Question

A cylinder is within the cube touching all the vertices face a cone is inside the cylinder if there height are the same with the same base find the ratio of their volume

Answer 1

a³ : 1/3πr²h : πr²h
(2r)³ : 1/3r² : r²
8r³ : 1/3 r² : r²

Answer 2

Given

• cylinder is within the cube touching all the vertices faces.
• a cone is inside the cylinder if there height are the same with the same base.

Explanation:

Let the length of each edge of the cube be x

then,

$\maltese \ {\boxed{\sf{ Volume_{(Cube)} = x^3 \cdots(1) }}}$

☞ We know that the cylinder lies within the cube and touches all its vertical faces.

So, the radius of the base of the cylinder be ${\sf{ \dfrac{x}{2} }}$ and height of the cylinder = x

$\maltese \ {\boxed{\sf{ Volume_{(Cүlinder)} = πr^2h }}}$

$\colon\implies{\sf{ \left[ \dfrac{22}{7} \times \left( \dfrac{x}{2} \right)^2 \times x \right] }} \\ \\ \\ \colon\implies{\sf{ \left[ \dfrac{11x^3}{14} \right] \cdots(2) }}$

☞ We also Know that A cone is drawn inside the cylinder such that both have the same base and same height.

${\sf{ Radius_{(Cone)} = \dfrac{x}{2} }} \\ {\sf{ Height_{(Cone)} = x }} \\ \\ \maltese \ {\boxed{\sf{ Volume_{(Cone)} = \dfrac{1}{3} πr^2h }}} \\ \\ \\ \colon\implies{\sf{ \left[ \dfrac{1}{3} \times \dfrac{22}{7} \times \dfrac{x^2}{4} \times x \right] }} \\ \\ \\ \colon\implies{\sf{ \left[ \dfrac{11x^3}{42} \right] \cdots(3) }}$

Now Finally, We can find the Ratio of their Volumes :-

$\colon\implies{\sf{ V_1 \colon V_2 \colon V_3 }} \\ \\ \\ \colon\implies{\sf{ x^3 \colon \dfrac{11x^3}{14} \colon \dfrac{11x^3}{42} }} \\ \\ \\ \colon\implies{\sf{ \dfrac{42x^3}{42} \colon \dfrac{33x^3}{42} \colon \dfrac{11x^3}{42} }} \\ \\ \\ \colon\implies{\sf{ \dfrac{x^3}{42} \left[ 42 \colon 33 \colon 11 \right] }} \\ \\ \\ \colon\implies{\boxed{\mathfrak\pink{ 42 \colon 33 \colon 11 }}}$

Hence,

• The Ratio of their Volumes is 42:33:11