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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
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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 QPRofΔPQR . Prove that QSSR=PQPR

1635933619787

Answer:

1635933634308

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

Given: PS is the bisector of QPRofΔPQR .

QPS=SPR.....................................1

By construction,

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

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

From the above equations, we get

PRT=QTR

PT=PR

By construction, PS||TR

In QTR, by Thales theorem,

QSSR=QPPT

QSSR=PQPR

Hence proved.

Q2(1) In Fig. 6.57, D is a point on hypotenuse AC of triangle ABC, such that BD AC, DM BC and DN AB. Prove that : DM2=DN.MC

1635933650488

Answer:

1635933658497

Join BD

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

CDB=90

2+3=90.............................1

In CDM, 1+2+DMC=180

1+2=90.......................2

In DMB, 3+4+DMB=180

3+4=90.......................3

From equation 1 and 2, we get 1=3

From equation 1 and 3, we get 2=4

In DCMandBDM,

1=3

2=4

DCMBDM, (By AA)

BMDM=DMMC

DNDM=DMMC (BM=DN)

DM2=DN.MC

Hence proved

Q2 (2) In Fig. 6.57, D is a point on hypotenuse AC of D ABC, such that BD AC, DM BC and DN AB. Prove that: DN2=DM.AN

1635933672815

Answer:

1635933684708

In DBN,

5+7=90.......................1

In DAN,

6+8=90.......................2

BD AC, ADB=90

5+6=90.......................3

From equation 1 and 3, we get 6=7

From equation 2 and 3, we get 5=8

In DNAandBND,

6=7

5=8

DNABND (By AA)

ANDN=DNNB

ANDN=DNDM (NB=DM)

DN2=AN.DM

Hence proved.

Q3 In Fig. 6.58, ABC is a triangle in which ABC > 90° and AD CB produced. Prove that AC2=AB2+BC2+2BC.BD.

1635933712397

Answer:

In ADB, by Pythagoras theorem

AB2=AD2+DB2.......................1

In ACD, by Pythagoras theorem

AC2=AD2+DC2.......................2

AC2=AD2+(BD+BC)2

AC2=AD2+(BD)2+(BC)2+2.BD.BC

AC2=AB2+BC2+2BC.BD. (From 1)

Q4 In Fig. 6.59, ABC is a triangle in which ABC < 90° and AD BC. Prove that AC2=AB2+BC22BC.BD.

1635933729274

Answer:

In ADB, by Pythagoras theorem

AB2=AD2+DB2

AD2=AB2DB2...........................1

In ACD, by Pythagoras theorem

AC2=AD2+DC2

AC2=AB2BD2+DC2 (From 1)

AC2=AB2BD2+(BCBD)2

AC2=AB2BD2+(BC)2+(BD)22.BD.BC

AC2=AB2+BC22BC.BD.

Q5 (1) In Fig. 6.60, AD is a median of a triangle ABC and AM BC. Prove that : AC2=AD2+BCDM+(BC2)2

1635933758550

Answer:

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

In AMD, by Pythagoras theorem

AD2=AM2+MD2.......................1

In AMC, by Pythagoras theorem

AC2=AM2+MC2

AC2=AM2+(MD+DC)2

AC2=AM2+(MD)2+(DC)2+2.MD.DC

AC2=AD2+DC2+2DC.MD. (From 1)

AC2=AD2+(BC2)2+2(BC2).MD. (BC=2 DC)

AC2=AD2+BCDM+(BC2)2

Q5 (2) In Fig. 6.60, AD is a median of a triangle ABC and AM BC. Prove that : AB2=AD2BC.DM+(BC2)2

1635933791582

Answer:

In ABM, by Pythagoras theorem

AB2=AM2+MB2

AB2=(AD2DM2)+MB2

AB2=(AD2DM2)+(BDMD)2

AB2=AD2DM2+(BD)2+(MD)22.BD.MD

AB2=AD2+(BD)22.BD.MD

AB2=AD2+(BC2)22(BC2).MD.=AC2 (BC=2 BD)

AD2+(BC2)2BC.MD.=AC2

Q5 (3) In Fig. 6.60, AD is a median of a triangle ABC and AM BC. Prove that: AC2+AB2=2AD2+12BC2

1635933807753

Answer:

In ABM, by Pythagoras theorem

AB2=AM2+MB2.......................1

In AMC, by Pythagoras theorem

AC2=AM2+MC2 ..................................2

Adding equation 1 and 2,

AB2+AC2=2AM2+MB2+MC2

AB2+AC2=2AM2+(BDDM)2+(MD+DC)2

AB2+AC2=2AM2+(BD)2+(DM)22.BD.DM+(MD)2+(DC)2+2.MD.DC

AB2+AC2=2AM2+2.(DM)2+BD2+(DC)2+2.MD.(DCBD) AB2+AC2=2(AM2+(DM)2)+(BC2)2+(BC2)2+2.MD.(BC2BC2) AB2+AC2=2(AM2+(DM)2)+(BC2)2+(BC2)2

