# A cube is divided into two cuboids

A cube is divided into two cuboids. The surfaces of those cuboids are in the ratio $7: 5$. Calculate the ratio of the volumes.

How can I calculate this?

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Without loss of generality we can let the sides of the original cube be $1$.

The larger of the two cuboids has four of its sides equal to say $x$. (The other $8$ sides are $1$.) Then the smaller cuboid has four of its sides equal to $1-x$.

The surface area of the larger cuboid is $2+4x$. (Two $1\times 1$ faces, and four $x\times 1$ faces). Similarly, the surface area of the smaller cuboid is $2+4(1-x)=6-4x$.

We are told that $$\frac{2+4x}{6-4x}=\frac{7}{5}.$$ Solve. We get $x=\frac{2}{3}$, making $1-x=\frac{1}{3}$. So the volumes are in the ratio $2:1$.

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Suppose the original cube had side $a$. And let dimensions of the resulting cuboids be $a \times a \times b$ and $a \times a \times c$, where $a = b+c$.

The surface areas of the resulting cuboids is $2\left(a \cdot a + a \cdot b + a \cdot b\right) = 2 a(a+2 b)$ and $2a(a+2 c)$, and volumes $a^2 b$ and $a^2 c$ respectively. It is known that: $$\frac{7}{5} = \frac{a+2b}{a+2c} \quad b+c = a$$ Solving this gives $b = \frac{2}{3} a$ and $c=\frac{a}{3}$, hence the ratio of volumes is $\frac{a^2 b}{a^2 c} = \frac{b}{c} = 2$.

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You have a cube with say side length $s$. So the volume is $s^3$. You cut this cube into two cuboids. So one cuboid will have side lengths $s$, $s$, and $s - x$ for some $x$. The other cube will have side lengths $s$, $s$, and $x$. Now

• Find expressions that give you the surface area of each of the cuboids. These expressions will be functions of $x$ (and the constant $s$).
• You know that when dividing these two expressions by each other you get $\frac{7}{5}$. So you solve for x.
• Now that you have $x$, you can find the volumes of each of the cuboids. Then you can find their ratio.
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