If an area is contained by a rational (straight line) and a fifth binomial (straight line) then the square root of the area is the irrational (straight line which is) called the square root of a rational plus a medial (area) . * For let the area $AC$ be contained by the rational (straight line) $AB$ and the fifth binomial (straight line) $AD$, which has been divided into its (component) terms at $E$, such that $AE$ is the greater term. * So I say that the square root of area $AC$ is the irrational (straight line which is) called the square root of a rational plus a medial (area) .
If the rational straight line has unit length then this proposition states that the square root of a fifth binomial straight line is the square root of a rational plus a medial area: i.e., a fifth binomial straight line has a length \[\alpha\,(\sqrt{1+\beta}+1)\] whose square root can be written \[\rho\,\sqrt{\frac{\sqrt{1+\delta^{2}}+\delta}{2\,(1+\delta^{2})}}+\rho\,\sqrt{\frac{\sqrt{1+\delta^{2}}-\delta}{2\,(1+\delta^{2})}},\] where \[\rho=\sqrt{\alpha\,(1+\delta^{2})}\quad\text{ and }\quad\delta^{2}=\beta.\] This is the length of the square root of a rational plus a medial area (see [Prop. 10.40]), since $\rho, \delta,\alpha,\beta$ are all positive rational numbers.
Proofs: 1
