NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.6 Solutions

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NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.6 Solutions are based on the Volume of a cylinder.

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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.
    If for a data, median is 5 and mode is 4, then mean is equal to :

      • 7
      • 11
      • \(\frac{11}{2}\)
      • \(\frac{14}{3}\)

    • 2.
      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

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

      • 3.
        Which of the following graphs represents a polynomial with both zeroes being positive?


        • 4.
          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.

          • 5.
            For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

              • \(0\)
              • \(\frac{80}{81}\)
              • \(1\)
              • \(\frac{82}{80}\)

            • 6.
              It is given that \(\triangle ABC \sim \triangle EDF\). Which of the following is not true?

                • \(\frac{\text{Perimeter of } \triangle ABC}{\text{Perimeter of } \triangle EDF} = \frac{AB}{ED}\)
                • \(\frac{AB}{ED} = \frac{AC}{EF}\)
                • \(\angle A = \angle D, \angle C = \angle F\)
                • \(\frac{AB + BC}{AC} = \frac{ED + DF}{EF}\)

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