NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 12: Surface Areas & Volumes

Exercise 12.2

Two identical solid hemispheres of equal base radius \(r\) cm are stuck together along their bases. The total surface area of the combination is \(6\pi r^2\).

A solid cylinder of radius \(r\) and height \(h\) is placed over another cylinder of same height and radius. The total surface area of the shape so formed is \(4\pi rh + 4\pi r^2\).

A solid cone of radius \(r\) and height \(h\) is placed over a solid cylinder having same base radius and height as that of the cone. The total surface area of the combined solid is \(\pi r[\sqrt{r^2+h^2}+3r+2h]\).

A solid ball is exactly fitted inside the cubical box of side \(a\). The volume of the ball is \(\dfrac{4}{3}\pi a^3\).

The volume of the frustum of a cone is \(\dfrac{1}{3}\pi h[r_1^2+r_2^2-r_1r_2]\), where \(h\) is vertical height of the frustum and \(r_1, r_2\) are the radii of the ends.

The capacity of a cylindrical vessel with a hemispherical portion raised upward at the bottom (see Fig. 12.7) is \(\dfrac{\pi r^2}{3}[3h-2r]\).

The curved surface area of a frustum of a cone is \(\pi l(r_1+r_2)\), where \(l=\sqrt{h^2+(r_1+r_2)^2}\), \(r_1,r_2\) are radii and \(h\) is height.

An open metallic bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base. The surface area of the metallic sheet used is equal to curved surface area of frustum + area of circular base + curved surface area of cylinder.