Answer:
[tex]L = 210[/tex], [tex]W=200[/tex]
[tex]W=210[/tex], [tex]L = 200[/tex]
Step-by-step explanation:
By definition, the perimeter of a rectangle is:
[tex]P = 2L + 2W[/tex]
Where:
P is the perimeter, L is the length and W is the width
Also, the area of a rectangle is:
[tex]A = LW[/tex]
Where L is the length of the base and W is the width.
We know that for this rectangle:
[tex]P = 2L + 2W = 820 ft\\\\A = LW = 42,000 ft ^ 2[/tex]
Now we have two equations and two unknowns (L and W)
Then we solve the system.
[tex]L = \frac{42,000}{W}[/tex]
Now we substitute this relation in the perimeter equation.
[tex]2(\frac{42,000}{W}) + 2W = 820\\\\\frac{42,000}{W} + W = 410\\\\42000 + W ^ 2 = 410W\\\\W ^ 2 -410W + 42000 = 0\\\\(W - 210)(W - 200) = 0\\\\W = 210\\\\W = 200[/tex]
Then for W=210:
[tex]L = \frac{42000}{W}\\\\L = \frac{42000}{210}\\\\L = 200[/tex]
And for W=200
[tex]L = \frac{42000}{200}\\\\L = 210[/tex]
Answer
[tex]210 \: by \: 200[/tex]
EXPLANATION
Let l and w be the dimensions of the parking lot
The perimeter of the parking lot is given by
[tex]p =2( l + w)[/tex]
This implies that
[tex]820 = 2(l + w)[/tex]
Dividing both sides by 2
[tex]l + w = \frac{820}{2} [/tex]
[tex]l + w = 410......eq1[/tex]
Area of the of the parking lot is given by
[tex]l \times w = 42000...........eqn2[/tex]
putting eqn 1 into ran 2
[tex]410l - {l}^{2} = 42000[/tex]
[tex] 0= {l}^{2} - 410l + 42000[/tex]
[tex] {l}^{2} - 210l - 200l+ 42000 = 0[/tex]
[tex] ( {l}^{2} - 210l )-1( 200l - 42000 )= 0[/tex]
[tex]l ( {l} - 210l )- 200(l - 210)= 0[/tex]
[tex]( {l} - 210l )(l - 200)= 0[/tex]
[tex]l = 210 \: or \: 200[/tex]
if
[tex]l = 200[/tex]
then
[tex]w = 210[/tex]
and also if
[tex]l = 210[/tex]
then
[tex]w = 200[/tex]
Hence the dimensions of the parking lot is
[tex]210 \: by \: 200[/tex]
