Question

A solid metallic sphere of radius 6 cm is melted to make two cones whose radius are in the ratio of 3 : 4 keeping the height same as 12 cm. Then, the ratio of their slant heights, is

Answer 1

Answer:

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Answer 2

Given,

Radius of sphere$\[ = 6cm\]$

The ratio of radii of cones$\[ = 3:4\]$

The height of the cone$\[ = 12\,{\rm{cm}}\]$

Solution,

The Formula used, Volume of the sphere$\[ = \frac{4}{3}\pi {r^3}\]$

The Formula used, Volume of the cone$\[ = \frac{1}{3}\pi {r^2}h\]$

The Formula used, Slant height of the cone$\[ = \sqrt {{r^2} + {h^2}} \]$

Assume that the height of the cones is $\[3x\,{\rm{and}}\,4x\]$respectively.

Kow that the volume always remains the same.

Therefore,

$\[\begin{array}{l} \Rightarrow \frac{4}{3} \times \frac{{22}}{7} \times 6 \times 6 \times 6 = \frac{1}{3} \times \frac{{22}}{7} \times 3x \times 3x \times 12 + \frac{1}{3} \times \frac{{22}}{7} \times 4x \times 4x \times 12\\ \Rightarrow 216 = 27{x^2} + 48{x^2}\\ \Rightarrow 216 = 75{x^2}\\ \Rightarrow x = 1.7\end{array}\]$

So, the radius of the cones is $\[5.1\,cm\,{\rm{and}}\,6.8\,cm\]$.

Therefore, the ratio of the slant height of the cones

$\[\begin{array}{l} \Rightarrow \frac{{\sqrt {{{\left( {5.1} \right)}^2} + {{\left( {12} \right)}^2}} }}{{\sqrt {{{\left( {6.8} \right)}^2} + {{\left( {12} \right)}^2}} }}\\ \Rightarrow \sqrt {\frac{{170.01}}{{190.24}}} \\ \Rightarrow \sqrt {0.89} \\ \Rightarrow \frac{{189}}{{200}}\end{array}\]$

Hence, the ratio of the slant height of the cone is $\[\frac{{189}}{{200}}\].$