NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.2 Solutions

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NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.2 Solutions are based on the concept of Surface area of a right circular 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.
    The natural number 1 is :

      • a prime number.
      • a composite number.
      • prime as well as composite.
      • neither prime nor composite.

    • 2.
      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.


        • 3.
          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

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

          • 4.
            Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)


              • 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.
                    The HCF of 960 and 432 is :

                      • 48
                      • 54
                      • 72
                      • 36

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