Respuesta :
Let the width be W and length be L
LW = 35
L = 35/W
Perimeter = 2 ( Length + Width )
27 = 2 ( 35/W + W )
13.5 = (35+ W^2) / W
W^2 - 13.5W + 35 = 0
W^2 - 10W - 3.5W +35 = 0
W ( W -10) - 3.5 ( W - 10) = 0
( W-3.5) ( W-10) = 0
W = 3.5 or 10
ANSWER Width = 3.5' Length = 10'
LW = 35
L = 35/W
Perimeter = 2 ( Length + Width )
27 = 2 ( 35/W + W )
13.5 = (35+ W^2) / W
W^2 - 13.5W + 35 = 0
W^2 - 10W - 3.5W +35 = 0
W ( W -10) - 3.5 ( W - 10) = 0
( W-3.5) ( W-10) = 0
W = 3.5 or 10
ANSWER Width = 3.5' Length = 10'
The dimension of the section is 3.5 by 10 feet
Part I : Solve for W
The equation is given as;
P = 2(L + W).
Substitute 27 for P
2(L + W) = 27
Divide through by 2
L + W = 13.5
Subtract L from both sides
W = 13.5 - L
Part II: Write the equation of area
We have:
A = LW
Substitute W = 13.5 - L
A = (13.5 - L) * L
Expand
A = 13.5L - L^2
A = 35.
So, we have:
13.5L - L^2 = 35
Multiply through by 2
27L - 2L^2 = 70
Rewrite as:
2L^2 - 27L + 70 = 0
Part III: Solve for L using quadratic formula
We have:
2L^2 - 27L + 70 = 0
This means that:
a = 2, b = -27 and c = 70
The quadratic formula is:
[tex]L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}[/tex]
So, we have:
[tex]L = \frac{27 \pm \sqrt{(-27)^2 - 4*2*70}}{2*2}}[/tex]
Evaluate
[tex]L = \frac{27 \pm 13}{4}}[/tex]
Split
L = (27 + 13)/4 and (27 - 13)/4
Solve
L = 10 and 3.5
Part IV: Calculate W
We have:
W = 13.5 - L
This means that:
W = 13.5 - 10 and W = 13.5 - 3.5
Evaluate
W = 3.5 and 10
Hence, the dimension of the section is 3.5 by 10 feet
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