Problem

As shown in the figure, there is an object with $5$ floors and a total volume of $335.625 \ m^3$. Each floor is square, $0.5 \ m$ high, and the floor immediately below is $0.5 \ m$ wider on all sides than the floor above. Find the length of one side of the top surface (the fifth floor surface).

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Solution

When one side of the surface of the 5th floor is $x \ m$, the volume of the 5th floor is

$$x^2×\frac{1}{2}=\frac{x^2}{2} \ m^3.$$

Since one side of the 4th floor is $x+1 \ m$, its volume is

$$(x+1)^2×\frac{1}{2}=\frac{x^2+2x+1}{2} \ m^3.$$

Since one side of the third floor is $x+2 \ m$, its volume is

$$(x+2)^2×\frac{1}{2}=\frac{x^2+4x+4}{2} \ m^3.$$

Since one side of the second floor is $x+3 \ m$, its volume is

$$(x+3)^2×\frac{1}{2}=\frac{x^2+6x+9}{2} \ m^3.$$

Since one side of the first floor is $x+4 \ m$, its volume is

$$(x+4)^2×\frac{1}{2}=\frac{x^2+8x+16}{2} \ m^3.$$

Therefore, the total volume is

$$\frac{x^2}{2}+\frac{x^2+2x+1}{2}+\frac{x^2+4x+4}{2}+\frac{x^2+6x+9}{2}+\frac{x^2+8x+16}{2} =\frac{5x^2+20x+30}{2} \ m^3.$$

 And since we know this is $335.625 \ m^3$,

$$\frac{5x^2+20x+30}{2}=335.625,$$

$$x^2+4x-128.25=0,$$

$$∴ \quad x=-2+\sqrt{4+128.25}=-2+11.5=9.5 \ m.$$

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Reference

Yoshikazu Yamakawa, ed. (1997) Okayama ken no Sangaku (Sangaku in Okayama Prefecture), p.53; p.296.