NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.3 Solutions

Collegedunia Team logo

Collegedunia Team

Content Curator

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volume Exercise 13.3 Solutions are based on the following concepts:

  • Height of the cylinder based on given condition
  • Radius of the resulting sphere
  • Number of cones for a given situation.
  • Number of coins formed by melting a cuboid structured object.

Download PDF NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.3 Solutions

Check out the solutions of NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.3 Solutions

Read More: NCERT Solutions For Class 10 Maths Chapter 13 Surface Areas and Volume

Exercise Solutions of Class 10 Maths Chapter 13 Surface Areas and Volume

Also check other Exercise Solutions of Class 10 Maths Chapter 13 Surface Areas and Volume

Also Check:

Also check:

CBSE X Related Questions

  • 1.
    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.


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


          • 3.
            If \( \alpha, \beta \) are the zeroes of the quadratic polynomial \( px^2 + qx + r \), then find the value of \( \alpha^3\beta + \beta^3\alpha \).


              • 4.
                A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is \(3/5\), then find the number of yellow balls.


                  • 5.
                    In a class test, the sum of Anamika's marks obtained in Maths and Science is 30. Had she got 2 marks more in Maths and 3 marks less in Science, the product of the marks would have been 210. Find the marks she got in the two subjects.


                      • 6.
                        D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.

                          Comments


                          No Comments To Show