(diagram 1) 11. Again, let AG be a balance and its middle B, and let KDTR be a trapezoid having sides KD, TR converging to G, with perpendiculars DR, KT to BG, and let DR fall on B, and let trapezoid DKTR have this ratio to L, that which AB has to BH, and let trapezoid DKTR be suspended from the balance at B, H and Z at A, and let Z incline equally to trapezoid DKRT holding as it is now positioned. In a way similar to those earlier, area Z will be proved smaller than L similarly to those previously.

In this sample proof, I repeat Archimedes argument from QP 10, with such changes as are necessary.

(diagram 2) For let BH be cut at E in such a way that DE have this ratio to EH, that which the double of DR and KT have a ratio to the double of KT and DR. (diagram 3) (Note that the center of weight of weight at Q will be the intersection of the line joining the bisections of DR, KT and the parallel to DR from E, or in other words, the perpendicular to BG at E).

And after ENS being drawn through E parallel to BR, let SN be bisected at Q. Q is, in fact, the center of the weight of trapezoid DKTR. (diagram 2) And so if trapezoid DKTR is suspended at E and if it is released from points B, H, it maintains the same situation for the same (reasons) as those before and inclines equally to area Z. (diagram 4) And so since trapezoid BDHK when suspended at E inclines equally to area Z when suspended at A, as AB will be to BE trapezoid BDHK will be to area Z. Therefore, trapezoid BDHK will have a larger ratio to Z than to L, since AB will also have a larger ratio to BE than to BH. Thus Z will be smaller than L.

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