|
Call the shorter edge of the corn field C, the shorter edge of the hay field B and the longer egde A.
The problem states c*(c+1) = 3*ab [area] and 3*2*(c+c+1) = 2*(a+b).
The second equation can be written as c = (a+b-3)/6.
Substituting into the first equation yields ((a+b-3)/6)*((a+b+3)/6) = 3*a*b
This equation can be solved for a as follows:
(a+b)^2 - 9 = 108*a*b
a^2 + 2*a*b + b^2 - 108*a*b = 9
(a - 53*b)^2 = 9 + 2808*b^2
a = 53*b +/- sqrt(9 + 2808*b^2)
Testing values of b less than 100, as specified in the problem, gives only one value of b: b=3.
If b=3 then a = 109 +/- 159, a = -50 or 318. Since a must be positive, a = 318. c then equals 53
The corn field is 53x54 and the hay field is 3x318. |
