If an area be contained by a rational straight line and a fifth apotome, the “side” of the area is a straight line which produces with a rational area a medial whole.
若一面积由一条有理线段和一条第五余线围成，则该面积的“边”是一条与有理面积产生中项全体的直线。
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For let the area AB be contained by the rational straight line AC and the fifth apotome AD; I say that the “side” of the area AB is a straight line which produces with a rational area a medial whole. For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, the annex GD is commensurable in length with the rational straight line AC set out, and the square on the whole AG is greater than the square on the annex DG by the square on a straight line incommensurable with AG. [X. Deff. III. 5] Therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into incommensurable parts. [X. 18] Let then DG be bisected at the point E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is incommensurable in length with FG.
设面积AB由有理线段AC和第五余线AD围成，取AD的附加线段DG，则AG、GD为仅平方可通约的有理线段，GD与AC长度可通约，且AG上的正方形比DG上的正方形大一个与AG不可通约的线段上的正方形。
Now, since AG is incommensurable in length with CA, and both are rational, therefore AK is medial. [X. 21] Again, since DG is rational and commensurable in length with AC, DK is rational. [X. 19] Now let the square LM be constructed equal to AI, and let the square NO equal to FK and about the same angle, the angle LPM, be subtracted; therefore the squares LM, NO are about the same diameter.
在AG上作一个等于DG上正方形四分之一且缺一正方形的平行四边形，将其分为AF、FG，则AF与FG长度不可通约。由于AG与CA长度不可通约且均为有理，故AK为中项面；又因DG为有理且与AC长度可通约，故DK为有理。
[VI. 26] Let PR be their diameter, and let the figure be drawn. Similarly then we can prove that LN is the “side” of the area AB. I say that LN is the straight line which produces with a rational area a medial whole. For, since AK was proved medial and is equal to the squares on LP, PN, therefore the sum of the squares on LP, PN is medial.
作正方形LM等于AI，减去与LM同角LPM的正方形NO等于FK，则LM、NO共直径。设PR为直径，作图，可证LN为面积AB的边。
Again, since DK is rational and is equal to twice the rectangle LP, PN, the latter is itself also rational. And, since AI is incommensurable with FK, therefore the square on LP is also incommensurable with the square on PN; therefore LP, PN are straight lines incommensurable in square which make the sum of the squares on them medial but twice the rectangle contained by them rational. Therefore the remainder LN is the irrational straight line called that which produces with a rational area a medial whole; [X. 77] and it is the “side” of the area AB.
由于AK为中项面且等于LP、PN上的正方形和，故LP、PN上的正方形和为中项；又DK为有理且等于二倍矩形LP、PN，故该矩形为有理；且AI与FK不可通约，故LP上的正方形与PN上的正方形不可通约，因此LP、PN为平方不可通约的直线，其平方和为中的，而二倍矩形为有理，故LN为与有理面积产生中项全体的无理直线，即面积AB的边。