Berlin Papyrus 6619/Examples/Problem 1/Historic 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.

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

The method of solution used is the rule of false position.

Assume that:

$\ds x$

$=$

$\ds \dfrac 1 2 + \dfrac 1 4 = \dfrac 3 4$

$\ds y$

$=$

$\ds 1$

$\ds \leadsto \ \ $

$\ds x^2 + y^2$

$=$

$\ds 1 + \dfrac 9 {16}$

But the result should be:

$100 = 64 \times \paren {1 + \dfrac 9 {16} }$

Therefore, our two squares must be $64$ times bigger.

Therefore their sides must be $8$ times bigger.

So the result is:

$\ds x$

$=$

$\ds \dfrac 3 4 \times 8$

$\ds $

$=$

$\ds 6$

$\ds y$

$=$

$\ds 1 \times 8$

$\ds $

$=$

$\ds 8$

$\ds \leadsto \ \ $

$\ds x^2 + y^2$

$=$

$\ds 100$

$\blacksquare$