Question

2 cones have their height in the ratio6:7 and thier ratio of radius of the base is 7:6 what is the ratio of thier volume​

Answer 1

Answer:-

$\sf{The \ ratio \ of \ their \ volumes }$

$\sf{is \ 7:6}$

Given:

• Two cones have their height in the ratio of 6:7

• Their ratio of radius is 7:6

To find:

• The ratio of volumes of cones.

Solution:

$\sf{For \ 1^{st} \ cone,}$

$\sf{Height=h_{1},}$

$\sf{Radius=r_{1}}$

$\sf{For \ 2^{nd} \ cone,}$

$\sf{Height=h_{2},}$

$\sf{Radius=r_{2}}$

$\sf{According \ to \ the \ given \ condition}$

$\sf{\dfrac{h_{1}}{h_{2}}=\dfrac{6}{7}}$

$\sf{\therefore{6h_{2}=7h_{1}}}$

$\sf{\longmapsto{\therefore{h_{2}=\dfrac{7h_{1}}{6}...(1)}}}$

$\sf{\dfrac{r_{1}}{r_{2}}=\dfrac{7}{6}}$

$\sf{\therefore{7r_{2}=6r_{1}}}$

$\sf{\longmapsto{\therefore{r_{2}=\dfrac{6r_{1}}{7}...(2)}}}$

$\boxed{\sf{Volume \ of \ cone=\dfrac{1}{3}\times\pi \times \ r^{2}\times \ h}}$

$\sf{\therefore{Ratio \ of \ two \ comes}}$

$\sf{=\dfrac{\dfrac{1}{3}\times\pi\times \ r_{1}^{2}\times \ h_{1}}{\dfrac{1}{3}\times\pi\times \ r_{2}^{2}\times \ h_{2}}}$

$\sf{=\dfrac{r_{1}^{2}\times \ h_{1}^{2}}{r_{2}^{2}\times \ h_{2}^{2}}}$

$\sf{...from \ (1) \ and \ (2), \ we \ get}$

$\sf{=\dfrac{r_{1}^{2}\times \ h_{1}^{2}}{(\dfrac{6r_{1}}{7})^{2}\times\dfrac{7h_{1}}{6}}}$

$\sf{=\dfrac{r_{1}^{2}\times \ h_{1}\times7\times7\times6}{r_{1}^{2}\times \ h_{1}\times6\times6\times7}}$

$\sf{=\dfrac{7}{6}}$

$\sf{=7:6}$

$\sf\purple{\tt{\therefore{The \ ratio \ of \ their \ volumes }}}$

$\sf\purple{\tt{is \ 7:6}}$