The perimeter of a rectangle is 58yd , and the area of the rectangle is 54yd . Find the dimensions of the rectangle.
Length:
Width:
looks like 27x2 to me. 54 doesn't have that many factors.
Which of these combinations could form the required perimeter?
2, 26
3, 18
4, 13
6, 9
2,26
To find the dimensions of a rectangle given its perimeter and area, we can set up a system of equations using the information provided.
Let's call the length of the rectangle "L" and the width "W".
We know the formula for the perimeter of a rectangle is: P = 2L + 2W
And we know the formula for the area of a rectangle is: A = L * W
Given that the perimeter is 58 yards, we can set up the equation: 58 = 2L + 2W
And given that the area is 54 yards, we can set up the equation: 54 = L * W
Now we have a system of two equations with two variables. We can solve this system to find the values for L and W.
Let's solve for L in terms of W in the first equation:
58 = 2L + 2W
2L = 58 - 2W
L = (58 - 2W)/2
L = 29 - W
Now we substitute this expression for L into the second equation:
54 = L * W
54 = (29 - W) * W
54 = 29W - W^2
Rearranging the equation:
W^2 - 29W + 54 = 0
We can now solve this quadratic equation to find the possible values for W. We can factor or use the quadratic formula.
Factoring the equation, we get:
(W-2)(W-27) = 0
So W = 2 or W = 27
Now let's substitute these values of W back into the equation L = 29 - W to find the corresponding values of L.
For W = 2, L = 29 - 2 = 27
For W = 27, L = 29 - 27 = 2
Therefore, the dimensions of the rectangle are:
Length: 27 yards
Width: 2 yards