# Q.E.D.

18 19 InsIde out two proofs in wedges Archimedes demonstrated how to mathematically kill two birds with one stone by using an ingenious idea to relate the insides and outsides of circles and spheres. Here is a sketch of his argument. First he dissected the circle of radius r into a number of equal wedges opposite, top and arranged them into a roughly rectangular slab. He then observed that this can be done with an ever increasing number of wedges, and that as the number of wedges grows, the slab becomes indistinguishable from a rectangle whose short side has length r and whose long side is half the circumference of the circle. Therefore the area of the rectangle coincides with that of the circle giving the formula area of circle 1 2 . circumference of circle . r. We arrive at the same result by calculating the area of the sawtooth figure just note that one of the triangles has area 1 2 . base length . r, and that the bases of the triangles sum to the circumference. To derive a similar formula for a sphere of radius r, Archimedes dissected it into triangular cones whose common vertex is the centre of the sphere and whose bases are contained in the surface of the sphere lower, opposite. These cones can play the role of the triangles in the sawtooth figure, and since from page 12 the volume of one of these cones is 1 3 . base area . r we obtain the formula volume of sphere 1 3 . surface area of sphere . r. As the grand finale, we plug the formul for the circumference of a circle of radius r and that of the volume of a sphere of radius r see previous page into our new relationships to conclude that the area of the circle is p r 2 and that the surface area of the sphere is 4 p r 2.