The dimensions of the rectangle are 6 and 16.
Perimeter of rectangle
[tex]\rm{ Perimeter\ of\ the\ rectangle = 2(length +width)[/tex]
Given to us
- the perimeter of the rectangle = 44 m
- The length is 4 m more than twice the width.
Assumption
Let's assume that the width of the rectangle is x.
From statement 2,
The length is 4 m more than twice the width.
[tex]length = 4 + (2\times width)\\length = 4 +(2 \times x)\\length = 4+2x[/tex]
From statement 1,
the perimeter of the rectangle = 44 m
[tex]\rm{ Perimeter\ of\ the\ rectangle = 2(length +width)[/tex]
substituting values we get,
[tex]44 = 2 [ (4+2x) +x]\\\\\dfrac{44}{2} = [4+2x+x]\\\\22 = 4+3x\\22 -4=3x\\18 = 3x\\3x =18\\\\x=\dfrac{18}{3}\\\\x =6[/tex]
Dimensions of the rectangle
Substituting the value of x to get the dimensions of the rectangle,
[tex]width = x = 6[/tex]
[tex]\rm{ Length = 4+2x = 4+2(6) = 4+12 = 16[/tex]
Hence, the dimensions of the rectangle are 6 and 16.
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