Arectangle has a perimeter of 68 meters. the length is eight meters more than the width. find the area of the rectangle.

First we are given that the rectangle's perimeter is 68 meters. Now the perimeter is just adding the length and width together, twice over, in other words counting all sides and adding them together.

It then goes on to say that the length is "eight meters more than the width". This gives us that the width, being the smaller number, is now just 'x' which we will have to solve. Then we are given that the length is eight meters more, basically being 'x + 8'. In other words:

The length is eight meters more than the width. Width = x, Length = x + 8.
The final step is including that both the width and length are doubled, with that meaning just make sure to add a 2 beside them as shown:

2x + 2(x + 8) = 68 | First, multiply.
2x + 2x + 16 = 68 | Next, add like terms.
4x + 16 = 68 | Use the subtraction property of equality by subtracting both sides by 16.
4x = 52 | Use the division property of equality by dividing both sides by 4.
x = 13

We have now solved that 'x', the width, is 13. Now, add 8 to this to get the length. Doing so gives us 21, so the length is 21. Let's check before continuing:

13 + 13 + 21 + 21 = 68
26 + 42 = 68
68 = 68

It checks out, to finalize we now have to solve for the area, the area is length times width only. Such as:
l * w = A [l = length, w = width, A = Area]

Input the variables we know:
21 * 13 = A | Simply multiply.
273 is the Area.

Conclusion: Width = 13, Length = 21, Perimeter = 68, Area = 273.

I hope this helps, have a great rest of your day! ^ ^
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