NCERT Solutions Class 10 Mathematics Solutions for Surface Areas and Volumes - Exercise 13.1 in Chapter 13 - Surface Areas and Volumes

Question 6 Surface Areas and Volumes - Exercise 13.1

A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.


The diagram is as follows:

Now, the diameter of hemisphere = Edge of the cube = l

So, the radius of hemisphere = l/2

∴ The total surface area of solid = surface area of cube + CSA of hemisphere – Area of base of hemisphere

TSA of remaining solid = 6 (edge)2+2πr2-πr2

= 6l^2+ πr^2\\ = 6l^2+π(\frac{l}{2})^2\\ = 6l^2+π\frac{l^2}{4}\\ = \frac{l^2}{4}(24+π)\ \text{sq. units}

Connect with us on social media!
2022 © Quality Tutorials Pvt Ltd All rights reserved