The area of a rectangle is 72 square inches and the perimeter is 44 inches. Find its dimensions.
let the length and width be x and y respectively
then
2x + 2y = 44
x + y = 22
y = 22-x
area = xy = 72
x(22-x) = 72
22x - x^2 - 72 = 0
x^2 - 22x + 72 = 0
(x-4)(x-18) = 0
x = 4 or x = 18
if x = 4, then y = 22-4 = 18
if x = 18, then y = 22-18 = 4
the dimensions are 4 by 18 inches
One way to solve this is to find the factors of 72. Then see which of these will fit for the perimeter.
To find the dimensions of the rectangle, we can set up a system of equations based on the area and perimeter.
Let's denote the length of the rectangle as L and the width as W.
Given that the area of the rectangle is 72 square inches, we can write the equation:
L * W = 72 --------(1)
Given that the perimeter of the rectangle is 44 inches, the equation for the perimeter is:
2L + 2W = 44 --------(2)
To solve this system of equations, we can make use of the substitution method.
From equation (2), we can simplify it to:
L + W = 22 --------(3)
Now, we can use equation (3) to solve for either L or W in terms of the other variable. Let's solve for W in terms of L:
W = 22 - L --------(4)
We can substitute equation (4) into equation (1):
L * (22 - L) = 72
Expanding the equation:
22L - L^2 = 72
Now, rearranging the equation to quadratic form:
L^2 - 22L + 72 = 0
To factorize this quadratic equation, we can look for two numbers whose sum is -22 and whose product is 72. After some calculation, we find that the factors are -6 and -12.
Therefore, the equation can be factored as:
(L - 6)(L - 12) = 0
Setting each factor equal to zero and solving for L:
L - 6 = 0 gives L = 6
L - 12 = 0 gives L = 12
Since the length cannot be negative, we discard L = 12 as a valid solution.
Thus, the length of the rectangle is L = 6 inches.
Substituting this value back into equation (3) to solve for W:
6 + W = 22
W = 22 - 6
W = 16
Therefore, the dimensions of the rectangle are L = 6 inches and W = 16 inches.
To find the dimensions of the rectangle, we can set up equations based on the given information.
Let's consider the length as L and the width as W.
According to the problem, the area of the rectangle is given as 72 square inches. The formula for the area of a rectangle is A = L * W. So, we have the equation:
L * W = 72 --- Equation (1)
The problem also gives us the perimeter of the rectangle as 44 inches. The formula for the perimeter is P = 2L + 2W. So, we have the equation:
2L + 2W = 44 --- Equation (2)
Now, we have a system of two equations (Equation 1 and Equation 2) that we can solve simultaneously to find the values of L and W.
To solve this system of equations, we can use substitution or elimination. Let's use the substitution method.
From Equation (1), we can isolate L:
L = 72 / W
Now, we substitute this value of L into Equation (2):
2(72 / W) + 2W = 44
Simplify the equation:
144 / W + 2W = 44
Multiply both sides of the equation by W to eliminate the fraction:
144 + 2W^2 = 44W
Rearrange the equation:
2W^2 - 44W + 144 = 0
Divide the entire equation by 2 to simplify:
W^2 - 22W + 72 = 0
Factorize the quadratic equation:
(W - 18)(W - 4) = 0
Now, set each factor equal to zero:
W - 18 = 0 or W - 4 = 0
If we solve the two equations, we find two possible values for W:
W = 18 or W = 4
Now, substitute these values back into Equation (1) to find the corresponding values of L:
For W = 18:
L * 18 = 72
L = 4
For W = 4:
L * 4 = 72
L = 18
So, we have two possible sets of dimensions for the rectangle:
1) Length = 4 inches, Width = 18 inches
2) Length = 18 inches, Width = 4 inches