A rectangle field has an area of 300 square meters and a perimeter of 80 meters. What are the length and width of the field?
I really need help I don't know the answer.
2 L + 2 w = 80 or L + w = 40
L w = 300 so w = 300/L
so
L + 300/L = 40
L^2 - 40 L + 300 = 0
(L-30)(L-10) = 0
try
L = 30 and w = 10
Factors of 300 include 30 by 10.
Would those dimensions give you a perimeter of 80?
To find the length and width of the rectangle field, we can set up a system of equations using the given information.
Let's assume the length of the rectangle is L meters and the width is W meters.
We know that the area of a rectangle is given by the formula: A = L * W.
So, we have the equation: L * W = 300 -- (Equation 1)
We also know that the perimeter of a rectangle is given by the formula: P = 2L + 2W.
Given that the perimeter is 80 meters, we have the equation: 2L + 2W = 80 -- (Equation 2)
Now, we need to solve this system of equations to find the length and width.
We can rearrange Equation 2 to get: 2L = 80 - 2W
Divide both sides by 2: L = 40 - W
Substitute this value of L into Equation 1:
(40 - W) * W = 300
Expanding the equation: 40W - W^2 = 300
Rearrange the equation to a quadratic form: W^2 - 40W + 300 = 0
Now, we can solve the quadratic equation to find the possible values of W (width) and then find the corresponding values of L (length).
Using the quadratic formula: W = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = -40, and c = 300.
W = (-(-40) ± √((-40)^2 - 4*1*300)) / (2*1)
= (40 ± √(1600 - 1200)) / 2
= (40 ± √400) / 2
= (40 ± 20) / 2
So, we have two possible values for W:
1. W = (40 + 20) / 2 = 60 / 2 = 30 meters
2. W = (40 - 20) / 2 = 20 / 2 = 10 meters
Substituting these values into Equation 1, we can find the corresponding values of L:
1. L = 40 - 30 = 10 meters
2. L = 40 - 10 = 30 meters
Therefore, the length and width of the rectangle field are either 30 meters by 10 meters or 10 meters by 30 meters.
To find the length and width of the rectangle field, we can set up a system of equations using the given information.
Let's assume that the length of the rectangle is denoted by "L" and the width is denoted by "W".
We are given two pieces of information:
1. The area of the rectangle is 300 square meters, which means L * W = 300.
2. The perimeter of the rectangle is 80 meters, which implies 2L + 2W = 80.
Now we have a system of equations:
L * W = 300 ----- Equation 1
2L + 2W = 80 ----- Equation 2
To solve this system of equations, we can use a method called substitution or elimination. In this case, let's solve it using substitution.
Let's solve Equation 1 for L:
L = 300 / W
Substitute this value of L into Equation 2:
2(300 / W) + 2W = 80
Now we can simplify the equation:
600 / W + 2W = 80
To eliminate the fraction, we can multiply everything by W:
600 + 2W^2 = 80W
Rearrange the equation to form a quadratic equation:
2W^2 - 80W + 600 = 0
Simplifying further:
W^2 - 40W + 300 = 0
Now we can solve this quadratic equation by factoring, completing the square, or by using the quadratic formula. Factoring the equation gives us:
(W - 20)(W - 15) = 0
From this equation, we have two potential solutions for W:
1. W - 20 = 0 -> W = 20
2. W - 15 = 0 -> W = 15
If we substitute these values of W back into Equation 1, we can find the respective values of L:
If W = 20, L = 300 / 20 = 15
If W = 15, L = 300 / 15 = 20
Therefore, the length and width of the field can be either 15 meters by 20 meters or 20 meters by 15 meters.