠ of a quadrilateral]
Therefore, quadrilateral BMDN is a rectangle, (i) In ABMDZ1 + Z2 + ZDMB = 180°
⇒ ∠1 + ∠2 + 90° = 180°
⇒ ∠1 + ∠2 = 180° – 90°
⇒ ∠1 + ∠2 = 90° ………………..(1)
In ∆CMD
∠3 + ∠4 + ∠CMD = 180°
⇒ ∠3 + ∠4 + 90° = 180°
⇒ ∠3 + ∠4 = 180° – 90°
⇒ ∠3 + ∠4 = 90° ……………….(2)
BD ⊥ AC
∴ ∠1 + ∠3 = 90° ……………….(3)
From (1) and (3), we get
∠1 + ∠2 = ∠1 + ∠3
⇒ ∠2 = ∠3 ……………….(4)
From (2), and (3), we get
∠3 + ∠4 = ∠1 + ∠3
⇒ ∠1 = ∠4 …………….(5)
Now in ∆BMD and ∆DMC,
∠2 = ∠3 [from (4)]
∠1 = ∠4 [from (5)]
∆BMD ~ ∆DMC [By AA similarity criterion]
⇒ \(\frac{B M}{D M}=\frac{M D}{M C}\)
⇒ \(\frac{D N}{D M}=\frac{D M}{M C}\) [∵ BM = DN]
⇒ DM2 = DN.MC.
Hence Proved.

(ii) Similarly, ∆BND ~ ∆DNA
⇒ \(\frac{B N}{D N}=\frac{N D}{N A}\)
⇒ \(\frac{\mathrm{DM}}{\mathrm{DN}}=\frac{\mathrm{DN}}{\mathrm{AN}}\)
DN2 = DM.AN
Hence Proved.

Haryana Board Solutions for 10th Class Maths Chapter 6 Triangles Ex 6.6 width=

Question 3.
In the given figure, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that :
AC2 = AB2 + BC2 + 2BC.BD.

Haryana Board 10th Class Maths Solutions Chapter 6 Triangles Ex 6.6 4

Solution:
See this proof in result (1) of some results related to Pythagoras theorem.

Question 4.
In figure, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that AC2 = AB2 + BC2 – 2BC.BD.

Haryana Board 10th Class Maths Solutions Chapter 6 Triangles Ex 6.6 5

Solution:
See this prove in result (2) of some results related to Pythagoras theorem.

Question 5.
In given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that
(i) AC2 = AD2 + BC.DM + \(\left(\frac{\mathrm{BC}}{2}\right)^2\)
(ii) AB2 = AD2 – BC.DM + \(\left(\frac{\mathrm{BC}}{2}\right)^2\)
(iii) AC2 + AB2 = 2AD2 + \(\frac{1}{2}\) BC2

Haryana Board 10th Class Maths Solutions Chapter 6 Triangles Ex 6.6 6

Solution:
See this proof in result (3) of some results related to Pythagoras theorem.

Haryana Board Solutions for 10th Class Maths Chapter 6 Triangles Ex 6.6 width=

Question 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 with diagonals AC and BD.

Haryana Board 10th Class Maths Solutions Chapter 6 Triangles Ex 6.6 7

To Prove: Sum of the squares of the diagonals of ||gm is equal to sum of the squares of its sides. i.e., AC2 + BD2 = AB2 + BC2 + CD2 + AD2
Construction :
Draw AM ⊥ DC and BN ⊥ DC [Produced].
Proof : In right triangles AMD and BNC. AD = BC [Opposite sides of ]|gm]
AM = BN
[Perpendiculars between two || lines are same]
∠AMD = ∠BNC [Each is 90°]
∆AMD ≅ ∆BNC [By RHS congruence]
MD = NC [CPCT]
Now, in right ∆BND,
BD2 = BN2 + DN2 [By Pythagoras theorem]
⇒ BD2 = BN2 + (DC + CN)2
⇒ BD2 = BN2 + DC2 + CN2 + 2DC.CN
⇒ BD2 = (BN2 + CN2) + DC2 + 2DC.CN
⇒ BD2 = BC2 + DC2 + 2DC.CN …………………….(1)
[∵ In right ∆BNC, BC2 = BN2 + CN2]
In right ∆AMC, AC2 = AM2 + MC2
[By Pythagoras theorem]
AC2 = AM2 + (DC – DM)2
AC2 = AM2 + DC2 + DM2 – 2DC.DM
AC2 = (AM2 + DM2) + DC2 – 2DC.DM
AC2 = AD2 + DC2 – 2DC.DM
[∵ In right ∆AMD AD2 = AM2 + MD2]
AC2 = AD2 + AB2 – 2DC.CN ………………….(2)
[DC = AB and DM = CN]
Adding (1) and (2), we get
BD2 + AC2 = BC2 + DC2 + 2DC.CN + AD2 + AB2 – 2DC.CN
⇒ AC2 + BD2 = AB2 + BC2 + CD2 + AD2
Hence, sum of square of diagonals is equal to sum of square of its sides.
Hence Proved.

