Question 1207290
as far as i can tell, the enclosed region has 2 lengths and 3 widths.
2 lengths and 2 widths are on the outside of the enclosed region.
one width splits the enclosed region into two smaller regions that are equal in area.


i get the following two possibilities.


the length is 27 and the width is 38.
the length is 18 and the width if 57.


length * width = 1026 in both cases.
2 lengths and 3 widths = 168 in both cases.


my equations were:


length * width = 1026
2 * length + 3 * width = 168.


i solved for length * width from length * width = 1026.
this led to length = width / 1026.


i substituted for length in 2 * length + 3 * width = 168 to get:
2 * 1026 / width + 3 * width = 168.


i multiplied both sides of taht equation by width to get:
2 * 1026 + 3 * width squared = 168 * width.


i subtracted 168 * width from both sides of the equation to get:
2 * 1026 + 3 * width squared - 168 * width = 0
i simplified and order in descending order of degree to get:
3 * width squared - 168 * width + 2052 = 0


i factored this quadratic equation to get width = 38 or 18.


when width was 38, length was 27.
when width was 18, length was 57.


length * width = 1026 in both equations.
2 * length + 3 * width = 168 in both equtions.


those appeared to be two feasible solutions to the problem as i understood it.
my diagram looked like this:


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