NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.2 Solutions

Collegedunia Team logo

Collegedunia Team

Content Curator

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.

Download PDF: NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.2 Solutions

Check out NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.2 Solutions

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

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

Also Check:

Also Check:

CBSE X Related Questions

  • 1.
    If roots of the quadratic equation \(x^2 - k\sqrt{3}x + 2 = 0\) are real and equal, then value of k is

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

    • 2.
      For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

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

      • 3.
        It is given that \(\triangle ABC \sim \triangle EDF\). Which of the following is not true?

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

        • 4.
          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

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

          • 5.
            In an A.P., if $a_{14} - a_8 = 24$, then the common difference of the A.P. is

              • $6$
              • $4$
              • $\pm 4$
              • $3$

            • 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

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

              Comments


              No Comments To Show