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------------------------------------------------ The length of a rectangle is 5 metres longer than its width. If the perimeter of the rectangle is 42 metres, determine the dimensions of the rectangle.

Question

High School

Answer :

RyanHarmon181 RyanHarmon181 · Answer 1 · 2025-03-26T06:32:23-04:00

To solve this problem, let's assign some variables to the dimensions of the rectangle.

Let:

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:

These dimensions satisfy both the condition for the length being 5 meters longer than the width and the given perimeter of 42 meters.