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

Exercise 6.6

1. In the given figure, PS is the bisector of $\angle QPR$of $\Delta PQR$. Prove that

$\frac{QS}{SR}=\frac{PQ}{RQ}$

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Ans: 

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Let us have a construction, draw TR parallel to PS intersecting extended PQ

It is given that $\angle QPS=\angle SPR$

$\angle SPR=\angle PRT$

$\angle QPS=\angle QTR$

Hence we have $PT=PR$……($1$)

(by construction)

Since 

$PS\parallel TR$

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

$\frac{QS}{SR}=\frac{QP}{PR}$ (from equation ($1$))

2. In the given figure, D is a point on hypotenuse AC of $\Delta ABC$, such that $BD\bot AC$, $DM\bot BC$and $DN\bot AB$, 

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Prove that:

(i) $D{{M}^{2}}=DN.MC$

Ans: It is given in the question that $DN\parallel CB,DM\parallel AB$

Also $DM\bot CB,DN\bot AB$

Hence we get DNMB a rectangle

Now let us join DB as shown

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In this case D is the foot of perpendicular drawn from B to AC

Hence $\angle CDM+\angle MDB={{90}^{\circ }}$

Also $\angle DCM+\angle CDM={{90}^{\circ }}$

Hence $\angle MDB=\angle DCM$

Also $\angle MDB+\angle DBM={{90}^{\circ }}$

Hence we found $\angle CDM=\angle DBM$

Now in $\Delta DCM,\Delta BDM$

$\angle MDB=\angle DCM$

$\angle CDM=\angle DBM$

Hence by AA criterion \[\Delta DCM\sim \Delta BDM\]

$\Rightarrow \frac{BM}{DM}=\frac{DM}{CM}$

$\Rightarrow \frac{DN.CM}{1}=\frac{D{{M}^{2}}}{1}$ proved (since $BM=DN$

(ii) $D{{N}^{2}}=DM.AN$

Ans: 

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In right triangle DBN

$\angle BDN+\angle NBD={{90}^{\circ }}$

Also $\angle BDN+\angle NDA={{90}^{\circ }}$

Hence $\angle NBD=\angle NDA$

Also $\angle NDA+\angle DAN={{90}^{\circ }}$

Hence we have $\angle BDN=\angle DAN$

Now in $\Delta DNA,\Delta BND$

$\angle NBD=\angle NDA$

$\angle BDN=\angle DAN$

Hence by AA criterion $\Delta DNA\sim \Delta BND$

$\Rightarrow \frac{AN}{DN}=\frac{DN}{NB}$ hence proved

3. In the given figure, ABC is a triangle in which $\angle ABC>{{90}^{\circ }}$and $AD\bot CB$are produced. Prove that $A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}+2BC.BD$

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Ans: By Pythagoras theorem we have

$A{{B}^{2}}=A{{D}^{2}}+D{{B}^{2}}$……($1$)

$A{{C}^{2}}=A{{D}^{2}}+D{{C}^{2}}$

$\Rightarrow A{{C}^{2}}=A{{D}^{2}}+{{\left( DB+BC \right)}^{2}}$

$\Rightarrow A{{C}^{2}}=A{{D}^{2}}+D{{B}^{2}}+B{{C}^{2}}+2DB.BC$

Now from the equation ($1$) we have 

$A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}+2DB.BC$ hence proved

4. In the given figure, ABC is a triangle in which $\angle ABC<{{90}^{\circ }}$and $AD\bot CB$.

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Prove that

$A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}-2BC.BD$

Ans: By Pythagoras theorem 

$A{{C}^{2}}=A{{D}^{2}}+C{{D}^{2}}$

$\Rightarrow A{{C}^{2}}=A{{D}^{2}}+{{\left( BC-BD \right)}^{2}}$

$\Rightarrow A{{C}^{2}}=A{{D}^{2}}+B{{C}^{2}}+B{{D}^{2}}-2BC.BD$

Also we have $A{{B}^{2}}=A{{D}^{2}}+B{{D}^{2}}$

Hence $A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}-2BC.BD$

5. In the given figure, AD is a median of a triangle ABC and $AM\bot CB$. Prove that:

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(i) $A{{C}^{2}}=A{{D}^{2}}+{{\left( \frac{BC}{2} \right)}^{2}}+BC.DM$

Ans: 

By Pythagoras theorem 

$A{{M}^{2}}+M{{D}^{2}}=A{{D}^{2}}$……($1$)

$A{{M}^{2}}+M{{C}^{2}}=A{{C}^{2}}$

$\Rightarrow A{{M}^{2}}+{{\left( MD+DC \right)}^{2}}=A{{C}^{2}}$

$\Rightarrow A{{D}^{2}}+D{{C}^{2}}+2MD.DC=A{{C}^{2}}$(from equation ($1$))

