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Two cubes have surface areas of 81 square inches and 144 square inches. What is the ratio of the volume of the small cube to the volume of the large cube?

Respuesta :

We know, surface area of cube = 6a²

So, side length of 1st cube = a² = 81/6 
a = 3.67

Side length of 2nd cube = a² = 144/6
a = 4.90

Now, volume of cube = a³

Volume of 1st cube = 3.67³ = 49.43

Volume of 2nd cube = 4.90³ = 117.64

So, Ratio = 49.43 / 117.64 = 1/2.38

In short, Your Answer would be: 1: 2.38

Hope this helps!
Cetacea

The ratio of the volume of the small cube to the volume of the large cube is = [tex]\frac{27}{64}[/tex].

For a cube:

Let side length = x

So, surface area = [tex]6x^2[/tex].

Volume = [tex]x^3[/tex]

Let the side lengths of the cubes are x and y.

Given that: Two cubes have surface areas of 81 square inches and 144 square inches.

So, [tex]6x^2=81 , 6y^2=144[/tex] for the cubes.

So,

[tex]\frac{6x^{2}}{6y^{2}}=\frac{81}{144}\\\frac{x^{2}}{y^{2}}=\frac{81}{144}\\\left(\frac{x}{y}\right)^{2}=\left(\frac{9}{12}\right)^{2}\\\frac{x}{y}=\frac{9}{12}\\\frac{x}{y}=\frac{3}{4}\\\left(\frac{x}{y}\right)^{3}=\left(\frac{3}{4}\right)^{3}\\\frac{x^{3}}{y^{3}}=\frac{27}{64}[/tex]

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