Answer :
To solve this problem, let's assign some variables to the dimensions of the rectangle.
Let:
- [tex]w[/tex] be the width of the rectangle in meters.
- [tex]l[/tex] be the length of the rectangle in meters.
According to the problem, the length is 5 meters longer than the width. This can be written as:
[tex]l = w + 5[/tex]
The perimeter [tex]P[/tex] of a rectangle is given by the formula:
[tex]P = 2l + 2w[/tex]
We are told that the perimeter is 42 meters, so we can write:
[tex]2l + 2w = 42[/tex]
Now, substitute the expression for [tex]l[/tex] from the first equation into the second equation:
[tex]2(w + 5) + 2w = 42[/tex]
Expanding the equation, we get:
[tex]2w + 10 + 2w = 42[/tex]
Combine the [tex]w[/tex] terms:
[tex]4w + 10 = 42[/tex]
Subtract 10 from both sides to solve for [tex]w[/tex]:
[tex]4w = 32[/tex]
Divide both sides by 4:
[tex]w = 8[/tex]
Now that we have the width, we can find the length using the relation [tex]l = w + 5[/tex]:
[tex]l = 8 + 5 = 13[/tex]
Therefore, the dimensions of the rectangle are:
- Width: 8 meters
- Length: 13 meters
These dimensions satisfy both the condition for the length being 5 meters longer than the width and the given perimeter of 42 meters.
