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NCERT Solutions for Exercise 6.6 Class 10 Maths Chapter 6 - Triangles

NCERT Solutions for Exercise 6.6 Class 10 Maths Chapter 6 - Triangles

Edited By Ramraj Saini | Updated on Nov 25, 2023 08:29 PM IST | #CBSE Class 10th
Ongoing Event
CBSE Class 10th  Exam Date : 01 Jan' 2025 - 14 Feb' 2025

NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.6

NCERT Solutions for Exercise 6.6 Class 10 Maths Chapter 6 Triangles are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. Class 10 maths ex 6.6 help us to understand how to apply all the concepts till now we have learned through all the exercises in the chapter. The notion of triangle similarity, criteria for triangle similarity, areas of similar triangles, and Pythagoras Theorem are covered in this exercise.

This Story also Contains
  1. NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.6
  2. More About NCERT Solutions for Class 10 Maths Exercise 6.6
  3. Benefits of NCERT Solutions for Class 10 Maths Exercise 6.6
  4. NCERT Solutions of Class 10 Subject Wise
  5. Subject Wise NCERT Exemplar Solutions

10th class Maths exercise 6.6 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Triangles Class 10 Chapter 6 Exercise: 6.6

Q1 In Fig. 6.56, PS is the bisector of \angle QPR \: \: of\: \: \Delta PQR . Prove that \frac{QS }{SR } = \frac{PQ }{PR }

1635933619787

Answer:

1635933634308

A line RT is drawn parallel to SP which intersect QP produced at T.

Given: PS is the bisector of \angle QPR \: \: of\: \: \Delta PQR .

\angle QPS=\angle SPR.....................................1

By construction,

\angle SPR=\angle PRT.....................................2 (as PS||TR)

\angle QPS=\angle QTR.....................................3 (as PS||TR)

From the above equations, we get

\angle PRT=\angle QTR

\therefore PT=PR

By construction, PS||TR

In \triangle QTR, by Thales theorem,

\frac{QS}{SR}=\frac{QP}{PT}

\frac{QS }{SR } = \frac{PQ }{PR }

Hence proved.

Q2(1) In Fig. 6.57, D is a point on hypotenuse AC of triangle ABC, such that BD \perp AC, DM \perp BC and DN \perp AB. Prove that : DM^2 = DN . MC

1635933650488

Answer:

1635933658497

Join BD

Given : D is a point on hypotenuse AC of D ABC, such that BD \perp AC, DM \perp BC and DN \perp AB.Also DN || BC, DM||NB

\angle CDB=90 \degree

\Rightarrow \angle 2+\angle 3=90 \degree.............................1

In \triangle CDM, \angle 1+\angle 2+\angle DMC=180 \degree

\angle 1+\angle 2=90 \degree.......................2

In \triangle DMB, \angle 3+\angle 4+\angle DMB=180 \degree

\angle 3+\angle 4=90 \degree.......................3

From equation 1 and 2, we get \angle 1=\angle 3

From equation 1 and 3, we get \angle 2=\angle 4

In \triangle DCM\, \, and\, \, \triangle BDM,

\angle 1=\angle 3

\angle 2=\angle 4

\triangle DCM\, \, \sim \, \, \triangle BDM, (By AA)

\Rightarrow \frac{BM}{DM}=\frac{DM}{MC}

\Rightarrow \frac{DN}{DM}=\frac{DM}{MC} (BM=DN)

\Rightarrow DM^2 = DN . MC

Hence proved

Q2 (2) In Fig. 6.57, D is a point on hypotenuse AC of D ABC, such that BD \perp AC, DM \perp BC and DN \perp AB. Prove that: DN^2 = DM . AN

1635933672815

Answer:

1635933684708

In \triangle DBN,

\angle 5+\angle 7=90 \degree.......................1

In \triangle DAN,

\angle 6+\angle 8=90 \degree.......................2

BD \perp AC, \therefore \angle ADB=90 \degree

\angle 5+\angle 6=90 \degree.......................3

From equation 1 and 3, we get \angle 6=\angle 7

From equation 2 and 3, we get \angle 5=\angle 8

In \triangle DNA\, \, and\, \, \triangle BND,

\angle 6=\angle 7

\angle 5=\angle 8

\triangle DNA\, \, \sim \, \, \triangle BND (By AA)

\Rightarrow \frac{AN}{DN}=\frac{DN}{NB}

\Rightarrow \frac{AN}{DN}=\frac{DN}{DM} (NB=DM)

\Rightarrow DN^2 = AN . DM

Hence proved.

Q3 In Fig. 6.58, ABC is a triangle in which \angle ABC > 90° and AD \perp CB produced. Prove that AC^2 = AB^2 + BC^2 + 2 BC . BD.

