[ 555 ] , 
The area of the triangle, AHB, is equal to^the 
areas of the three feftors A TO, BOT, H'^'T, aoded 
to the interftitial area TOT. But the triangle APIB 
is equilateral. Therefore each of the feftors A TO, 
EOT, H T T is of the circle to which it belongs : 
and, the circles being equal, the three fedbors are equal 
to the half of any one of the circles. Therefore, the 
area of the triangle AHB is equal to | the area 
of one circle ^as of A) added to the intcilditial aiea 
yOT. Tlierefore, from the area „of the triangle 
A H B take | the area of the circle A, and there 
will remain the interftitial area TOT. 
Now, if the sadius AO be put zr; i, each iide of 
the triangle AHB will be 2. 
Therefore, the area of the triangle AHB| _ ^ - q- 
will be . . -> 
But the radius being J, | the area of thed^^^^^g 
circle A is 
The difference is 0,1612 
And this is, the interditial area TOT, the half area 
of the circle A being Therefore, the femi- 
circle is to the interftice as 1,5708 to 0,1612, or as 
9,74 to I, or as 39 to 4, very nearly. 
Corollary. 
If a parcel of equal circles be fo difpofed upon a 
plane furface of any figure whatfoever, that the centers 
of every three adjacent circles are fituated at the angles 
of equal equilateral triangles, having Tides greater than 
the diameters of the circles, but greater in a finite 
proportion, the ultimate proportion of the fpace 
^ 4 B 2 covered 
