If the surface area of a sphere and a cube are equal then how do you show that their volumes are in the ratio $\sqrt{7} : \sqrt{5}$ $sqrt(7) : sqrt(5)$ ?

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If the surface area of a sphere and a cube are equal then how do you show that their volumes are in the ratio $\sqrt{7} : \sqrt{5}$ $sqrt(7) : sqrt(5)$ ?

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Answer 1 · 2016-03-21T20:10:15

This proposition is false...

Explanation:

Suppose the sphere has radius $r$ $r$ and the cube side $t$ $t$ .

Since the surface areas are equal, we have:

$4 π r^{2} = 6 t^{2}$ $4pi r^2 = 6t^2$

So ${(\frac{r}{t})}^{2} = \frac{3}{2 π}$ $(r/t)^2 = 3/(2pi)$ and ${(\frac{r}{t})}^{3} = {(\frac{3}{2 π})}^{\frac{3}{2}}$ $(r/t)^3 = (3/(2pi))^(3/2)$

Then:

$\frac{V_{sphere}}{V_{cube}} = \frac{\frac{4}{3} π r^{3}}{t^{3}} = \frac{4}{3} π {(\frac{r}{t})}^{3} = \frac{4}{3} π {(\frac{3}{2 π})}^{\frac{3}{2}} \approx 1.38197659788534191701$ $V_"sphere" / V_"cube"=(4/3 pi r^3)/t^3=4/3pi (r/t)^3 = 4/3pi (3/(2pi))^(3/2) ~~ 1.38197659788534191701$

Whereas:

$\frac{\sqrt{7}}{\sqrt{5}} \approx 1.18321595661992320851$ $sqrt(7)/sqrt(5) ~~ 1.18321595661992320851$

So the proposition is false.

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If the surface area of a sphere and a cube are equal then how do you show that their volumes are in the ratio $\sqrt{7} : \sqrt{5}$ $sqrt(7) : sqrt(5)$ ?

1 Answer

Explanation:

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If the surface area of a sphere and a cube are equal then how do you show that their volumes are in the ratio $\sqrt{7} : \sqrt{5}$ $sqrt(7) : sqrt(5)$ ?