NCERT Solutions for Class 9 Maths Exercise 10.6 Circles updated for new academic session 2022-2023. Class 9 Exercise 10.6 Solutions are given here in Hindi and English Medium to download in PDF or videos. All the questions are explained step by step process to make the solutions easier.

## NCERT Solutions for Class 9 Maths Exercise 10.6

### Theorems for Exercise 10.6 in Class 9 Maths

Exercise 10.6 the extension of theorem 10.8 gives rise to another theorem 10.9 which says that angles are in the same segment of the circle are equal. Now, this very easy to prove.

Theorem 10 says that if a line segment joining 2 points of a circle subtends equal angles at two other point lying on the same side of the line containing the line segments. The four points are con-cyclic that is the four points lie on the same circle.

#### Apply the right Theorem in Questions

You will not find theorems and questions difficult, if you practice questions relating the correct theorem. Theorem 10.11 says that in case of cyclic quadrilateral the sum of either pair of opposite angles is 180o.

The theorem 10.12 is the converse of theorem 10.11 and it says that sum of the pair of opposite angles of a quadrilateral is 180o. The quadrilateral must be cyclic.

##### Questions of Class 9 Maths Exercise 10.6

Exercise 10.6 contains 8 questions out of which number 1, 4, 5, 6, 7, and 8 are questions requiring proof of certain statement.

Whereas, question number 2 and 3 require finding out the length of either the radius of the segment of a circle. After practicing once or twice, questions given in exercise looks easier to solve.

###### How to solve Class 9 Maths Exercise 10.6

Before attempting this exercise it advised to go through all theorems in this chapter from theorem 1 to 12. Particularly, the following two important ones.

Angle subtended by an arc in the centre is double the angle subtended by it at any point on the remaining part of the circle (theorem 8). The sum of either pair of opposite angles of a cyclic quadrilateral is 180o (theorem 11).