Answer :
Let the width be x.
Width = x
Length = x + 20
The area is 200 ft²
x(x + 20) = 200
x² + 20x - 200 = 0
a = 1, b = 20, c = -200
[tex]x = \dfrac{-b\pm \sqrt{b^2-4ac} }{2a} [/tex]
[tex]x = \dfrac{-20 \pm \sqrt{20^2-4(1)(-200)} }{2(1)} [/tex]
[tex]x = \dfrac{-20 \pm \sqrt{400+800} }{2} [/tex]
[tex]x = \dfrac{-20 \pm \sqrt{1200} }{2} [/tex]
[tex]x = -27.32 \text { (rejected, length cannot be negative) } \or \ x = 7.32[/tex]
Width = 7.32
Length = 7.32 + 20 = 27.32
Answer: The pen is 7.32 ft by 27.32 ft
Width = x
Length = x + 20
The area is 200 ft²
x(x + 20) = 200
x² + 20x - 200 = 0
a = 1, b = 20, c = -200
[tex]x = \dfrac{-b\pm \sqrt{b^2-4ac} }{2a} [/tex]
[tex]x = \dfrac{-20 \pm \sqrt{20^2-4(1)(-200)} }{2(1)} [/tex]
[tex]x = \dfrac{-20 \pm \sqrt{400+800} }{2} [/tex]
[tex]x = \dfrac{-20 \pm \sqrt{1200} }{2} [/tex]
[tex]x = -27.32 \text { (rejected, length cannot be negative) } \or \ x = 7.32[/tex]
Width = 7.32
Length = 7.32 + 20 = 27.32
Answer: The pen is 7.32 ft by 27.32 ft
The dimensions of the pen are given by the formula for the area of a
rectangle.
The width of the pen is (10·√3 - 10) feet.
The length of the pen is (10·√3 + 10) feet.
Reasons:
The given parameters are;
The area of the rectangular pen, A = 200 ft.²
The length, L, of the pen = The width, W + 20 ft.
Therefore;
A = L × W = (W + 20) × W
200 = W² + 20·W
W² + 20·W - 200 = 0
Using the quadratic formula, for the equation of the form;
y = a·x² + b·x + c, we have;
[tex]x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}[/tex]
Which gives;
[tex]W = \dfrac{-20\pm \sqrt{20^{2}-4\times 1 \times (-200)}}{2\times 1} = \dfrac{-20 \pm 20 \cdot \sqrt{3} }{2}[/tex]
W = -10 + 10·√3, or x = -10 - 10·√3
Therefore;
- The width of the pen, W = 10·√3 - 10 feet
- The length of the pen, L = 10·√3 - 10 + 20 = 10·√3 + 10 feet
Learn more here:
https://brainly.com/question/12932084