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  • The volume of the frustum of a cone is where h is vertical height of the frustum and r _1 ,r_ 2 are the radii of the ends

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.

Answers (1)

Answer False

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.

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infoexpert21

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