I really need help with this one

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 the width be w inches

then the length is w+64

so
2w + 2(w+64) = 224

solve, let me know what you get.

7inx32in

To find the length and width of the rectangle, we can use the information given in the problem. Let's denote the width as "x" inches.

According to the problem, the length exceeds the width by 64 inches. This means that the length is equal to the width plus 64 inches, or (x + 64) inches.

The perimeter of a rectangle is calculated by adding the lengths of all four sides. In this case, we have two side lengths that are equal to the width (x inches) and two side lengths that are equal to the length, which is (x + 64) inches. Therefore, the perimeter is given by the equation:

Perimeter = 2(width) + 2(length)

Now we can substitute the given perimeter value:

224 inches = 2(x) + 2(x + 64)

Simplifying the equation, we have:

224 inches = 2x + 2x + 128
224 inches = 4x + 128
96 inches = 4x

Dividing both sides of the equation by 4, we find:

x = 24

So, the width of the rectangle is 24 inches.

Now, to find the length, we substitute the value of the width (x) back into the expression we obtained earlier:

Length = x + 64
Length = 24 + 64
Length = 88

Therefore, the length of the rectangle is 88 inches.

To summarize:

The length is 88 inches.
The width is 24 inches.