x^2 + y^2 = diagonal^2 = d^2
x y = 4000 so y = 4000/x
d = x+10 so d^2 = x^2 + 20 x + 100
x^2 + 16*10^6/x^2 = x^2 + 20 x + 100
16*10^6 = 20 x^3 + 100 x^2
5 x^3 + 25 x^2 - 4*10^6 = 0
I do not see an easy way to solve that. Use iteration, x and 5x^3+25 x^2
x = 10^2 --> 5,250,000 too big
x = 80 --> 2,720,000 too small
x = 90 --> 3,847,500 getting close
x = 92 --> 4,333,568 very close
x = 91 --> 3,974,880 between 91 and 92
x = 91.5-> 4,039,610 between 91 and 91.5
so 91 to nearest foot
91^2 + y^2 = (91+10))^2
8281 +y^2 = 10201
y = 43.8 or 44
91 by 44
let one side be x
then the diagonal is x+10
let the third side by y
x^2 + y^2 = (x+10)^2
x^2 + y^2 = x^2 + 20x + 100
y^2 = 20x + 100
y = √(20x+100)
so the area of the triangle is half the rectangle area
xy = 4000
x√(20x+100) = 4000
x^2(20x+100) = 16000000
20x^3 + 100x^2 - 16000000 = 0
x^3 + 5x^2 - 800000 = 0
I must admit that I used Wolfram to solve this
and I got x = 91.1947
then y = 43.8622
and the diagonal would be 101.1947
area of rectangle = xy = (91.1947)(43.8622) = 4000.00017 (close enough)
I will let you check if Pythagoras also works out, it does.
I don't know what method you would have to solve that cubic, since it does not factor, and therefore does not have rational roots.
Do you know Newton's Method ?
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