1635933712397

Answer:

In \triangle ADB, by Pythagoras theorem

AB^2=AD^2+DB^2.......................1

In \triangle ACD, by Pythagoras theorem

AC^2=AD^2+DC^2.......................2

AC^2=AD^2+(BD+BC)^2

\Rightarrow AC^2=AD^2+(BD)^2+(BC)^2+2.BD.BC

AC^2 = AB^2 + BC^2 + 2 BC . BD. (From 1)

Q4 In Fig. 6.59, ABC is a triangle in which \angle ABC < 90° and AD \perp BC. Prove that AC^2 = AB^2 + BC^2 - 2 BC . BD.

1635933729274

Answer:

In \triangle ADB, by Pythagoras theorem

AB^2=AD^2+DB^2

AD^2=AB^2-DB^2...........................1

In \triangle ACD, by Pythagoras theorem

AC^2=AD^2+DC^2

AC^2=AB^2-BD^2+DC^2 (From 1)

\Rightarrow AC^2=AB^2-BD^2+(BC-BD)^2

\Rightarrow AC^2=AB^2-BD^2+(BC)^2+(BD)^2-2.BD.BC

AC^2 = AB^2 + BC^2 - 2 BC . BD.

Q5 (1) In Fig. 6.60, AD is a median of a triangle ABC and AM \perp BC. Prove that : AC ^2 = AD ^2 + BC DM + \left ( \frac{BC}{2} \right ) ^2

1635933758550

Answer:

Given: AD is a median of a triangle ABC and AM \perp BC.

In \triangle AMD, by Pythagoras theorem

AD^2=AM^2+MD^2.......................1

In \triangle AMC, by Pythagoras theorem

AC^2=AM^2+MC^2

AC^2=AM^2+(MD+DC)^2

\Rightarrow AC^2=AM^2+(MD)^2+(DC)^2+2.MD.DC

AC^2 = AD^2 + DC^2 + 2 DC . MD. (From 1)

AC^2 = AD^2 + (\frac{BC}{2})^2 + 2(\frac{BC}{2}). MD. (BC=2 DC)

AC ^2 = AD ^2 + BC DM + \left ( \frac{BC}{2} \right ) ^2

Q5 (2) In Fig. 6.60, AD is a median of a triangle ABC and AM \perp BC. Prove that : AB ^2 = AD ^2 - BC .DM + \left ( \frac{BC}{2} \right ) ^2

1635933791582

Answer:

In \triangle ABM, by Pythagoras theorem

AB^2=AM^2+MB^2

AB^2=(AD^2-DM^2)+MB^2

\Rightarrow AB^2=(AD^2-DM^2)+(BD-MD)^2

\Rightarrow AB^2=AD^2-DM^2+(BD)^2+(MD)^2-2.BD.MD

\Rightarrow AB^2=AD^2+(BD)^2-2.BD.MD

\Rightarrow AB^2 = AD^2 + (\frac{BC}{2})^2 -2(\frac{BC}{2}). MD.=AC^2 (BC=2 BD)

\Rightarrow AD^2 + (\frac{BC}{2})^2 -BC. MD.=AC^2

Q5 (3) In Fig. 6.60, AD is a median of a triangle ABC and AM \perp BC. Prove that: AC ^2 + AB ^2 = 2 AD^2 + \frac{1}{2} BC ^2

1635933807753

Answer:

In \triangle ABM, by Pythagoras theorem

AB^2=AM^2+MB^2.......................1

In \triangle AMC, by Pythagoras theorem

AC^2=AM^2+MC^2 ..................................2

Adding equation 1 and 2,

AB^2+AC^2=2AM^2+MB^2+MC^2

\Rightarrow AB^2+AC^2=2AM^2+(BD-DM)^2+(MD+DC)^2

\Rightarrow AB^2+AC^2=2AM^2+(BD)^2+(DM)^2-2.BD.DM+(MD)^2+(DC)^2+2.MD.DC

\Rightarrow AB^2+AC^2=2AM^2+2.(DM)^2+BD^2+(DC)^2+2.MD.(DC-BD) \Rightarrow AB^2+AC^2=2(AM^2+(DM)^2)+(\frac{BC}{2})^2+(\frac{BC}{2})^2+2.MD.(\frac{BC}{2}-\frac{BC}{2}) \Rightarrow AB^2+AC^2=2(AM^2+(DM)^2)+(\frac{BC}{2})^2+(\frac{BC}{2})^2

AC ^2 + AB ^2 = 2 AD^2 + \frac{1}{2} BC ^2

Q6 Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Answer:

1635933824435

In parallelogram ABCD, AF and DE are altitudes drawn on DC and produced BA.

