NCERT Solutions for Class 10 Maths Chapter 6: Triangles - Exercise 6.2

Exercise 6.2

1. (i) From the figure (i) , if \[\text{DE  }\!\!|\!\!\text{  }\!\!|\!\!\text{  BC}\]. Find \[\text{EC}\].

Ans: Let us assume that \[\text{EC = x cm}\]

Given that $\,\text{DE  }\!\!|\!\!\text{  }\!\!|\!\!\text{  BC}$

But from basic proportionality theorem, we know that

$\dfrac{\text{AD}}{\text{DB}}$ $=$ $\dfrac{\text{AE}}{\text{EC}}$

$\dfrac{\text{1}\text{.5}}{\text{3}}$ $=$ $\dfrac{\text{1}}{\text{x}}$

\[\text{x = }\dfrac{\text{3 x 1}}{\text{1}\text{.5}}\]

\[x\text{ }=\text{ }2\]

\[\therefore \]\[\text{EC = 2 cm}\]

(ii) From the figure (ii) , if \[\text{DE  }\!\!|\!\!\text{  }\!\!|\!\!\text{  BC}\]. \[\text{AD}\] in (ii).

Ans: 

Let us assume that \[\text{AD = x cm}\]

Given that \[\text{DE  }\!\!|\!\!\text{  }\!\!|\!\!\text{  BC}\text{.}\]

But from basic proportionality theorem we know that

$\dfrac{\text{AD}}{\text{DB}}$ $\text{=}$ $\dfrac{\text{AE}}{\text{EC}}$

$ \dfrac{\text{x}}{\text{7}\text{.2}}\text{ = }\dfrac{\text{1}\text{.8}}{\text{5}\text{.4}} $

$ \text{x = }\dfrac{\text{1}\text{.8 x 7}\text{.2}}{\text{5}\text{.4}} $

$ \text{x = 2}\text{.4} $

\[\therefore \text{AD = 2}\text{.4}\]$\text{cm}$

2. (i) In a $\text{ }\!\!\Delta\!\!\text{ PQR,}$ \[\text{E}\] and \[\text{F}\] are any two points on the sides \[\text{PQ}\] and \[\text{PR}\] respectively. State whether \[\text{EF  }\!\!|\!\!\text{  }\!\!|\!\!\text{  QR}\] for \[\text{PE = 3}\text{.9 cm, EQ = 3 cm, PF = 3}\text{.6 cm}\] and \[\text{FR = 2}\text{.4 cm}\]

Ans:

Given, \[\text{PE = 3}\text{.9 cm, EQ = 3 cm, PF = 3}\text{.6 cm}\],\[\text{FR = 2}\text{.4 cm}\]

$\dfrac{\text{PF}}{\text{EQ}}$ $\text{=}$$\dfrac{\text{3}\text{.9}}{\text{3}}$\[\text{ }\!\!~\!\!\text{ = 1}\text{.3}\]

$\dfrac{\text{PF}}{\text{FR}}$ \[\text{=}\] $\dfrac{\text{3}\text{.6}}{\text{2}\text{.4}}$ \[\text{= 1}\text{.5}\]

Hence, $\dfrac{\text{PE}}{\text{EQ}}$ \[\ne \] $\dfrac{\text{PF}}{\text{FR}}$

Therefore , \[\text{EF}\] is parallel to \[\text{QR}\].

