If the length of a rod is 56 cm when it is bent into a rectangle then the perimeter of rectangle = 56cm a) Perimeter of rectangle = [tex]2(l+b)[/tex]
[tex]2(l+b) = 56\\l+b = 28[/tex]
Hence, the sum of the length and breadth of the rectangle is 28cm
b) length of diagonal = 20cm
applying pythagoras theorem;
[tex]l^{2}+b^{2} = 20^{2} \\[/tex]
as we know ,
[tex](l-b)^{2} = (l+b) ^{2} -4ab[/tex]
[tex](l-b)^{2} = (28)^{2} -4ab[/tex]
also,
[tex](l+b)^{2} = 28^{2} \\l^{2}+ b^{2} +2ab = 784[/tex]
[tex]400 +2ab = 784\\2ab = 384 \\ab = 192[/tex]
[tex](l-b)^{2} = 784 - 4(192)\\(l-b)^{2} = 16[/tex]
[tex](l-b) = 4[/tex]
Now , solving
[tex](l+b) = 28\\(l-b) = 4[/tex]
we get ,
[tex]l= 16 \\b= 12[/tex]
So , length of rectangle = 16cm and breadth = 12cm