Answer:
Given that,
the sum of the lengths of three sides of a rectangle is 55cm
each side is a whole number
the length is 8 cm more than the width
we get,
Let l be the length and b be the width,
[tex]l=b+8[/tex]Also we have either
[tex]55=l+2b-----(1)[/tex]or,
[tex]55=2l+b------(2)[/tex]Substitute the value of l, in the equation (1), we get
[tex]55=8+b+2b[/tex][tex]55=3b+8[/tex][tex]3b=55-8[/tex][tex]3b=47[/tex][tex]b=\frac{47}{3}[/tex]we get, b is not a whole number, Hence this is not possible.
Consider the equation (2), we get
[tex]55=2l+b[/tex][tex]55=2(b+8)+b[/tex][tex]55=2b+16+b[/tex][tex]3b=55-16[/tex][tex]3b=39[/tex][tex]b=13[/tex]Perimeter of the rectangle is,
[tex]2(l+b)[/tex]we get,
[tex]=2(b+8+b)=2(2b+8)[/tex][tex]=4b+16[/tex]substitute b=13 cm, we get
[tex]=(4\times13)+16[/tex][tex]=68[/tex]Perimeter of the rectangle is 68 cm.
