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NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6

December 26, 2018 by Dattu

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 6 Triangles Class 10 NCERT Solutions Ex 6.6.

  • Triangles Class 10 Ex 6.1
  • Triangles Class 10 Ex 6.2
  • Triangles Class 10 Ex 6.3
  • Triangles Class 10 Ex 6.4
  • Triangles Class 10 Ex 6.5
BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 6
Chapter NameTriangles
ExerciseEx 6.6
Number of Questions Solved10
CategoryNCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6

NCERT Solutions for Class 10 Maths

Question 1. In the given figure, PS is the bisector of ∠QPR of ∆PQR. Prove that
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 107

Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 108
Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T.
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 109
Question 2. In the given figure, D is a point on hypotenuse AC of  ∆ABC, DM ⊥ BC and DN ⊥ AB, Prove that:

(i). DM2 = DN.MC
(ii). DN2 = DM.AN
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 110

Solution:
(i).  Let us join DB.
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 111
DN || CB
DM || AB
So, DN = MB
DM = NB
The condition to be proved is the case when  DNBM is a square or D is the  midpoint of side AC.
Then ∠CDB = ∠ADB = 90°
∠2 + ∠3 = 90°        (1)
In ∆CDM
∠ 1 + ∠ 2 + ∠ DMC = 180°
∠ 1 + ∠ 2 = 90°        (2)
In ∆DMB
∠ 3 + ∠ DMB + ∠ 4 = 180°
∠ 3 + ∠ 4 = 90°        (3)
From equation (1) and (2)
∠ 1 = ∠ 3
From equation (1) and (3)
∠ 2 = ∠ 4
∆BDM ~ ∆ DCM
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 112
(ii).  Similarly in ∆DBN
∠ 4 + ∠ 3 = 90°            (4)
In ∆DAN
∠ 5 + ∠ 6 = 90°            (5)
In ∆DAB
∠4 + ∠ 5 = 90°            (6)
From equation (4) and (6)
∠3 = ∠ 5
From equation (5) and (6)
∠4 = ∠ 6
∆DNA   ~ ∆BND
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 113

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.
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 114

Solution:
In ∆ADB applying Pythagoras theorem
AB2 = AD2 + DB2        (1)
In ∆ACD applying Pythagoras theorem
AC2 = AD2 + DC2
AC2 = AD2 + (DB + BC)2
AC2 = AD2 + DB2 + BC2 + 2DB  x  BC
Now using equation (1)
AC2 = AB2 + BC2 + 2BC . BD

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

NCERT Solutions for Class 10 Maths Chapter 6 Triangles 115

Solution:
In ∆ADB applying Pythagoras theorem
AD2 + DB2 = AB2
AD2 = AB2 – DB2                (1)
In ∆ ADC applying Pythagoras theorem
AD2 + DC2 = AC2                (2)
Now using equation (1)
AB2 –  BD2 + DC2 = AC2
AB2 –  BD2 + (BC –  BD)2 = AC2
AC2 = AB2 –  BD2 + BC2 + BD2 – 2BC. BD
= AB2 + BC2 – 2BC. BD

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

NCERT Solutions for Class 10 Maths Chapter 6 Triangles 116
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 117

Solution:
(i).  In ∆AMD
AM2 + MD2 = AD2            (1)
In ∆AMC
AM2 + MC2 = AC2            (2)
AM2 + (MD + DC)2 = AC2
(AM2 + MD2) + DC2 + 2MD.DC = AC2
Using equation (1) we may get
AD2 + DC2 + 2MD.DC = AC2
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 118
(ii).  In ∆ABM applying Pythagoras theorem
AB2 = AM2 + MB2
= (AD2 – DM2) + MB2
= (AD2 –  DM2) + (BD –  MD)2
= AD2 –  DM2 + BD2 + MD2 – 2BD.MD
= AD2 + BD2 – 2BD.MD
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 119(iii).  In ∆AMB
AM2 + MB2 = AB2        (1)
In ∆ AMC
AM2 + MC2 = AC2        (2)
Adding equation (1) and (2)
2AM2 + MB2 + MC2 = AB2 + AC2
2AM2 + (BD – DM)2 + (MD + DC)2 = AB2 + AC2
2AM2+BD2 + DM2 – 2BD.DM + MD2 + DC2 + 2MD.DC = AB2 + AC2
2AM2 + 2MD2 + BD2 + DC2 + 2MD (-BD + DC) = AB2 + AC2
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 120

