Volume of a truncated pyramid and squaring the circle
Frustum reckoned at 80; a triangle set equal to the semicircle
Leonardo works the cubic content of a truncated pyramid (upper side 2, base side 4, height 8): the inner prism is 32, the outer prism 128, their difference 96, half of which is 48, plus 32 gives 80 for the whole solid, with 32 for its lateral surface. He then continues his squaring-the-circle arguments: a wedge a b as a forty-fourth of the circle, the triangle a b c set equal to the remainder of semicircle b o n (half-diameter 7, perimeter 22, 22 times 22 equal 154 after Archimedes), and the square a b c d taken as the quadrature less a nearly flat millionth-portion of curve, claimed nearer the truth than Archimedes. Figures at the right show two truncated pyramids marked 8 and 4, a triangle, and circle and semicircle diagrams.
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Volume and surface of a truncated pyramid
The frustum has upper side 2, base side 4, and height 8. Building an inner and an outer prism, subtracting, halving the difference and adding the inner prism gives 80 for the solid; the lateral surface, from a parallelogram plus two triangles, comes to 32.
Working the frustum by multiplication
2 times 2 is 4, times the height 8 makes 32, the lesser prism; 4 times 4 is 16, times 8 makes 128, the greater prism; 128 less 32 is 96, whose half 48 added to 32 makes 80. For the surface, 2 times 8 gives 16, and the two triangles give 16, totalling 32.
Triangle equal to the remainder of the semicircle
The rectilinear triangle a b c is set precisely equal to the remainder of the semicircle b o n, with half-diameter 7 and perimeter 22. The square a b c d is then taken as the quadrature of the circle less the nearly flat curved portion a o b, and the triangle a b p, doubled, squares the semicircle.