Babylonian Mathematics/Examples/Sum of Squares

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Example of Babylonian Mathematics

An area $A$, consisting of the sum of $2$ squares, is $1000$.

The side of one square is $10$ less than $\dfrac 2 3$ of the other square.

What are the sides of the squares?


Solution

The lengths of the sides of the $2$ squares are $10$ and $30$.


Proof

Let $x$ be the length of the side of the smaller square.

Let $y$ be the length of the side of the larger square.

Thus:

\(\text {(1)}: \quad\) \(\ds x^2 + y^2\) \(=\) \(\ds 1000\)
\(\text {(2)}: \quad\) \(\ds x\) \(=\) \(\ds \dfrac {2 y} 3 - 10\)
\(\ds y\) \(=\) \(\ds \dfrac {3 \paren {x + 10} } 2\)
\(\ds \leadsto \ \ \) \(\ds x^2 + \paren {\dfrac {3 \paren {x + 10} } 2}^2\) \(=\) \(\ds 1000\) substituting for $y$ in $(2)$
\(\ds \leadsto \ \ \) \(\ds 4 x^2 + 9 \paren {x^2 + 20 x + 100}\) \(=\) \(\ds 4000\) simplifying
\(\ds \leadsto \ \ \) \(\ds 13 x^2 + 180 x - 3100\) \(=\) \(\ds 0\) simplifying
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \dfrac {-180 \pm \sqrt {180^2 + 4 \times 13 \times 3100} } {26}\) Solution to Quadratic Equation
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \dfrac {-180 \pm 440} {26}\) simplifying
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds 10 \text { or } x = - \dfrac {310} {13}\) performing the arithmetic


We can dispose of the negative value, leaving us with:

\(\ds x\) \(=\) \(\ds 10\)
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds 3 \paren {10 + 10} 2\)
\(\ds \) \(=\) \(\ds 30\)

and we see that:

\(\ds 10^2 + 30^2\) \(=\) \(\ds 100 + 900\)
\(\ds \) \(=\) \(\ds 1000\)

as required.

$\blacksquare$


Sources

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