Put $DC=\frac{BC}{2}$ in above as shown

$A{{D}^{2}}+{{\left( \frac{BC}{2} \right)}^{2}}+2MD.\frac{BC}{2}=A{{C}^{2}}$ hence proved

(ii) $A{{B}^{2}}=A{{D}^{2}}+{{\left( \frac{BC}{2} \right)}^{2}}-BC.DM$

Ans: By Pythagoras theorem

$A{{B}^{2}}=A{{M}^{2}}+M{{B}^{2}}$

$\Rightarrow \left( A{{D}^{2}}-D{{M}^{2}} \right)+M{{B}^{2}}=A{{B}^{2}}$

Put $DC=\frac{BC}{2}$ in above as shown

$A{{D}^{2}}+{{\left( \frac{BC}{2} \right)}^{2}}-BC.MD=A{{B}^{2}}$ hence proved

(iii) $A{{C}^{2}}+A{{B}^{2}}=2A{{D}^{2}}+2{{\left( \frac{BC}{2} \right)}^{2}}$

Ans: It is clearly observable that 

$A{{B}^{2}}+A{{C}^{2}}=A{{D}^{2}}+{{\left( \frac{BC}{2} \right)}^{2}}-BC.MD+A{{D}^{2}}+{{\left( \frac{BC}{2} \right)}^{2}}+2MD.\frac{BC}{2}$

$A{{B}^{2}}+A{{C}^{2}}=2A{{D}^{2}}+\left( \frac{B{{C}^{2}}}{2} \right)$

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

Ans: 

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Let ABCD be a parallelogram 

Let us have a construction, draw perpendiculars BE to extended AD and AF on BC

Now by Pythagoras theorem

$B{{E}^{2}}+A{{E}^{2}}=A{{B}^{2}}$……($1$

$D{{E}^{2}}+B{{E}^{2}}=D{{B}^{2}}$

$\Rightarrow {{\left( AE+AD \right)}^{2}}+B{{E}^{2}}=D{{B}^{2}}$

$\Rightarrow A{{D}^{2}}+A{{B}^{2}}+2AD.AE=D{{B}^{2}}$

Also by Pythagoras theorem

$B{{F}^{2}}+A{{F}^{2}}=A{{B}^{2}}$

$C{{F}^{2}}+A{{F}^{2}}=A{{C}^{2}}$

$\Rightarrow {{\left( BC-BF \right)}^{2}}+A{{F}^{2}}=A{{C}^{2}}$

$\Rightarrow {{\left( BC-BF \right)}^{2}}+A{{F}^{2}}=A{{C}^{2}}$

$\Rightarrow B{{C}^{2}}+A{{B}^{2}}-2BC.BF=A{{C}^{2}}$……($3$)

Also we have $AB=CD,AD=BC$……($4$)

Clearly  $\Delta BEA\sim \Delta ABF$

$\Rightarrow AE=BF$….($5$)

Adding equations $1$and $3$and putting values from equations $4$ and $5$ we get

$A{{B}^{2}}+B{{C}^{2}}+C{{D}^{2}}+D{{A}^{2}}=A{{C}^{2}}+B{{D}^{2}}$ hence proved

7. In the given figure, two chords AB and CD intersect each other at the point P. prove that:

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(i) $\Delta APC\sim \Delta DPB$

Ans:

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Let us have a construction, draw BC 

$\Delta APC,\Delta DPB$

$\angle CAP=\angle BDP$ (angle subtended by same chord is equal)

$\angle CPA=\angle BPD$(vertically opposite angle)

Hence By AA criterion $\Delta APC\sim \Delta DPB$

(ii) $AP.PB=CP.DP$

Now from above proved result

$\frac{AP}{DP}=\frac{CP}{BP}$ hence $AP.PB=CP.DP$ proved

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

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(i) $\Delta PAC\sim \Delta PDB$

Ans: 

In $\Delta PAC,\Delta PDB$

$\angle APC=\angle BPD$ (common angle)

$\angle PAC=\angle PBD$ (exterior angle of a cyclic quadrilateral equal to the interior angle)

Hence by AA criterion $\Delta PAC\sim \Delta PDB$

(ii) $PA.PB=PC.PD$ 

Ans: Now as proved earlier $\Delta PAC\sim \Delta PDB$

$\Rightarrow \frac{PA}{PD}=\frac{PC}{BP}$

Hence $PA.BP=PC.PD$proved

9. In the given figure, D is a point on side BC of $\Delta ABC$such that $\frac{BD}{CD}=\frac{AB}{AC}$ Prove that AD is the bisector of $\angle BAC$.