In \triangle DEA, by Pythagoras theorem

DA^2=DE^2+EA^2.......................1

In \triangle DEB, by Pythagoras theorem

DB^2=DE^2+EB^2

DB^2=DE^2+(EA+AB)^2

DB^2=DE^2+(EA)^2+(AB)^2+2.EA.AB

DB^2=DA^2+(AB)^2+2.EA.AB ....................................2

In \triangle ADF, by Pythagoras theorem

DA^2=AF^2+FD^2

In \triangle AFC, by Pythagoras theorem

AC^2=AF^2+FC^2=AF^2+(DC-FD)^2

\Rightarrow AC^2=AF^2+(DC)^2+(FD)^2-2.DC.FD

\Rightarrow AC^2=(AF^2+FD^2)+(DC)^2-2.DC.FD

\Rightarrow AC^2=AD^2+(DC)^2-2.DC.FD.......................3

Since ABCD is a parallelogram.

SO, AB=CD and BC=AD

In \triangle DEA\, and\, \triangle ADF,

\angle DEA=\angle AFD\, \, \, \, \, \, \, (each 90 \degree)

\angle DAE=\angle ADF (AE||DF)

AD=AD (common)

\triangle DEA\, \cong \, \triangle ADF, (ASA rule)

\Rightarrow EA=DF.......................6

Adding 2 and, we get

DA^2+AB^2+2.EA.AB+AD^2+DC^2-2.DC.FD=DB^2+AC^2 \Rightarrow DA^2+AB^2+AD^2+DC^2+2.EA.AB-2.DC.FD=DB^2+AC^2

\Rightarrow BC^2+AB^2+AD^2+2.EA.AB-2.AB.EA=DB^2+AC^2 (From 4 and 6)

\Rightarrow BC^2+AB^2+CD^2=DB^2+AC^2 \

Q7 (1) In Fig. 6.61, two chords AB and CD intersect each other at point P. Prove that : \Delta APC \sim \Delta DPB

1635933839234

Answer:

1635933848323

Join BC

In \triangle APC\, \, and\, \triangle DPB,

\angle APC\, \, = \angle DPB ( vertically opposite angle)

\angle CAP\, \, = \angle BDP (Angles in the same segment)

\triangle APC\, \, \sim \triangle DPB (By AA)

Q7 (2) In Fig. 6.61, two chords AB and CD intersect each other at point P. Prove that : AP . PB = CP . DP

1635933860992

Answer:

1635933929931

Join BC

In \triangle APC\, \, and\, \triangle DPB,

\angle APC\, \, = \angle DPB ( vertically opposite angle)

\angle CAP\, \, = \angle BDP (Angles in the same segment)

\triangle APC\, \, \sim \triangle DPB (By AA)

\frac{AP}{DP}=\frac{PC}{PB}=\frac{CA}{BD} (Corresponding sides of similar triangles are proportional)

\Rightarrow \frac{AP}{DP}=\frac{PC}{PB}

\Rightarrow AP.PB=PC.DP

Q8 (1) In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that \Delta PAC \sim \Delta PDB

1635933941642

Answer:

In \Delta PAC \,and \,\Delta PDB,

\angle P=\angle P (Common)

\angle PAC=\angle PDB (Exterior angle of a cyclic quadrilateral is equal to opposite interior angle)

So, \Delta PAC \sim \Delta PDB ( By AA rule)

Q8 (2) In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that PA. PB = PC. PD

1635933976947

Answer:

In \Delta PAC \,and \,\Delta PDB,

\angle P=\angle P (Common)

\angle PAC=\angle PDB (Exterior angle of a cyclic quadrilateral is equal to opposite interior angle)

So, \Delta PAC \sim \Delta PDB ( By AA rule)

24440 \frac{AP}{DP}=\frac{PC}{PB}=\frac{CA}{BD} (Corresponding sides of similar triangles are proportional)

\Rightarrow \frac{AP}{DP}=\frac{PC}{PB}

\Rightarrow AP.PB=PC.DP

Q9 In Fig. 6.63, D is a point on side BC of D ABC such that \frac{BD }{CD} = \frac{AB}{AC} Prove that AD is the bisector of \angle BAC.

1635933987426

Answer:

1635933998131

Produce BA to P, such that AP=AC and join P to C.

\frac{BD }{CD} = \frac{AB}{AC} (Given )

\Rightarrow \frac{BD }{CD} = \frac{AP}{AC}

Using converse of Thales theorem,

AD||PC \Rightarrow \angle BAD=\angle APC............1 (Corresponding angles)

\Rightarrow \angle DAC=\angle ACP............2 (Alternate angles)

By construction,

AP=AC

\Rightarrow \angle APC=\angle ACP............3

From equation 1,2,3, we get

\Rightarrow \angle BAD=\angle APC

Thus, AD bisects angle BAC.

