If an area be contained by a rational straight line and a sixth apotome, the “side” of the area is a straight line which produces with a medial area a medial whole.
若一面积由一条有理线段和一条第六余线围成，则该面积的“边”是一条与中面积产生中整线的线段。
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For let the area AB be contained by the rational straight line AC and the sixth apotome AD; I say that the “side” of the area AB is a straight line which produces with a medial area a medial whole. For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, neither of them 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 in length with AG. [X. Deff. III. 6] Since then the square on AG is greater than the square on GD by the square on a straight line incommensurable in length with AG, 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 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. But, as AF is to FG, so is AI to FK.
设面积AB由有理线段AC和第六余线AD围成，取AD的附加线DG，则AG、GD为仅平方可通约的有理线段，且均与AC长度不可通约，AG上的正方形比DG上的正方形大一个与AG长度不可通约的线段上的正方形。
[VI. 1] therefore AI is incommensurable with FK. [X. 11] And, since AG, AC are rational straight lines commensurable in square only, AK is medial. [X. 21] Again, since AC, DG are rational straight lines and incommensurable in length, DK is also medial. [X. 21] Now, since AG, GD are commensurable in square only, therefore AG is incommensurable in length with GD. But, as AG is to GD, so is AK to KD; [VI. 1] therefore AK is incommensurable with KD.
由于AG上的正方形比GD上的正方形大一个与AG长度不可通约的线段上的正方形，故在AG上作等于DG上正方形四分之一且缺一正方形的平行四边形，将其分为不可通约部分。设DG中点为E，在AG上作等于EG上正方形且缺一正方形的矩形AF、FG，则AF与FG长度不可通约。
[X. 11] Now let the square LM be constructed equal to AI, and let NO equal to FK, and about the same angle, be subtracted; therefore the squares LM, NO are about the same diameter. [VI. 26] Let PR be their diameter, and let the figure be drawn. Then in manner similar to the above we can prove that LN is the “side” of the area AB. I say that LN is a straight line which produces with a medial 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.
由AF比FG等于AI比FK，得AI与FK不可通约。因AG、AC为仅平方可通约的有理线段，故AK为中线段；又AC、DG为有理线段且长度不可通约，故DK也为中线段。因AG与GD仅平方可通约，故AG与GD长度不可通约，从而AK与KD不可通约。
Again, since DK was proved medial and is equal to twice the rectangle LP, PN, twice the rectangle LP, PN is also medial. And, since AK was proved incommensurable with DK, the squares on LP, PN are also incommensurable with twice the rectangle LP, PN. 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, twice the rectangle contained by them medial, and further the squares on them incommensurable with twice the rectangle contained by them. Therefore LN is the irrational straight line called that which produces with a medial area a medial whole; [X. 78] and it is the “side” of the area AB.
作正方形LM等于AI，减去等于FK且同角的正方形NO，则两正方形同直径。设PR为直径，作图。类似可证LN为面积AB的边。因AK为中线段且等于LP、PN上的正方形和，故LP、PN上的正方形和为中量；又DK为中线段且等于LP、PN所成矩形的二倍，故该二倍也为中量；且AK与DK不可通约，故两正方形和与二倍矩形不可通约；又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LN为称为“与中面积产生中整线”的无理线段，即为面积AB的边。