(ii) In a $\text{ }\!\!\Delta\!\!\text{ PQR,}$ \[\text{E}\] and \[\text{F}\] are any two points on the sides \[\text{PQ}\] and \[\text{PR}\] respectively. State whether \[\text{EF  }\!\!|\!\!\text{  }\!\!|\!\!\text{  QR}\] for \[\text{PE = 4 cm, QE = 4}\text{.5 cm, PF = 8 cm}\] and \[\text{RF = 9 cm}\]

 Ans:

\[\text{PE = 4 cm,QE = 4}\text{.5 cm,PF = 8 cm,RF = 9 cm}\]

$\dfrac{\text{PE}}{\text{EQ}}\text{ = }\dfrac{\text{4}}{\text{4}\text{.5}}\text{ = }\dfrac{\text{8}}{\text{9}} $

$ \dfrac{\text{PF}}{\text{FR}}\text{ = }\dfrac{\text{8}}{\text{9}} $

Hence, $\dfrac{\text{PE}}{\text{EQ}}\text{ = }\dfrac{\text{PF}}{\text{FR}}$

Therefore, \[\text{EF}\] is parallel to \[\text{QR}\].

(iii) In a $\Delta PQR,$ \[\text{E}\] and \[\text{F}\] are any two points on the sides \[\text{PQ}\] and \[\text{PR}\] respectively. State whether \[\text{EF  }\!\!|\!\!\text{  }\!\!|\!\!\text{  QR}\] for \[\text{PQ = 1}\text{.28 cm, PR = 2}\text{.56 cm, PE = 0}\text{.18 cm}\] and \[\text{PF = 0}\text{.63 cm}\]

Ans:

\[\text{PQ = 1}\text{.28 cm,PR = 2}\text{.56 cm,PE = 0}\text{.18 cm,PF = 0}\text{.36 cm}\]

$\dfrac{\text{PE}}{\text{PQ}}\text{ = }\dfrac{\text{0}\text{.18}}{\text{1}\text{.28}}\text{ = }\dfrac{\text{18}}{\text{128}}\text{ = }\dfrac{\text{9}}{\text{64}} $

$\dfrac{\text{PF}}{\text{PR}}\text{ = }\dfrac{\text{0}\text{.36}}{\text{2}\text{.56}}\text{ = }\dfrac{\text{9}}{\text{64}} $

Hence, $\dfrac{\text{PE}}{\text{PQ}}\text{ = }\dfrac{\text{PF}}{\text{PR}}$

Therefore, \[\text{EF}\] is parallel to \[\text{QR}\].

3. In the figure given below, if sides \[\text{LM  }\!\!|\!\!\text{  }\!\!|\!\!\text{  CB}\] and \[\text{LN  }\!\!|\!\!\text{  }\!\!|\!\!\text{  CD,}\]Show that $\dfrac{\text{AM}}{\text{AB}}\text{ = }\dfrac{\text{AN}}{\text{AD}}$

Ans:

Given that in the figure, \[\text{LM  }\!\!|\!\!\text{  }\!\!|\!\!\text{  CB}\]

But from basic proportionality theorem, we know that

$\dfrac{\text{AM}}{\text{AB}}\text{ = }\dfrac{\text{AL}}{\text{AC}}\text{ }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{ (i)}$

Also, \[\text{LN  }\!\!|\!\!\text{  }\!\!|\!\!\text{  CD}\]

$\therefore \dfrac{\text{AN}}{\text{AD}}\text{ = }\dfrac{\text{AL}}{\text{AC}}\text{ }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{ (ii)}$

From (i) and (ii), we get

$\dfrac{\text{AM}}{\text{AB}}\text{ = }\dfrac{\text{AN}}{\text{AD}}$

4. In the figure given below, if sides $\text{DE  }\!\!|\!\!\text{  }\!\!|\!\!\text{  AC}$ and $\text{DF  }\!\!|\!\!\text{  }\!\!|\!\!\text{  AE}\text{.}$Show that $\dfrac{\text{BF}}{\text{FE}}\text{ = }\dfrac{\text{BE}}{\text{EC}}$

Ans:

In

$\text{ }\!\!\Delta\!\!\text{ ABC,DE  }\!\!|\!\!\text{  }\!\!|\!\!\text{  AC}$ 

$\therefore \dfrac{\text{BD}}{\text{DA}}\text{ = }\dfrac{\text{BE}}{\text{EC}} $

(By Basic proportionality theorem)

