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

Collegedunia Team logo

Collegedunia Team

Content Curator

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.

Download PDF: NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.1 Solutions

Check out NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.1 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.
    The value of k for which sum of the zeroes of the polynomial \(p(x) = 3x^2 - kx + 6\) is 2, is

      • 2
      • \(-6\)
      • \(-2\)
      • 6

    • 2.
      Which of the following graphs represents a polynomial with both zeroes being positive?


      • 3.
        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

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

        • 4.
          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

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

          • 5.
            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

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

            • 6.
              If for a data, median is 5 and mode is 4, then mean is equal to :

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

              Comments


              No Comments To Show