BOOK V. 11. 9-10 



But if the area is to be less than a semicircle, we 

 shall measure the arc as follows : let there be an arc 

 the base of which measures 16 feet and the depth 

 4 feet. I add the base to the depth, which together 

 make 20 feet. This I multiply by 4, making 80, of 

 which the half is 40. Again, the half of 16 feet, which 



16 feet 



form the base, is 8. This I multiply by itself, making 

 61. I then take a fourteenth part of this, which 

 make 4 feet and a little more.'* This you will add to 

 40, and together they make a total of 44. This I 

 declare to be the number of square feet in the arc, 

 which is equivalent to half a scripulum (576^ of a 

 mgerum) less ^ of a scripulum.^ 



If the area has six angles, it is reduced to square 10 

 feet in the following manner. Let there be a 

 30 feet 



30 feet 



30 feet 



Total area 



2340 

 square feet 



30 feet 



30 feet 

 hexagon, each side of which measures 30 feet. I 

 multiply one side by itself: 30 times 30 makes 900. 

 Of this sum I take one-third, which is 300, a tenth part 

 of which is 90 : total 390. This must be multiplied 



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