Berlin Papyrus 6619/Examples/Problem 1/Modern Proof

Problem 1 from Berlin Papyrus $\mathit { 6619 }$

If it is said to thee:

the area of a square of $100$ is equal to that of two smaller squares.

The side of one is $\frac 1 2 + \frac 1 4$ the side of the other.

Let me know the sides of the two unknown squares.

Solution

The sides of the two unknown squares are $6$ and $8$.

Proof

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

We have:

\(\ds x^2 + y^2\)

\(=\)

\(\ds 100\)

Pythagoras's Theorem

\(\ds \leadsto \ \ \)

\(\ds x\)

\(=\)

\(\ds \dfrac 3 4 y\)

Definition of $x$

\(\ds \leadsto \ \ \)

\(\ds x^2\)

\(=\)

\(\ds \dfrac 9 {16} y^2\)

squaring both sides

\(\ds \leadsto \ \ \)

\(\ds \dfrac 9 {16} y^2 + y^2\)

\(=\)

\(\ds 100\)

\(\ds \leadsto \ \ \)

\(\ds \paren {9 + 16} y^2\)

\(=\)

\(\ds 1600\)

multiplying both sides by $16$

\(\ds \leadsto \ \ \)

\(\ds 25 y^2\)

\(=\)

\(\ds 1600\)

\(\ds \leadsto \ \ \)

\(\ds y^2\)

\(=\)

\(\ds 64\)

dividing both sides by $25$

\(\ds \leadsto \ \ \)

\(\ds y\)

\(=\)

\(\ds \pm 8\)

taking square root of both sides

As this is the side of an actual square we are investigating here, we can dispose of the negative square root.

Thus we have:

$y = 8$

and so:

$x = \dfrac 3 4 \times 8 = 6$

$\blacksquare$