Question 353659
Let length is 'a' and width is 'b'.

according to question,

perimeter      2(a+b)= 34 
 
    or, a+b = 17 ...........(1)

     

 a^2 + b^2 = (13)^2


=> a^2 + b^2 = 169


now putting  a = 17-b from (1) in above equation

 => (17-b)^2 + b^2 = 169


 => 289 + b^2 - 34b + b^2 = 169


  => 2b^2 -34b + 120 = 0


 => b^2 - 17b + 60 = 0   (dividing by 2)


=>  b^2 -12b -5b +60 =0


=> b(b-12)-5(b -12) = 0


=> ( b-5) (b-12) = 0


so, b= 5 or b =12


if b= 5, a = 12


        if b = 12 , a= 5



so length and width are 5 and 12.