AC2+AB2=2AD2+12BC2

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 DEA, by Pythagoras theorem

DA2=DE2+EA2.......................1

In DEB, by Pythagoras theorem

DB2=DE2+EB2

DB2=DE2+(EA+AB)2

DB2=DE2+(EA)2+(AB)2+2.EA.AB

DB2=DA2+(AB)2+2.EA.AB ....................................2

In ADF, by Pythagoras theorem

DA2=AF2+FD2

In AFC, by Pythagoras theorem

AC2=AF2+FC2=AF2+(DCFD)2

AC2=AF2+(DC)2+(FD)22.DC.FD

AC2=(AF2+FD2)+(DC)22.DC.FD

AC2=AD2+(DC)22.DC.FD.......................3

Since ABCD is a parallelogram.

SO, AB=CD and BC=AD

In DEAandADF,

DEA=AFD(each90)

DAE=ADF (AE||DF)

AD=AD (common)

DEAADF, (ASA rule)

EA=DF.......................6

Adding 2 and, we get

DA2+AB2+2.EA.AB+AD2+DC22.DC.FD=DB2+AC2 DA2+AB2+AD2+DC2+2.EA.AB2.DC.FD=DB2+AC2

BC2+AB2+AD2+2.EA.AB2.AB.EA=DB2+AC2 (From 4 and 6)

BC2+AB2+CD2=DB2+AC2 \

Q7 (1) In Fig. 6.61, two chords AB and CD intersect each other at point P. Prove that : ΔAPCΔDPB

1635933839234

Answer:

1635933848323

Join BC

In APCandDPB,

APC=DPB ( vertically opposite angle)

CAP=BDP (Angles in the same segment)

APCDPB (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 APCandDPB,

APC=DPB ( vertically opposite angle)

CAP=BDP (Angles in the same segment)

APCDPB (By AA)

APDP=PCPB=CABD (Corresponding sides of similar triangles are proportional)

APDP=PCPB

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 ΔPACΔPDB

1635933941642

Answer:

In ΔPACandΔPDB,

P=P (Common)

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

So, ΔPACΔPDB ( 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 ΔPACandΔPDB,

P=P (Common)

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

So, ΔPACΔPDB ( By AA rule)

24440 APDP=PCPB=CABD (Corresponding sides of similar triangles are proportional)

APDP=PCPB

AP.PB=PC.DP

Q9 In Fig. 6.63, D is a point on side BC of D ABC such that BDCD=ABAC Prove that AD is the bisector of BAC.

1635933987426

Answer:

1635933998131

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

BDCD=ABAC (Given )

BDCD=APAC

Using converse of Thales theorem,

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

DAC=ACP............2 (Alternate angles)

By construction,

AP=AC

APC=ACP............3

From equation 1,2,3, we get

BAD=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 ABC, using Pythagoras theorem

AC2=AB2+BC2

AC2=(1.8)2+(2.4)2

AC2=3.24+5.76

AC2=9.00

AC=3m

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.

= 12×5=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 ADB,

AB2+BD2=AD2

(1.8)2+BD2=(2.4)2

BD2=5.763.24=2.52m2

BD=1.587m

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
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Also, See:

JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

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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Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Hello

Since you are a domicile of Karnataka and have studied under the Karnataka State Board for 11th and 12th , you are eligible for Karnataka State Quota for admission to various colleges in the state.

1. KCET (Karnataka Common Entrance Test): You must appear for the KCET exam, which is required for admission to undergraduate professional courses like engineering, medical, and other streams. Your exam score and rank will determine your eligibility for counseling.

2. Minority Income under 5 Lakh : If you are from a minority community and your family's income is below 5 lakh, you may be eligible for fee concessions or other benefits depending on the specific institution. Some colleges offer reservations or other advantages for students in this category.

3. Counseling and Seat Allocation:

After the KCET exam, you will need to participate in online counseling.

You need to select your preferred colleges and courses.

Seat allocation will be based on your rank , the availability of seats in your chosen colleges and your preferences.

4. Required Documents :

Domicile Certificate (proof that you are a resident of Karnataka).

Income Certificate (for minority category benefits).

Marksheets (11th and 12th from the Karnataka State Board).

KCET Admit Card and Scorecard.

This process will allow you to secure a seat based on your KCET performance and your category .

check link for more details

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Hope this helps you .

Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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