NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13 1 Solutions

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NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.1 Solutions are based on the concept of Surface area of a cuboid and a cube.

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Read More: NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas and Volumes

Exercise Solutions of Class 9 Maths Chapter 13 Surface Areas and Volumes

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CBSE X Related Questions

  • 1.
    In the given figure, $PQ$ is tangent to the circle with centre $O$. $S$ is a point on the circle such that $\angle SQT = 55^\circ$. The $m\angle QPS$ is

      • $55^\circ$
      • $20^\circ$
      • $35^\circ$
      • $70^\circ$

    • 2.
      If the zeroes of the polynomial \(p(x) = 2x^2 - 7x + 6\) are \(\alpha\) and \(\beta\), then the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\) is

        • \(\frac{7}{2}\)
        • \(\frac{6}{7}\)
        • \(\frac{-7}{6}\)
        • \(\frac{7}{6}\)

      • 3.
        If roots of the quadratic equation \(x^2 - k\sqrt{3}x + 2 = 0\) are real and equal, then value of k is

          • \(-2\)
          • \(\sqrt{\frac{8}{3}}\)
          • \(1\)
          • \(2\)

        • 4.
          Mean and Median of a frequency distribution are 43 and 40 respectively. The value of mode is

            • 34
            • 43
            • 38.5
            • 41.5

          • 5.
            Assertion (A) : H.C.F. \((36 m^{2}, 18 m) = 18 m\), where \(m\) is a prime number.
            Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.

              • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
              • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
              • Assertion (A) is true, but Reason (R) is false.
              • Assertion (A) is false, but Reason (R) is true.

            • 6.
              If \( \alpha \) and \( \beta \) are two zeroes of a polynomial \( f(x) = px^2 - 2x + 3p \) and \( \alpha + \beta = \alpha\beta \), then value of p is :

                • \( -\frac{2}{3} \)
                • \( \frac{2}{3} \)
                • \( \frac{1}{3} \)
                • \( -\frac{1}{3} \)

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