NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.3 Solutions

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NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.3 Solutions are based on two theorems, (i) The perpendicular from the center of a circle to a chord is seen to bisect the chord. (ii) The line drawn through the center of a circle to bisect a chord is seen to be perpendicular to it.

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Check out NCERT Solutions for Class 9 Maths Chapter 10 Exercise 10.3 Solutions

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Exercise Solutions of Class 9 Maths Chapter 10 Circles

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CBSE X Related Questions

  • 1.
    In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).


      • 2.
        In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.


          • 3.
            D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.


              • 4.
                PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.


                  • 5.
                    There are two sections A and B of Grade X. There are 28 students in Section A and 30 students in Section B. What is the minimum number of books you will acquire for the class library so that they can be distributed equally among students of Section A or Section B ?

                      • 144
                      • 2
                      • 420
                      • 272

                    • 6.
                      If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is

                        • \(4 \text{ cm}\)
                        • \(4\sqrt{2} \text{ cm}\)
                        • \(8 \text{ cm}\)
                        • \(2\sqrt{2} \text{ cm}\)

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