What is the answer to,"if the area of a rectangle is 24 and the perimeter is 28, what are the length and width?"
Factors of 24:
1, 24
2, 12
3, 8
4, 6
Which of those two possible measurements of the area will give a perimeter of 28?
P = 2L + 2W
Thx 2 and 12
Right.
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To find the lengths and widths of a rectangle given its area and perimeter, we can use a system of equations. Let's assume the length of the rectangle is 'L' and the width is 'W'.
We are given two pieces of information:
1. The area of the rectangle is 24. The formula for the area of a rectangle is A = L * W.
2. The perimeter of the rectangle is 28. The formula for the perimeter of a rectangle is P = 2L + 2W.
Let's set up the equations:
Equation 1: A = L * W (Area equation)
Equation 2: P = 2L + 2W (Perimeter equation)
Substituting the given values into the equations:
Equation 1: 24 = L * W
Equation 2: 28 = 2L + 2W
Now, we have a system of equations. We can solve them simultaneously to find the values of L and W.
To solve the system of equations, we can use either the substitution method or the elimination method. I will use the substitution method in this case:
From Equation 1: 24 = L * W
We can rewrite Equation 1 as: L = 24/W
Next, substitute this value of L into Equation 2:
28 = 2L + 2W
28 = 2(24/W) + 2W
Now, simplify this equation and solve for W:
28 = 48/W + 2W
Multiply the equation by W to eliminate the denominator:
28W = 48 + 2W^2
Rearrange the equation, and bring everything to one side:
2W^2 - 28W + 48 = 0
Now we have a quadratic equation. We can solve this by factoring, completing the square, or using the quadratic formula. Let's solve this quadratic equation using factoring:
2W^2 - 28W + 48 = 0
2(W^2 - 14W + 24) = 0
Now let's factor the quadratic expression:
2(W - 2)(W - 12) = 0
Setting each factor equal to zero:
W - 2 = 0 or W - 12 = 0
Solving for W:
If W - 2 = 0, then W = 2
If W - 12 = 0, then W = 12
Now that we have two possible values for W, we can find the corresponding lengths (L) using Equation 1:
If W = 2, then L = 24/2 = 12
If W = 12, then L = 24/12 = 2
Therefore, the possible solutions are:
Length (L) = 12, Width (W) = 2
or
Length (L) = 2, Width (W) = 12
Both of these combinations satisfy the given conditions: an area of 24 and a perimeter of 28.