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(diagram 3 = general diagram)
(diagram 1) Again, let AG be a balance and its middle B, and let DEKH be a trapezoid having right angles at points E, H and lines KD, EH converging to G, and the ratio that AB has to BH, let trapezoid DKEH have this to M, and let trapezoid DKEH have this ratio to L, that which AB has to BE, and let trapezoid DKEH be suspended from the balance at E, H, and let area Z be suspended at A, and let it incline equally to the trapezoid holding as it is now positioned. I say that Z is larger than L, but less than M.
(diagram 2) For I took the center of the weight of trapezoid DKEH. Let it be Q. It will be taken similarly to that previously. And I draw QI parallel to DE. (diagram 3) And so if the trapezoid is suspended from the balance at I, but is released from EH, it will maintain the same situation and will incline equally to Z for the same reasons as those previously. (diagram 4) But since the trapezoid when suspended at I inclines equally to Z when suspended at A, the trapezoid has the same ratio to Z which AB has to BI. And so it is clear that AKEH has a larger ratio to L than to Z, but a smaller to M than to Z. Thus Z is larger than L and smaller than M.