$\text{In}$ 

$\text{ }\!\!\Delta\!\!\text{ BAE,DF  }\!\!|\!\!\text{  }\!\!|\!\!\text{  AE} $

$ \therefore \dfrac{\text{BD}}{\text{DA}}\text{ = }\dfrac{\text{BE}}{\text{FE}} $

By Basic proportionality theorem

From (i) and (ii),we get

$\dfrac{\text{BE}}{\text{EC}}\text{ = }\dfrac{\text{BF}}{\text{FE}}$

5. In the figure given below, if sides $\text{DE  }\!\!|\!\!\text{  }\!\!|\!\!\text{  OQ}$ and $\text{DF  }\!\!|\!\!\text{  }\!\!|\!\!\text{  OR}$, Show that $\text{EF  }\!\!|\!\!\text{  }\!\!|\!\!\text{  QR}$

Ans:

$\text{In}$ 

$ \text{ }\!\!\Delta\!\!\text{ POQ,DE }\!\!|\!\!\text{  }\!\!|\!\!\text{ OQ} $

$ \therefore \dfrac{\text{PE}}{\text{EQ}}\text{=}\dfrac{\text{PD}}{\text{DO}} $              ……………………(i) By basic proportionality theorem$\text{In}$

$ \text{ }\!\!\Delta\!\!\text{ POR,DF  }\!\!|\!\!\text{  }\!\!|\!\!\text{  OR} $

$ \therefore \dfrac{\text{PF}}{\text{FR}}\text{=}\dfrac{\text{PD}}{\text{DO}}$

……………………(ii) By basic proportionality theorem

From (i) and (ii),we get

$\dfrac{\text{PE}}{\text{EQ}}\text{ = }\dfrac{\text{PF}}{\text{FR}} $

$ \therefore \text{EF  }\!\!|\!\!\text{  }\!\!|\!\!\text{  QR} $                             Converse of Basic proportionality theorem

6.In the figure given below, \[\text{A, Band C}\] are points on \[\text{OP, OQ and OR}\] respectively such that \[\text{AB  }\!\!|\!\!\text{  }\!\!|\!\!\text{  PQ}\] and \[\text{AC  }\!\!|\!\!\text{  }\!\!|\!\!\text{  PR}\]. Prove that \[\text{BC  }\!\!|\!\!\text{  }\!\!|\!\!\text{  QR}\].

Ans:

In

$\text{ }\!\!\Delta\!\!\text{ POQ,AB  }\!\!|\!\!\text{  }\!\!|\!\!\text{  PQ} $

$\therefore \dfrac{\text{OA}}{\text{OP}}\text{ = }\dfrac{\text{OB}}{\text{PQ}} $

$……………………(i) By basic proportionality theorem

$\text{In}$ 

$\text{ }\!\!\Delta\!\!\text{ POR,AC  }\!\!|\!\!\text{  }\!\!|\!\!\text{  PR} $

\[\therefore \dfrac{\text{OA}}{\text{OP}}\text{ = }\dfrac{\text{OC}}{\text{CR}}\]  ………………(ii) By basic proportionality theorem

From (i) and (ii),we get

$\dfrac{\text{OB}}{\text{BQ}}\text{ = }\dfrac{\text{OC}}{\text{CR}} $

$ \therefore \text{BC  }\!\!|\!\!\text{  }\!\!|\!\!\text{  CR} $

Converse of Basic proportionality theorem

7. By using Basic proportionality theorem, Show that a line passing through the mid-points of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX). 

Ans:

Let us assume in the given figure in which \[\text{PQ}\] is a line segment passing through the mid-point \[\text{P}\] of line \[\text{AB}\], such that \[\text{PQ  }\!\!|\!\!\text{  }\!\!|\!\!\text{  BC}\].

