NCERT Solutions For Class 10 Maths Chapter 6 Triangles Exercise 6.2 - 2025-26

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.

Conclusion

Class 10 Exercise 6.2  helps solidify the foundational knowledge of triangle similarity, a crucial concept in geometry. By working through various problems, students practice identifying similar triangles and using proportional reasoning to solve geometric problems. This exercise not only enhances problem-solving skills but also prepares students for more advanced topics in geometry and trigonometry. The understanding gained here is essential for progressing in mathematics, especially in class 10 ex 6.2 that require spatial reasoning and the properties of geometric figures.

Class 10 Maths Chapter 6: Exercises Breakdown

CBSE Class 10 Maths Chapter 6 Other Study Materials

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