a very careful examination.

struction of the diagram:

The following is the con

On the straight line A B, describe the right-angled triangle A B C, making B C, equal to three-fourths of AB. On the sides of the triangle A B C, describe the squares ABFG, BDE C, and A CHI. On A B, describe the square A BLK, and about it circumscribe the circle X. About the circle X, circumscribe the square M N O P; and about the square M N O P, circumscribe the circle Y. With the point B as centre, and A B as radius, describe the semi-circle LRASF, forming the inscribed quadrants BLRA B, and BASFB, in the squares A BLK and ABFG. With the point K as centre, and K L or K A as radius, draw the curved line AT L, describing the quadrant K ATL K, and the trapezium A TLRA, in the square ABLK. Draw A L, the diagonal of the square ABLK.

In the first place, I must direct attention to certain peculiar facts, with reference to this diagram.

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Let the side A B, of the right-angled triangle A B C, be 4. The side B C, is, by construction, equal to threefourths of A B, (4), = 2 × 4, 75 × 4, = 3; and √A B2 + B C2, = √√42 + 32, √= 16 + 9, = √25, = 5; will be the value of A C, the third side of the triangle A BC; and this triangle represents the first commensurable right-angled triangle, of which the values of the three sides can be given in integers or whole numbers.

Then, the value of A C is 5, and the value of B C is 3, and the arithmetical mean between these two numbers, equal to (5+3)= 4, is the given value of A B, the remaining side of the triangle ABC; and in every rightangled triangle, of which the two sides adjacent to the right angle, are in the ratio of 3 to 4, the value of the longer

of the two sides, adjacent to the right angle, is the arithmetical mean between the values of the other two sides.

Again, Let the value of the side A B, of the rightangled triangle A B C, be 1; and the sides B C, A B, adjacent to the right-angle, in the ratio of 3 to 4. Then, A B2 = 12 = 1, will be the area of square A B FG. 2 (1), = 2 × 1, = '75 × 1, = '75, will be the value of the I, X I, side B C, of the triangle A B C; and 7525625, will be the area of the square B D E C, described on the side B C, of the triangle A B C. And A B+ B C2, = √12 + ·752, = √1 + 5625, = = 1.25, will be the value of the side A C, of the triangle A B C; and 125 15625, will be the area of the square A CHI, described on A C, the third side of the triangle A B C.

=

2

√15625,

Again, Let the value of the side A B, of the rightangled triangle A B C, be √1; and the sides B C, A B, adjacent to the right angle, in the ratio of 3 to 4. Then, A B2, = √i, = , I, will be the area of the square A B F G, described on the side A B, of the triangle A B C; 2 ( √1), = √16 x I, = √5625 × 1, 5625, will be the value

9

2

2

of the side B C; and B C2 = √5625 = 5625, will be the area of the square B D E C, described on the side B C, of the triangle A BC; and A B2 + B C2, = √1 ̊+ √'5625, =1+·5625,=1.5625; will be the area of the square ACHI, described on AC, the third side of the triangle A B C; and whether we adopt I or √1, as the value of A B, of the triangle A B C, we arrive at the same values of the areas of the squares described on its sides; but we have worked out this result, by means of different arithmetical symbols. Then, √15625, 125, is the value of the side AC; and √562575, is the value of the side B C, in the triangle A B C; and the arithmetical mean between these

two numbers, equal to (1:25 +75) = 1, is the given value of A B, the remaining side of the triangle A B C.

The area of square A B F G, and the area of square BDE C, are together equal to the area of square A CHI. Hence, it is evident, that if the areas of any two geometrical figures, of whatever form, be together equal to the area of a third geometrical figure, the areas of the three figures will be severally equal to the areas of squares, described on the sides of a right-angled triangle; and the square root of the areas of the three figures, will give the value of the sides of such triangle; and if the value of the middle side, be the arithmetical mean between the values of the other two sides, the triangle will be one, of which the two sides adjacent to the right angle, are in the ratio of 3 to 4.

In my Pamphlet I have shewn, that from any two given numbers, a commensurable right-angled triangle may be obtained; but from commensurable right-angled triangles, so obtained, in no case will the two sides adjacent to the right angle be in the ratio of 3 to 4, with one exception. (The exception being, when the given numbers are 1 and 2, which produce the first commensurable right-angled triangle, of which the values of the sides can be given in integers or whole numbers.)

In treating of diagram No. 10, (see Fig. XIX.) I have demonstrated, and here assume it to be admitted, that the quadrants ARL B A, and A T L K A, inscribed in the square A K L B, are each equal in area, to the area of a circle inscribed in the same square.

Let A B, the side of square A K L B, be 4, and let the area of a circle of which the diameter is unity, be •78125.

Then, 416, will be the area of square AKL B;