Then, let the side L M, of the triangle L M N, in dia-
gram No. 2, be equal to a side of the equilateral triangle
A F G, in diagram No. I,= V48. The side N M, is to the
side L M, in the triangle L M N, in the ratio of 3 to 4, by
construction. Therefore,! ( ^48),= V^X48, ^5625x48,
= ^TJfy will be the value of the side N M ; * ( ^48^
= n/II X 48, = N/rs62S x~48, = ^75^ will be the value
of the side L N, in the triangle L M N ; and the following

results necessarily ensue :

— 2
The superficial area of square P, must be ^/27 = 27, and

is exactly equal to three and one-eighth times the super-
ficial area of the triangle B E D, in diagram No. i. The
superficial area of square R, must be ^/48 = 48, and is
exactly equal to three and one-eighth times the superficial
area of the triangle B E A, in diagram No. i. The

a

superficial area of square T, must be ^75 = 75> and is
exactly equal to three and one-eighth times the superficial
area of the triangle A B C, in diagram No. i.
Then, the area of square P, plus the area of square
R, plus the area of square T, equal to 27 + 48 + 75 =
150; and the squares P, R, and T, are together equal in
superficial area, to three times the superficial area of
circle X.
Again, let the diameter of a circle be equal to twice
the diameter of circle X = 16. Then, 16* = 256, will be



I $6 THE QUADRATURE OF THE CIRCLE.
