the perimeter of a rectangle is 224 incles. the length exceeds the width by 64 inces. find the length and the width.
the length is _____ inches
and
the width is _____ inches
Let
L=length
W=width = L-64
Perimeter = 224 = 2(L+W) = 2(L+L-64)
Solve for L.
A=1/2h(b1+b2)
solve for b
"A=1/2h(b1+b2)
solve for b"
I do not see the variable b in the expression: b1 and b2 are different variables than b.
In any case, I suggest you start a new thread for a new question.
To find the length and width of the rectangle, we can set up a system of equations based on the given information.
Let's denote the length of the rectangle as L and the width as W.
We know that the perimeter of a rectangle is calculated by adding all four sides. For a rectangle, this can be simplified to 2 times the length plus 2 times the width:
Perimeter = 2L + 2W
According to the problem, the perimeter is given as 224 inches. So we can write the equation:
2L + 2W = 224
The problem also states that the length exceeds the width by 64 inches. We can express this as:
L = W + 64
Now we have a system of two equations:
Equation 1: 2L + 2W = 224
Equation 2: L = W + 64
To solve this system, we can use substitution or elimination method. Let's use substitution:
We substitute the value of L from Equation 2 into Equation 1:
2(W + 64) + 2W = 224
2W + 128 + 2W = 224
4W + 128 = 224
4W = 224 - 128
4W = 96
Next, we solve for W by dividing both sides of the equation by 4:
W = 96 / 4
W = 24
Now that we have the value for W, we can substitute it back into Equation 2 to find L:
L = W + 64
L = 24 + 64
L = 88
Therefore, the length of the rectangle is 88 inches and the width is 24 inches.