Question

the height and base radius of a cone each is increased by 50% what is the ratio between the volume of the given cone and new cone​

Answer 1

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Answer 2

$\underline{\mathfrak{\huge{The\:Question\:asked:}}}$

The height and the base radius of a cone each is increased by 50%. What is the ratio between the Volume of the given cone and the new cone ?

$\underline{\mathfrak{\huge{Here's \:Your\:answer:}}}$

Let the base radius be = r

Let the Height of the cone be = h

Volume of the given cone = $\sf{\frac{1}{3} \pi r^{2} h}$

50% of increase in the Height and the base radius will make their values = $\sf{\frac{3r}{2}\:and\:\frac{3h}{2}}$

Volume of the new cone = $\sf{\frac{1}{3} \pi\times \frac{9r^{2}}{4} \times \frac{3h}{2}}$

Ratio = $\sf{\frac{Volume\:of\:the\:given\:cone}{Volume\:of\:new\:formed\:cone}}$

Ratio = $\sf{\frac{\pi r^{2} h\times 4\times 2\times 3}{3\times \pi \times 9r^{2} \times3h}}$

Ratio = $\sf{\frac{8}{27}}$

Ratio = $\tt{8:27}$