A rectanguler field with an area of 8000 meters squared is enclosed by 400 meters of fencing.Determine the dimessions of the field to the nearest tenth of a meter
x*y = 8000
x + y = 200
x * (200-x) = 8000
x^2 -200x +8000 = 0
Use the "quadratic equation" to solve that, and take the positive root.
x = (1/2) [200 +sqrt(40000 -32000)]
= 100 + 44.72
= 144.7
y = 55.3
The field is 144.7 x 55.3 meters
To determine the dimensions of the rectangular field, we can set up an equation based on the information given.
Let's consider the length of the field as "L" and the width as "W".
We know that the area of a rectangle is given by the formula: A = Length × Width.
Hence, we have the equation: A = L × W.
From the problem statement, we are given that the area of the field is 8000 square meters. Therefore, we can write: 8000 = L × W.
We are also given that the field is enclosed by 400 meters of fencing. Since the fencing goes around the perimeter of the field, we can equate this to the sum of all the sides of the rectangle: 400 = 2L + 2W.
Now, we have a system of two equations:
8000 = L × W Eq. (1)
400 = 2L + 2W Eq. (2)
To solve for the dimensions of the field, we can use the substitution method or the elimination method. However, in this case, we can solve for one variable in terms of the other and substitute that into the other equation.
From Eq. (2), we can rearrange it to obtain: W = (400 - 2L) / 2.
Substituting this expression for W into Eq. (1), we get:
8000 = L × ((400 - 2L) / 2).
Simplifying the equation, we have:
8000 = 200L - L².
Rearranging, we get a quadratic equation:
L² - 200L + 8000 = 0.
Now, we can solve this quadratic equation to find the values of L.
Using the quadratic formula, L = (-b ± √(b² - 4ac)) / 2a, where a = 1, b = -200, and c = 8000.
Substituting these values, we get:
L = (-(-200) ± √((-200)² - 4 × 1 × 8000)) / (2 × 1).
Simplifying further, we have:
L = (200 ± √(40000 - 32000)) / 2
L = (200 ± √(8000)) / 2
L = (200 ± 89.44) / 2.
Using the positive value, we have:
L = (200 + 89.44) / 2
L = 289.44 / 2
L = 144.72 meters (rounded to the nearest tenth).
Now, substitute this value of L back into Eq. (2) to find W:
400 = 2(144.72) + 2W
400 = 289.44 + 2W
110.56 = 2W
W = 55.28 meters (rounded to the nearest tenth).
Thus, the dimensions of the field, to the nearest tenth of a meter, are approximately 144.7 meters by 55.3 meters.