Question 1152451
  


let the length be {{{a}}} and the with {{{b}}}

if the perimeter of a rectangle is {{{70}}}, we have

{{{2(a+b)=70}}}

{{{a+b=35}}}

{{{a=35-b}}}


if its diagonal is {{{25}}}, we have

{{{d=25}}}

diagonal divides rectangle into two right triangles
   
in one right triangle, one leg  is the length {{{a}}}, the other leg is the width {{{b}}}

using Pythagorean theorem, we have:

{{{d^2=a^2+b^2}}}....substitute {{{a}}} and {{{d}}}

{{{25^2=(35-b)^2+b^2}}}.....solve for {{{b}}}

{{{625=b^2 - 70 b + 1225+b^2}}}

{{{2b^2 - 70 b + 1225-625=0}}}

{{{2b^2 - 70 b + 600=0}}}.....simplify

{{{b^2 - 35 b + 300=0}}}

{{{(b - 20) (b - 15) = 0}}}

=>{{{ b=20 }}}or{{{ b=15 }}}

since {{{b}}} is width, go with {{{b=15}}} and the other ({{{20}}}) will be the length {{{a}}}

or calculate:
{{{a=35-15}}}
{{{a=20 }}}

answer:

Its length is {{{20 }}}

Its width is {{{15 }}}

where length > width