Assume that the width is the side parallel to the wall.Then the perimeter of the fence is:`P=2l+w` Since the total length of the three sides is 400m, thenthe value of perimeter is 400.`400= 2l+w` Then, solve for w.`400-2l=w` (Let this be EQ1.) Next, use...

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Assume that the width is the side parallel to the wall.

Then the perimeter of the fence is:

`P=2l+w`

Since the total length of the three sides is 400m, then

the value of perimeter is 400.

`400= 2l+w`

Then, solve for w.

`400-2l=w` (Let this be EQ1.)

Next, use the given area to set-up the second equation.

`A=l*w`

`450=l*w`

Substitute EQ1.

`450=l*(400-2l)`

`450=400l-2l^2`

Set one side equal to zero.

`450-400l+2l^2=400l-2l^2-400l+2l^2`

`2l^2-400l+450=0`

To simplify, divide both sides by 2.

`(2l^2-400l+450)/2=0/2`

`l^2-200l+225=0`

Then, use the quadratic formula to solve for l.

`l=(-b+-sqrt(b^2-4ac))/(2a)=(-(-200)+-sqrt((-200)^2-(4*1*225)))/(2*1)`

`l=(200+-sqrt39100)/2`

So, values of l are:

`l=198.869` and `l = 1.131`

Next, solve for w. To do so, plug-in the values of l to EQ1.

`l=198.869` ,

`w=400-2l=400-2(198.869)=2.262`

`l=1.131` ,

`w=400-2l=400-2(1.131)=397.738` **Hence, the three sides of the rectangular yard can have either the following dimensions:****`198.869 xx 198.869 xx 2.262` meters or****`1.131xx 1.131xx 1.131` meters.**