From basic proportionality theorem, we know that

$\dfrac{\text{AQ}}{\text{QC}}\text{ = }\dfrac{\text{AP}}{\text{PB}} $

$ \dfrac{\text{AQ}}{\text{QC}}\text{ = 1} $

As \[\text{P}\] is the midpoint of \[\text{AB}\] ,\[\text{AP  =  PB}\]

\[\Rightarrow \text{AQ = QC}\]

Or

\[\text{Q}\] is the midpoint of \[\text{AC}\]

8. By using Converse of basic proportionality theorem, Show that the line joined by the midpoints of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX). 

Ans:

Let us assume that the given figure in which \[\text{PQ}\] is a line segment joined by the mid-points \[\text{P and Q}\] of lines \[\text{AB and AC}\] respectively. 

i.e., \[\text{AP  =  PB and AQ  =  QC}\]

Also it is clear that

$\dfrac{\text{AP}}{\text{PB}}\text{ = 1}$ and

$\dfrac{\text{AQ}}{\text{QC}}\text{ = 1} $

$ \therefore \dfrac{\text{AP}}{\text{PB}}\text{ = }\dfrac{\text{AQ}}{\text{QC}} $ 

Hence, using basic proportionality theorem, we get 

\[\text{PQ  }\!\!|\!\!\text{  }\!\!|\!\!\text{  BC}\]

9. If \[\text{ABCD}\] is a trapezium where \[\text{AB  }\!\!|\!\!\text{  }\!\!|\!\!\text{  DC}\] and its diagonals intersect each other at the point \[\text{O}\]. Prove that $\dfrac{\text{AO}}{\text{BO}}\text{ = }\dfrac{\text{CO}}{\text{DO}}$

Ans:

Draw a line  \[\text{EF}\] through point \[\text{O}\] , such that 

In \[\text{ }\!\!\Delta\!\!\text{ ADC}\], \[\text{EO  }\!\!|\!\!\text{  }\!\!|\!\!\text{  CD}\]

Using basic proportionality theorem, we get

$\dfrac{\text{AE}}{\text{ED}}\text{ = }\dfrac{\text{AO}}{\text{OC}}$____________________(i)

In \[\text{ }\!\!\Delta\!\!\text{ ABD}\]\[\text{, OE  }\!\!|\!\!\text{  }\!\!|\!\!\text{  AB}\] 

So, using basic proportionality theorem, we get

\[\frac{\text{AE}}{\text{ED}}\ =\ \frac{\text{BO}}{\text{DO}}\ \] ___________________(ii)

From equation (i) and (ii), we get 

$\frac{\text{AO}}{\text{CO}}\text{= }\frac{\text{BO}}{\text{DO}}$

$\therefore \ \frac{\text{AO}}{\text{BO}}\text{= }\frac{\text{CO}}{\text{DO}}$

10.  The diagonals of a quadrilateral \[\text{ABCD}\] intersect each other at the point \[\text{O}\] such that $\dfrac{\text{AO}}{\text{BO}}\text{ = }\dfrac{\text{CO}}{\text{DO}}$ Prove that \[\text{ABCD}\] is a trapezium. 

Ans: 

Let us assume the following figure for the given question.

Draw a line \[\text{OE  }\!\!|\!\!\text{  }\!\!|\!\!\text{  AB}\]

In \[\text{;ABD, OE  }\!\!|\!\!\text{  }\!\!|\!\!\text{  AB}\]

Using basic proportionality theorem, we get 

$\dfrac{\text{AE}}{\text{ED}}\text{ = }\dfrac{\text{BO}}{\text{OD}}\text{ }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{ (i)}$

However, it is given that 

$\frac{\text{AO}}{\text{BO}}\text{ = }\frac{\text{CO}}{\text{DO}} $

$ \therefore \text{ }\frac{\text{AO}}{\text{CO}}\text{ = }\frac{\text{BO}}{\text{DO}}\ \text{ }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ (ii)} $

From equations (i) and (ii), we get 

$\dfrac{\text{AE}}{\text{ED}}\text{ = }\dfrac{\text{AO}}{\text{OC}} $

$ \Rightarrow \text{EO  }\!\!|\!\!\text{  }\!\!|\!\!\text{  DC} $

By the converse of basic proportionality theorem

$\Rightarrow \text{ AB }\left| \left| \text{ OE } \right| \right|\text{ DC}  $

$\Rightarrow \text{AB  }\!\!|\!\!\text{  }\!\!|\!\!\text{  CD} $

\[\therefore \text{ ABCD}\] is a trapezium.