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:

NCERT Solutions for Class 10 Maths Chapter 6 Triangles 121
Let ABCD be a parallelogram
Let us draw perpendicular DE on extended side AB and AF on side DC.
In ∆DEA
DE2 + EA2 = DA2                      (i)
In ∆DEB
DE2 + EB2 = DB2
DE2 + (EA + AB)2 = DB2
(DE2 + EA2) + AB2 + 2EA. AB = DB2
DA2 + AB2 + 2EA.AB = DB2        (ii)
In ∆ADF
AD2 = AF2 + FD2
In ∆ AFC
AC2 = AF2 + FC2
= AF2 + (DC – FD)2
= AF2 + DC2 + FD2 – 2DC – FD
= (AF2 + FD2) + DC2 – 2DC . FD
AC2 = AD2 + DC2 – 2DC  FD        (iii)
Since ABCD is a parallelogram
AB = CD                 (iii)
And BC = AD           (iv)
In ∆DEA and ∆ADF
∠ DEA = ∠ AFD
∠ EAD = ∠ FDA        (EA || DF)
∠ EDA = ∠ FAD        (AF || ED)
AD is common in both triangles.
Since respective angles are same and respective sides are same
∆DEA ≅ ∆AFD
So EA = DF            (v)
Adding equation (ii) and (iii)
DA2 + AB2 + 2EA.AB + AD2 + DC2 – 2DC.FD = DB2 + AC2
DA2 + AB2 + AD2 + DC2 + 2EA.AB – 2DC.FD = DB2 + AC2
BC2 + AB2 + AD2 + DC2 + 2EA.AB-2AB.EA = DB2 + AC2
AB2 + BC2 + CD2 + DA2 = AC2 + BD2

Question 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
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 122

Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 123
Let us join CB
(i)    In ∆APC and ∆DPB
∠APC = ∠DPB    {Vertically opposite angles}
∠CAP = ∠BDP    {Angles in same segment for chord CB}
∆APC ~ ∆DPB    {BY AA similarly criterion}
(ii)    We know that corresponding sides of similar triangles are proportional
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 124
Question 8. In the given 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
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 125
Solution:
(i)    In ∆PAC and ∆PDB
∠ P = ∠ P        (common)
∠ PAC = ∠ PDB    (exterior angle of a cyclic quadrilateral is equal to opposite interior angle)
∠ PCA = ∠PBD
∆PAC ~ ∆PDB

(ii)    We know that corresponding sides of similar triangles are proportional.
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 126

Question 9. In the given figure, D is a point on side BC of ∆ ABC such that \(\frac { BD }{ CD } =\frac { AB }{ AC }\) Prove that AD is the bisector of ∠ BAC.
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 127

Solution:
In ∆ DBA and ∆DCA
\(\frac { BD }{ CD } =\frac { AB }{ AC }\)      (given)
AD = AD        (common)
So, ∆DBA ~ ∆DCA    (By SSS)
Now, corresponding angles of similar triangles will be equal.
∠ BAD = ∠ CAD
AD is angle bisector of ∠ BAC.
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 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?

NCERT Solutions for Class 10 Maths Chapter 6 Triangles 128
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 129
Let AB be the height of tip of fishing rod from water surface. Let BC be the horizontal distance of fly from the tip of fishing rod.
Then, AC is the length of string.
AC can be found by applying Pythagoras theorem in ∆ABC
AC2 = AB2 + BC2
AC2 = (1.8)2 + (2.4)2
AC2 = 3.24 + 5.76
AC2 = 9.00
AC = √9 = 3
Thus, length of string out is 3 m.
Now, she pulls string at rate of 5 cm per second.
So, string pulled in 12 seconds = 12 x 5 = 60 cm = 0.6 m
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 130
Let after 12 second Fly be at point D.
Length of string out after 12 second is AD
AD = AC – string pulled by Nazima in 12 seconds
= 3.00 – 0.6
= 2.4
In ∆ ADB
AB2 + BD2 = AD2
(1.8)2 + BD2 = (2.4)2
BD2 = 5.76 – 3.24 = 2.52
BD = 1.587
Horizontal distance of fly = BD + 1.2
= 1.587 + 1.2
= 2.787
= 2.79 m.

We hope the NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6, drop a comment below and we will get back to you at the earliest.

Filed Under: CBSE Tagged With: NCERT Solutions for Class 10 Maths

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