NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 6 **Triangles Class 10 NCERT Solutions Ex 6.2. **

- Triangles Class 10 Ex 6.1
- Triangles Class 10 Ex 6.3
- Triangles Class 10 Ex 6.4
- Triangles Class 10 Ex 6.5
- Triangles Class 10 Ex 6.6

Board | CBSE |

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 6 |

Chapter Name | Triangles |

Exercise | Ex 6.2 |

Number of Questions Solved | 10 |

Category | NCERT Solutions |

## NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2

NCERT Solutions for Class 10 Maths

Page No: 128

**Question 1. **(i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

**Solution: **(i)

**Let EC = x**

Since DE || BC.

Therefore, by basic proportionality theorem,

(ii)

Let AD = x

Since DE || BC,

Therefore by basic proportionality theorem,

**Question 2.** E and F are points on the sides PQ and PR respectively of a ΔPQR.

For each of the following cases, state whether EF || QR.

(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm

**Solution: (i) **Given that PE = 3.9, EQ = 3, PF = 3.6, FR = 2.4

Now,

**PE = 4, QE = 4.5, PF = 8, RF = 9**

(ii)

(ii)

**(iii)**

PQ = 1.28, PR = 2.56, PE = 0.18, PF = 0.36

**Question 3.** In the fig 6.18, if LM || CB and LN || CD, prove that

**Solution: **In the given figure

Since LM || CB,

Therefore by basic proportionality theorem,

**Question 4. **In figure , DE || AC and DF || AE. Prove that

**Solution: **In ∆ABC,

Since DE || AC

**Question 5.** In figure , DE || OQ and DF || OR, show that EF || QR.

**Solution: **In ∆POQ

Since DE || OQ

**Question 6.** In the figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

**Solution: **

**Question 7.** Using Basic proportionality theorem, prove that a line drawn 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).

**Solution: **Consider the given figure

PQ is a line segment drawn through midpoint P of line AB such that PQ||BC

i.e. AP = PB

Now, by basic proportionality theorem

i.e. AQ = QC

Or, Q is midpoint of AC.

**Question 8.** Using Converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

**Solution: **Consider the given figure

PQ is a line segment joining midpoints P and Q of line AB and AC respectively.

i.e. AP = PB and AQ = QC

Now, we may observe that

And hence basic proportionality theorem is verified

So, PQ||BC

**Question 9.** ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O.

Show that \(\frac { AO }{ BO } =\frac { CO }{ DO } \)

**Solution: **

**Question 10.** The diagonals of a quadrilateral ABCD intersect each other at the point O such that \(\frac { AO }{ BO } =\frac { CO }{ DO } \). Show that ABCD is a trapezium.

**Solution: **

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