Neha Tyagi and Amit Rastogi Solutions for Chapter: Surface Areas and Volumes, Exercise 2: Exercise

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

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

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

A solid ball is exactly fitted inside the cubical box of side $a.$ The volume of the ball is $\frac{4}{3}\mathrm{\pi }{a}^3.$

The volume of the frustum of a cone is $\frac{1}{3}\pi h\left[r_1^2+r_2^2-r_1{r}_2\right],$ where $h$ is vertical of frustum and $r_1,r_2$ are the radius of ends.

The capacity of a cylindrical vessel with a hemispherical portion raised upward at the bottom as shown in the figure is $\frac{\pi r^2}{3}\left[3h-2r\right].$

The curved surface area of a cone is $\pi l\left(r_1+r_2\right)$, where $l=\sqrt{h^2+{\left(r_1+r_2\right)}^2 }$, $r_1$ and $r_2$ are the radius of the two ends of the frustum and $h$ is the vertical height.

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