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

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

**Question 1.**

**In the given figure (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).**

**Solution:**

**Question 2.**

**E and F are points on the sides PQ and PR respectively of a ∆PQR. 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 = 0.36 cm

**Solution:**

**Question 3.**

**In the given figure, if LM || CB and LN || CD.**

**Solution:**

**Question 4.**

**In the given figure, DE || AC and DF || AE.**

**Solution:**

**Question 5.**

**In the given figure, DE || OQ and DF || OR. Show that EF || QR.**

**Solution:**

**Question 6.**

**In the given 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 B.P.T., prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.**

**Solution:**

**Given:** A ∆ABC in which D is the mid-point of AB and DE || BC

**Question 8.**

**Using converse of B.P.T., prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.**

**Solution:**

**Given:** A ΔABC in which D and E are mid-points of sides AB and AC respectively.

**To Prove:** DE || BC

**Question 9.**

**ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the**

**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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