1. A rectangular field with an area of 5000m^2 is enclosed by 300m of fencing. Find the dimensions of the field:
A. 100m x 50m
B. 75m by 75m
C. 150m x 150m
D. no solution
Please explain answer. Thanks!
Let the unknown dimensions be a and b. Solve the two equations
a b = 5000
2a + 2b = 300
Simplify to a + b = 150 and substitute 150-a for b in the first equation.
a(150-a) = 5000
a^2 - 150 a + 5000 = 0
You could either solve by factoring or just see which of the multiple choices satsifies the two equations. There IS a solution.
Well, let's see if we can solve this fencing puzzle! We know that the perimeter of the rectangular field is 300m, which means that the sum of all four sides should add up to 300m.
Now, let's assume that the length of the field is x meters. The width of the field would then be (300 - 2x)/2 meters, because we want the remaining fencing to be evenly divided between the two shorter sides.
The area of the field is given by the formula length x width. Plugging in the values, we have x * (300 - 2x)/2 = 5000.
Now, let's solve for x. Multiplying both sides of the equation by 2 to clear the fraction, we get x * (300 - 2x) = 10000.
Expanding the equation, we have 300x - 2x^2 = 10000.
Rearranging the terms, we have 2x^2 - 300x + 10000 = 0.
Now, if we try to solve this quadratic equation, we find that there are no real solutions. So the answer is D, which means there is no possible solution for the dimensions of the field.
Well, it seems like we're fencing in failure on this one! But hey, don't be too disappointed. Sometimes math doesn't cooperate, just like stubborn clowns trying to squeeze into tiny cars. Keep on trying, and you'll find the right dimensions eventually!
To find the dimensions of the rectangular field, we can use the formulas for area and perimeter.
The formula for the area of a rectangle is A = length × width.
The formula for the perimeter of a rectangle is P = 2(length + width).
From the given information, we know that the area of the field is 5000m^2 and the perimeter of the fence is 300m.
We can start by setting up two equations based on the given formulas:
Equation 1: A = length × width
Equation 2: P = 2(length + width)
Substituting the given values, we have:
5000 = length × width
300 = 2(length + width)
We can rearrange the second equation to get:
150 = length + width
Now, we have a system of two equations:
5000 = length × width
150 = length + width
We can solve this system of equations to find the dimensions of the field.
Option A: 100m × 50m
If we substitute these values into the first equation:
5000 = 100 × 50
5000 = 5000
Substituting into the second equation:
150 = 100 + 50
150 = 150
The dimensions in option A satisfy both equations, so option A is a valid solution.
Option B: 75m × 75m
If we substitute these values into the first equation:
5000 = 75 × 75
5000 = 5625 (not equal)
The dimensions in option B do not satisfy the first equation, so option B is not a valid solution.
Option C: 150m × 150m
If we substitute these values into the first equation:
5000 = 150 × 150
5000 = 22500 (not equal)
The dimensions in option C do not satisfy the first equation, so option C is not a valid solution.
Option D: No solution
Options B and C are not valid solutions, so the only possible solution is option A.
Therefore, the dimensions of the field are 100m × 50m.
To solve this problem, we need to use the formula for the perimeter of a rectangle, which is P = 2L + 2W, where L is the length and W is the width.
In this case, we are given that the perimeter is 300m. So we can write the equation as: 300 = 2L + 2W.
However, we are also given that the area of the rectangle is 5000m^2. The formula for the area of a rectangle is A = L x W.
We can use this information to rewrite the equation for the perimeter. We know that A = 5000, so dividing both sides of the equation by W, we get: L = 5000/W.
Now we can substitute this expression for L into the equation for the perimeter: 300 = 2(5000/W) + 2W.
Simplifying this equation, we get: 300 = 10000/W + 2W.
Multiplying both sides of the equation by W, we get: 300W = 10000 + 2W^2.
Rearranging the equation and setting it equal to zero, we get: 2W^2 - 300W + 10000 = 0.
Now we can use the quadratic formula to solve for W. The quadratic formula is: W = (-b ± √(b^2 - 4ac)) / (2a), where a = 2, b = -300, and c = 10000.
Plugging in the values, we get: W = (-(-300) ± √((-300)^2 - 4(2)(10000))) / (2(2)).
Simplifying this expression, we get: W = (300 ± √(90000 - 80000)) / 4.
Continuing to simplify, we get: W = (300 ± √10000) / 4.
Now we can calculate the two possible values for W: W₁ = (300 + √10000) / 4 and W₂ = (300 - √10000) / 4.
Calculating these values, we get: W₁ ≈ 75 and W₂ ≈ 25.
Since we can't have a negative width, we discard W₂ = 25.
Now we can use the equation L = 5000/W to find the length: L = 5000/75 ≈ 66.67.
So, the dimensions of the field are approximately 75m by 66.67m.
Therefore, the correct answer is B. 75m by 75m.