Let's assume
length of rectangle =L
width of rectangle =W
You enclose 3 sides of the garden with 40 feet of fencing
so, we get
[tex]L+2W=40[/tex]
now, we can solve for L
[tex]L=40-2W[/tex]
we know that
area of rectangle is
[tex]A=L*W[/tex]
[tex]100=L*W[/tex]
now, we can plug
[tex]100=(40-2W)*W[/tex]
now, we can solve for W
[tex]-2W^2+40W-100=0[/tex]
we can use quadratic formula
[tex]W=\frac{-40+\sqrt{40^2-4\left(-2\right)\left(-100\right)}}{2\left(-2\right)}[/tex]
[tex]W=\frac{-40-\sqrt{40^2-4\left(-2\right)\left(-100\right)}}{2\left(-2\right)}[/tex]
we can take anyone value ..because both are giving positive value
first dimensions:
[tex]W=2.929[/tex]
now, we can find L
[tex]L=40-2*2.929[/tex]
[tex]L=34.142[/tex]
so, length is 34.142feet
width is 2.929 feet
Second dimensions:
[tex]W=17.071[/tex]
now, we can find L
[tex]L=40-2*17.071[/tex]
[tex]L=5.858[/tex]
so, length is 5.858feet
width is 17.071 feet