NCERT Solutions Class 10 Maths – Triangles | Chapter 6 Exercise – 6.2 Question 10

Triangles | Chapter 6 Exercise 6.2 Question 10:  The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is trapezium. Solution: Detailed explanation is given in the below image and YouTube video.  

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NCERT Solutions Class 10 Maths – Triangles | Chapter 6 Exercise – 6.2 Question 9

Triangles | Chapter 6 Exercise 6.2 Question 9:  ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO. Solution: Detailed explanation is given in the below image and YouTube video. Next Post

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NCERT Solutions Class 10 Maths – Triangles | Chapter 6 Exercise – 6.2 Question 8

Triangles | Chapter 6 Exercise 6.2 Question 8:  Using Converse of basic proportionality theorem, prove that the line joining the midpoints of any two sides of a triangle is parallel to the third side. Solution: Detailed explanation is given in the below image and YouTube video. Next Post

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NCERT Solutions Class 10 Maths – Triangles | Chapter 6 Exercise – 6.2 Question 7

Chapter 6 Exercise 6.2 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. Solution: Detailed explanation is given in the below image and YouTube video. Next Post

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NCERT Solutions Class 10 Maths – Triangles | Chapter 6 Exercise – 6.2 Question 6

Chapter 6 Exercise 6.2 Question 6:  In the following 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: Detailed explanation is given in the below image and YouTube video. Next Post

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NCERT Solutions Class 10 Maths – Triangles | Chapter 6 Exercise – 6.2 Question 2

Chapter 6 Exercise 6.2 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, […]

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