Q10 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. 6.64)? 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?

1635934015475

Answer:

1635934024120

Let AB = 1.8 m

BC is a horizontal distance between fly to the tip of the rod.

Then, the length of the string is AC.

In \triangle ABC, using Pythagoras theorem

AC^2=AB^2+BC^2

\Rightarrow AC^2=(1.8)^2+(2.4)^2

\Rightarrow AC^2=3.24+5.76

\Rightarrow AC^2=9.00

\Rightarrow AC=3 m

Hence, the length of the string which is out is 3m.

If she pulls in the string at the rate of 5cm/s, then the distance travelled by fly in 12 seconds.

= 12\times 5=60cm=0.6m

Let D be the position of fly after 12 seconds.

Hence, AD is the length of the string that is out after 12 seconds.

Length of string pulled in by nazim=AD=AC-12

=3-0.6=2.4 m

In \triangle ADB,

AB^2+BD^2=AD^2

\Rightarrow (1.8)^2+BD^2=(2.4)^2

\Rightarrow BD^2=5.76-3.24=2.52 m^2

\Rightarrow BD=1.587 m

Horizontal distance travelled by fly = BD+1.2 m

=1.587+1.2=2.787 m

= 2.79 m

More About NCERT Solutions for Class 10 Maths Exercise 6.6

Class 10 Maths chapter 6 exercise 6: The questions in exercise 6.6 Class 10 Maths consist of many types of questions covering different types of theorems and formulas. Firstly we have a question in which we have to prove the left-hand side argument and right-hand side argument NCERT solutions for Class 10 Maths exercise 6.6 also have questions. Exercise 6.6 Class 10 Maths covers all types of questions that can be formed on the similarity of triangles. Students can also access Triangles Class 10 Notes here and use them for quickly revision of the concepts related to Triangles.

Benefits of NCERT Solutions for Class 10 Maths Exercise 6.6

  • Class 10 Maths chapter 6 exercise 6.6 broadly covers all kinds of questions that can be formed on the mixed concept of the triangle and gives a great amount of practice to conquer some amount of expertise in the triangle.
  • NCERT Class 10 Maths chapter 6 exercise 6.6, will be helpful in JEE Main (joint entrance exam) as triangles are a major part of some chapters.
  • Exercise 6.6 Class 10 Maths, is based on all the theorem, exercises and topics which have been covered in the whole chapter

Also, See:

NCERT Solutions of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

1. What are the basic concepts covered till NCERT solutions Class 10 Maths chapter 6 exercise 6.6?

  • similarity of triangle

  • Criteria for Similarity of Triangles 

  • Areas of Similar Triangles

  • Pythagoras Theorem

2. What is the triangle's similarity?

If the respective angles are congruent and the corresponding sides are proportional, two triangles are said to be similar.

3. How are NCERT solutions for Class 10 Maths exercise 6.6 is different from other exercises ?

NCERT solutions for Class 10 Maths  exercise 6.6 is different from other exercises as all other exercises are based on a single concept and theorem they are base building exercises but Class 10 Maths chapter 6 exercise 6.6 is based on all the concepts.

4. Is NCERT solutions for Class 10 Maths exercise 6.6 difficult from other exercises?

Yes  Class 10 Maths chapter 6 exercise 6.6  because it have questions which have mix concepts 

5. How many theorems are there which we require to solve NCERT solutions for Class 10 Maths 1 exercise 6.6 ?

There are 13 main  theorems are there which we require to solve  NCERT solutions for Class 10 Maths 1 exercise 6.6

6. How many questions are there in the Exercise 6.6 Class 10 Maths ?

There are ten  questions in exercise 6.6 Class 10 Maths  question 

7. How many types of questions are there in the exercise 6.6 Class 10 Maths and explain each type?

There are two  types of questions in exercise 6.6 Class 10 Maths  question one type is there in which we have to proof LHS and RHS these contains uses of theorem and other is real life application of triangle    

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CBSE Date Sheet 2025 Live: CBSE Class 10, 12 board exam timetable soon at cbse.gov.in; practical dates
November 14, 2024 - 03:21 PM IST
CBSE Date Sheet 2025 Live: CBSE to announce Class 10, 12 board exam dates soon at cbse.gov.in; sample paper
November 13, 2024 - 07:53 AM IST
CBSE Board Exam 2025: Class 10, 12 practical exam dates out at cbse.gov.in; passing marks
October 24, 2024 - 09:54 PM IST
CBSE urges schools to complete 100% registration of Class 9, 11 students by October 16; late fee for delays
October 11, 2024 - 08:56 AM IST
CBSE sample papers 2024-25 for Class 10, 12 out at cbseacademic.nic.in; marking scheme
September 05, 2024 - 07:38 PM IST

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