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.

Download PDF: NCERT Solutions for Class 9 Maths Chapter 10 Exercise 10.3 Solutions

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

Read More: NCERT Solutions For Class 9 Maths Chapter 10 Circles

Exercise Solutions of Class 9 Maths Chapter 10 Circles

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CBSE X Related Questions

  • 1.
    A circle centered at (2, 1) passes through the points A(5, 6) and B(-3, K). Find the value(s) of K. Hence find length of chord AB.


      • 2.
        The HCF of 960 and 432 is :

          • 48
          • 54
          • 72
          • 36

        • 3.
          In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).


            • 4.
              The graph of \(y = f(x)\) is given. The number of zeroes of \(f(x)\) is :

                • 0
                • 1
                • 3
                • 2

              • 5.
                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.


                  • 6.
                    If \( \alpha, \beta \) are the zeroes of the quadratic polynomial \( px^2 + qx + r \), then find the value of \( \alpha^3\beta + \beta^3\alpha \).

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