Haryana Board Solutions for 10th Class Maths Chapter 6 Triangles Ex 6.6 width=

Question 7.
In given figure, two chords AB and CD intersect each other at the point P. Prove that:
(i) ∆APC ~ ∆DPB
(ii) AP.PB = CP.DP

Haryana Board 10th Class Maths Solutions Chapter 6 Triangles Ex 6.6 8

Solution :
Given : Two chords AB and CD in-tersect each other at point P.
To Prove :
(i) ∆APC ~ ∆DPB
(ii) AP.PB = CP.DP.
Proof :
(i) In ∆APC and ∆DPB,
∠CAB = ∠BDC
[Angles in the same segment are equal]
⇒ ∠CAP = ∠BDP
and ∠APC = ∠BPD [Vertically opposite angles]
∴ ∆APC ~ ∆DPB [By AA similarity criterion]
Hence Proved.

(ii) ∆APC ~ ∆DPB (Proved above)
\(\)
[Corresponding sides of similar triangles are proportional]
⇒ AP.PB = CP.DP.
Hence proved.

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

Haryana Board 10th Class Maths Solutions Chapter 6 Triangles Ex 6.6 9

To Prove :
(i) ∆PAC ~ ∆PDB
(ii) PA.PB = PC.PD
Proof : (i) In ∆PAC and ∆PDB
∠ACP = ∠PBD [Exterior angle of the cyclic quadrilateral is equal to its interior opposite angle] ZP = ZP [Common]
∴ ∆PAC ~ ∆PDB [By AA similarity criterion]
Hence proved.

(ii) ∆PAC ~ ∆PDB (Proved above)
\(\frac{P A}{P D}=\frac{P C}{P B}\)
⇒ PA.PB = PC.PD.
Hence proved.

Haryana Board Solutions for 10th Class Maths Chapter 6 Triangles Ex 6.6 width=

Question 9.
In given figure, D is a point on side BC of ∆ABC such that \(\frac{\mathrm{BD}}{\mathrm{CD}}=\frac{\mathrm{AB}}{\mathrm{AC}}\). Prove that AD is the bisector of ∠BAC.

Haryana Board 10th Class Maths Solutions Chapter 6 Triangles Ex 6.6 10

Solution :
Given : A triangle ABC such that \(\frac{\mathrm{BD}}{\mathrm{CD}}=\frac{\mathrm{AB}}{\mathrm{AC}}\).
To Prove : AD is the bisector of ∠BAC.
Construction : Produce BA to E such that AE = AC and join CE.

Haryana Board 10th Class Maths Solutions Chapter 6 Triangles Ex 6.6 11

Proof: In ∆ABC AE = AC (By Construction)
∴ ∠1 = ∠2 ……………(1)
Now, \(\frac{\mathrm{BD}}{\mathrm{CD}}=\frac{\mathrm{AB}}{\mathrm{AC}}\) (Given)
⇒ \(\frac{\mathrm{BD}}{\mathrm{CD}}=\frac{\mathrm{AB}}{\mathrm{AE}}\) [∵ AC = AE]
∴ In ∆BCE
⇒ \(\frac{\mathrm{BD}}{\mathrm{CD}}=\frac{\mathrm{AB}}{\mathrm{AE}}\)
⇒ AD || CE [By the converse of BPT]
∠1 = ∠4 [Alternate interior angles] ……………(2)
∠3 = ∠2 [Corresponding angles] …………….(3)
From (1), (2) and (3), we get
∠3 = ∠4
Hence, AD is the bisector of ∠BAC.
Hence proved.

Haryana Board Solutions for 10th Class Maths Chapter 6 Triangles Ex 6.6 width=

Question 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?

Haryana Board 10th Class Maths Solutions Chapter 6 Triangles Ex 6.6 12

Solution:
Let AB be the distance from a point directly under the tip of the rod and BC be the tip of her fishing rod and AC be the length of string.
Now, in right ∆ABC,

Haryana Board 10th Class Maths Solutions Chapter 6 Triangles Ex 6.6 13

AC2 = AB2 + BC2 [By Pythagoras theorem]
AC2 = (2.4)2 + (1.8)2
AC2 = 5.76 + 3.24
AC2 = 9
AC = 3 m.
The length of string pulls in 12 seconds = 12 × 5 = 60 cm = 0.6 m.
Length of remaining string = 3 – 0.6 = 2.4 m.
In 2nd position,
String length (CD) = 2.4 m.

Haryana Board 10th Class Maths Solutions Chapter 6 Triangles Ex 6.6 14

Tip of her fishing rod (BC) = 1.8 m
Now In right ∆CBD,
CD2 = BD2 + BC2
⇒ 2.42 = BD2 + (1.8)2
⇒ 5.76 = BD2 + 3.24
⇒ BD2 = 5.76 – 3.24
⇒ BD2 = 2.52
⇒ BD = 1.59 (Approx)
So, horizontal distance of the fly from Nazima after 12 seconds = 1.59 + 1.2 = 2.79m. (Approx.)
Hence, Length of string = 3.0 m and
Horizontal distance of fly from Nazima = 2.79 m (Approx.)

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