NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.3 Solutions

Collegedunia Team logo

Collegedunia Team Content Curator

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.

    Given that $\sin \theta + \cos \theta = x$, prove that $\sin^4 \theta + \cos^4 \theta = \dfrac{2 - (x^2 - 1)^2}{2}$.


      • 2.

        The following data shows the number of family members living in different bungalows of a locality:
         

        Number of Members0−22−44−66−88−10Total
        Number of Bungalows10p60q5120


        If the median number of members is found to be 5, find the values of p and q.


          • 3.

            In the adjoining figure, $\triangle CAB$ is a right triangle, right angled at A and $AD \perp BC$. Prove that $\triangle ADB \sim \triangle CDA$. Further, if $BC = 10$ cm and $CD = 2$ cm, find the length of AD.


              • 4.

                Find the mean and mode of the following data:

                Class15--2020--2525--3030--3535--4040--45
                Frequency1210151175


                  • 5.
                    The given figure shows a circle with centre O and radius 4 cm circumscribed by \(\triangle ABC\). BC touches the circle at D such that BD = 6 cm, DC = 10 cm. Find the length of AE.
                     BC touches the circle at D such that BD = 6 cm


                      • 6.
                        OAB is sector of a circle with centre O and radius 7 cm. If length of arc \( \widehat{AB} = \frac{22}{3} \) cm, then \( \angle AOB \) is equal to

                          • \( \left(\frac{120}{7}\right)^\circ \)
                          • \( 45^\circ \)
                          • \( 60^\circ \)
                          • \( 30^\circ \)

                        Comments


                        No Comments To Show