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

## 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.

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

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.

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

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.

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)

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)

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

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

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

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)

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

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)

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

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

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

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

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

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

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

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.

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

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

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

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

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

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

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

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

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)

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)

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

$\Rightarrow AP.PB=PC.DP$

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)

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)

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

$\Rightarrow AP.PB=PC.DP$

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

$\Rightarrow \angle BAD=\angle APC$

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:

## Subject Wise NCERT Exemplar Solutions

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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