NCERT Chapter 6 Maths Class 10 Exercise 6.2

Triangles Class 10 Exercise 6.2 PDF Download

There are several questions available in the NCERT textbook which the students can solve to become thorough with their concepts. It is important to continually practise mathematics to get hold of a variety of questions and understand it better. We provide you with Class 10 ex 6.2 solutions in a PDF downloadable format. This allows the students to study whenever they want without having to worry about stable internet connection or other issues.

Class 10 Maths Exercise 6.2 Solution

This exercise of the chapter based on triangles consists of questions related to the similarity of triangles. It has a total of 10 questions with each one covering different fundamentals. To understand it better, let us walk you through the concepts covered in every question.

The Solution of Exercise  6.2 Class 10

Question 1:

The first question contains two triangles, each of them having the same condition as DE||BC. We need to find out the values of EC and AD in these respective figures. The students can find out the values easily by using the basic proportionality theorem and properties of similar triangles. For any further issues, you can check the solution of ex 6.2 Class 10.

Question 2:

The second question consists of a triangle having sides PQ and QR, respectively. The measurements of the sides are provided. Following this, the students need to find out whether EF||QR. The question can again be solved using properties of similar triangles. If the ratios of sides are similar, then, they are parallel to each other. Else, they are not.

Question 3:

In this question, the students are required to prove the similarity of ratios between the sides using the given conditions that LM||CB and LN||CD. They can easily do this by the basic proportionality theorem.

Question 4:

This question is quite similar to the previous one. The students can solve this in three steps by the properties of similarity of triangles. The given conditions in the question are DE||AC and DF||AE.

Question 5 and 6:

Both these questions in the exercise are based on the same concept of basic proportionality theorem and similarity of triangles. Some parallel conditions are given, and students need to prove that one of the lines in the triangle is parallel to the base of the triangle. Use the concepts wisely to prove the given questions.

Question 7:

While solving this question, the students need to recall the concept of mid-point theorem they learnt in Class 9. The question can easily be solved by combining both the mid-point theorem and the basic proportionality theorem.
 

Question 8:

This can be solved using theorem 6.2 which states that, ‘If a line divides any two sides of a triangle in the same ratio, then the given line is parallel to the third side.’ It can be solved by the converse of basic proportionality theorem and again the mid-point theorem learnt in Class 9.

Question 9:

This question is a little tricky. The students have to prove that in the given trapezium, the point at which diagonals intersect are divided in the same ratio. For this, the students must be well aware of the properties of trapezium as well as the triangle. In this question, the students need to draw a line parallel to AB and CD touching AD from O. Following this; they can use the basic proportionality theorem to prove the question.

Question 10:

Again, in this question, the students need to prove that the given quadrilateral is a trapezium. They can also use the concept from the previous question and consider the conditions in the question as well. The students can easily prove it using the basic proportionality theorem and its converse for the same.

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

NCERT Solutions for Class 10 Maths Chapter 6 Exercises

Key Features for Exercise 6.2 Class 10 NCERT Solutions

The students must understand the fundamentals and concepts of every exercise to be able to solve the questions easily. We have crafted the solutions to help the students understand the answers clearly. Some of the key features include:

• The solutions are well following the CBSE guidelines and patterns, allowing the students to prepare well for the examinations. 

• The solutions are provided in a streamlined manner, making it easier for the students to study and revise from the notes quickly. 

• They can also download the PDF to access the complete solutions anytime they want.