METHOD OF 



14 



If the field 

 have five sides, 

 the lines A C 

 and C E will 

 divide it into 

 three triangles, 

 and its content 

 is then found as 

 follows: Sup- 

 pose A C 402 

 yards, BG154, 

 E F 162, E C 

 303, and D H 

 137 Now, to 

 find the content 

 of the parts A 

 B, C E, we have 

 B G 154, and E P 162, the sum of which is 306, and the 

 half of this is 158 for a mean width ; then 



A. R. P. 



C 100 yards wide, is 8 1 9 

 402 yards long and 1 50 do 4 24 



/ 8 do 2 26 



Content of the parts A B, C E, 130 9 



And to find the content of the triangle E C D, we have 

 D H 137 yards, the half of which is 68 yards for a mear 

 Width, and the length E C 303 yards 



A. R. P. 



303 yards long and 60 yards wide 33 1 



...do do... 8 do 02 



Content of triangle E C D, 

 do. of the other part, 



4 1 

 13 



Whole content, 17 1 20 



