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#### Write ‘True’ or ‘False’ and justify your answer in the following: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 height of the frustum and $r_{1},r_{2}$ are the radii of the ends.

According to question

In this figure ABCE is a frustum of cone ABD, h is the height of frustum and $r_{1},r_{2}$  are the radii of the frustum.

From the figure

$\Delta ADF:\Delta EDF$

$\frac{DF}{DG}=\frac{AF}{EG}$

$\frac{h+h'}{h'}=\frac{r_{1}}{r_{2}}$

$\frac{h}{h'}+1=\frac{r_{1}}{r_{2}}$

$\frac{h}{h'}=\frac{r_{1}}{r_{2}}-1$

$\frac{h}{h'}=\frac{r_{1}-r_{2}}{r_{2}}$

$h' = \frac{r_{2}h}{r_{1}-r_{2}} \; \; \; \; \; \; \; \; \; \; \; \; \; ......(1)$

Volume of frustum ABCE = volume of ABD – volume of ECD

$=\frac{1}{3}\pi r^{2}\left [ h'+h \right ]=\frac{1}{3}\pi r{_{1}}^{2}h'$                          $\left ( \therefore \text {Volume of cone }=\frac{1}{3}\pi r^{2}h \right )$

$=\frac{1}{3}\pi r{_{1}}^{2}h'+\frac{1}{3}\pi r{_{1}}^{2}h'-\frac{1}{3}\pi r{_{2}}^{2}h'$

$=\frac{1}{3}\pi \left [ r{_{1}}^{2}h'+ \pi r{_{1}}^{2}h'-\pi r{_{2}}^{2}h' \right ]$

$=\frac{1}{3}\pi \left [ r{_{1}}^{2}h'+ h' \left ( r_{1}^{2} -r_{2}^{2}\right ) \right ]$

$=\frac{1}{3}\pi \left [ r{_{1}}^{2}h'+ h' \left ( r_{1} -r_{2}\right )\left ( r_{1}+r_{2} \right ) \right ]$                    $\left [ Q\left ( a^{2}-b^{2} \right )=(a-b)(a+b) \right ]$

$=\frac{1}{3}\pi \left [ r_{1}^{2}h+\frac{r_{2}h}{\left ( r_{1}-r_{2} \right )}\left ( r_{1}-r_{2} \right ) \left ( r_{1}+r_{2} \right ) \right ]$    [ using (1) ]

$=\frac{1}{3}\pi h\left [ r_{1}^{2}+r_{2}^{2}+r_{1}r_{2} \right ]$

Hence the volume of frustum of cone is not equal to  $=\frac{1}{3}\pi h\left [ r_{1}^{2}+r_{2}^{2}-r_{1}r_{2} \right ]$

Hence the given statement is false.