Question 1189850
<br>
Let the diagonal length be x
Then the length is x-1 and the width is x-2<br>
Apply the Pythagorean Theorem:<br>
{{{(x-1)^2+(x-2)^2=x^2}}}
{{{x^2-2x+1+x^2-4x+4=x^2}}}
{{{x^2-6x+5=0}}}
{{{(x-5)(x-1)=0}}}<br>
{{{x=5}}} or {{{x=1}}}<br>
x=1 doesn't work; it makes the length 0 and the width -1, which is nonsense.<br>
x=5 works.  The length is x-1=4 and the width is x-2=3; then 3^2+4^2=5^2<br>