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Ans: 

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Let us have a construction, extend BA to P such that $AP=AC$and join PC

Now it is given that

$\frac{BD}{CD}=\frac{AB}{AC}$

$\Rightarrow \frac{BD}{CD}=\frac{AB}{AP}$ (by construction)

Hence $AD\parallel PC$by converse of basic proportionality theorem

$\Rightarrow \angle BAD=\angle APC$(corresponding angles)

$\Rightarrow \angle DAC=\angle ACP$ (alternate interior angles)

By construction $\angle APC=\angle ACP$ 

Hence $\angle BAD=\angle DAC$ proved

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? 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?

Ans: 

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Let us have AB be the height of the tip from the water surface, BC be the horizontal distance of the fly and also AC is the length of the string

By Pythagoras theorem

$AC=\sqrt{A{{B}^{2}}+B{{C}^{2}}}$

$\Rightarrow AC=\sqrt{{{1.8}^{2}}+{{2.4}^{2}}}=3m$

 Also it is given that she pulls the string at $5cm{{s}^{-1}}$

Therefore the length of string pulled in $12$seconds is $12\times 5=0.6m$

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Also let the fly be at some point D after $12$seconds, hence length of the string is AD which is given by $3-0.6=2.4m$

Similarly by Pythagoras theorem

$BD=\sqrt{5.76-3.24}=1.587m$

Hence horizontal distance of fly is given by

$1.587+1.2=2.79m$

Exercise 6.6 Class 10 Maths NCERT Solutions

Basic Facts about Triangle

As per the definition of triangle, any closed shape forms with three lines that make three angles to each other is known as the triangle. In all cases, the total for the angles in a triangle is 180 degrees.

The CBSE syllabus prescribes sums on triangle for the students of the 10th standard as they learn many applications and experiments on it. The CBSE Class 10 Maths Exercise 6.6 is all about triangles and you should practice the sums with sheer dedication. 

Types of Triangles

The CBSE students can come across different types of triangles in the 10th standard maths book of NCERT. You can expect many sums to be present that are related to each type of triangle. Here are the types of the triangle and the notes about them:

Right-Angled Triangle: In a triangle, if one of the angles is 90 degrees, you can call it a right-angled triangle. The other two anglers can both be of 45 degrees or might vary.

Equilateral Triangle: If all sides of a particular triangle are equal, it is termed as the equilateral triangle. As per the theory mentioned above, you can only draw an equilateral triangle if all the angles are 60 degrees. 

Acute-Angled Triangle: Acute angle triangles have a couple of angles that are less than 90 degrees.   It might at times resemble an equilateral triangle but it is not.

Obtuse-Angled Triangle: In an obtuse angle, one of the angles is greater than 90 degrees and the other two angles might vary randomly. In the case of the obtuse angle, the length of the hypotenuse can the greatest of all the line segments. 

In the NCERT Solutions for Class 10 Maths Exercise 6.6, you can check out the sums and find the existence of all the types of triangles. 

Some Problems Related to Triangles

Here, the details of some problems related to triangles are mentioned that you may encounter as a student while solving the NCERT Class 10 Maths Ex 6.6. 

Concurrency: Concurrency is a special type of numerical that you will come across while solving the problems of the exercise of the triangle. These sums can carry a lot of marks if they come in the board examination and solving it with proper steps can fetch you marks. 

Triangle with the Parallel Line: There can be sums where a parallel line can intersect with the triangle and you will be asked to find the missing angle. These sums can help you to increase your knowledge regarding the mensuration and you can take up civil engineering in the higher studies. 

Finding the Missing Angle: In Exercise 6.6 of the Class 10 Ncert Maths book, you can find some complicated sums on the triangle where it is changed into a 3D structure and you have to find the angle at which the lines are intersecting it. 

Tips to Solve the Sums on Triangle Easily

Here are some basic tips to solve the sums on the triangle easily:

• Learn the  basic properties of triangle

• Go through the question several times and try to understand the question

• Note down everything that is already given in the sum

• Take the right approach to solve the sum as you have practised before

• Make sure writing all the steps of the sum while you are solving it

• Once you find the answer, you need to mark it properly on your paper

How to Practice Numerical Problems on Triangle 

As the Ex 6.6 Class 10 Maths deals with the sums based on triangles, you would need to follow the right approach to solve it. The most important thing that you will need to practise is drawing the triangle diagrams given in the book and label them properly. Once you are able to draw the triangle diagram correctly, it will be easier for you to derive the answer.

NCERT Solutions for Class 10 Maths Chapter 6 Other Exercises

NCERT Solutions for Class 10 Maths

• Chapter 1 - Real Numbers

• Chapter 2 - Polynomials

• Chapter 3 - Pair of Linear Equations in Two Variables

• Chapter 4 - Quadratic Equations

• Chapter 5 - Arithmetic Progressions

• Chapter 6 - Triangles

• Chapter 7 - Coordinate Geometry

• Chapter 8 - Introduction to Trigonometry

• Chapter 9 - Some Applications of Trigonometry

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Areas Related to Circles

• Chapter 13 - Surface Areas and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability