If an area is contained by a rational (straight line) and a sixth binomial (straight line) then the square root of the area is the irrational (straight line which is) called the square root of (the sum of) two medial (areas). * For let the area $ABCD$ be contained by the rational (straight line) $AB$ and the sixth 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 $AC$ is the square root of (the sum of) two medial (areas).
If the rational straight line has unit length then this proposition states that the square root of a sixth binomial straight line is the square root of the sum of two medial area: i.e., a sixth binomial straight line has a length \[\sqrt{\alpha}+\sqrt{\beta}\] whose square root can be written \[\alpha^{1/4}\left(\sqrt{\frac 12+\frac{\delta}{2\sqrt{1+\delta^2}}}+\sqrt{\frac 12-\frac{\delta}{2\sqrt{1+\delta^2}}}\right),\] where \[\delta^2=\frac{\alpha-\beta}\beta.\] This is the length of the square root of the sum of two medial area (see [Prop. 10.41]).
Proofs: 1
