Question 1033222
{{{a-b=5}}} --> {{{a=b+5}}}
and as {{{a*b=100}}} , as you said,
{{{(b+5)*b=100}}} <--> {{{b^2+5b=100}}} <--> {{{b^2+5b-100=0}}} .
That is a quadratic equation.
When you are lucky, you get a quadratic equation that can be solved by factoring.
No such luck in this case, because as you said no whole numbers that are factors of 100 have a difference of 5.
So, we have to solve that equation by completing the square or by using the quadratic formula.
Either way, the solutions are
{{{b=(-5 +- 5sqrt(17))/2=(5/2)(-1 +- Sqrt(17))}}} .
Since we need a positive {{{b}}} , {{{highlight(b=(-5 + 5sqrt(17))/2=(5/2)(-1 + sqrt(17)))}}} .
So, {{{highlight(a=5+(-5 + 5sqrt(17))/2=(5 + 5sqrt(17))/2=(5/2)(1 + sqrt(17)))}}} .
 
Accrording to the Pythagorean theorem, {{{c^2=a^2+b^2}}} , so
{{{c^2=((5/2)(sqrt(17)+1))^2+((5/2)(sqrt(17)-1))^2=(5/2)^2(sqrt(17)+1)^2+(5/2)^2(sqrt(17)-1)^2=(5/2)^2(17+2sqrt(17)+1+17-2sqrt(17)+1)=(5/2)^2*36=(5/2)^2*6^2}}} .
So, {{{c=sqrt((5/2)^2*6^2)=(5/2)*6=highlight(15)}}} .
Side {{{c}}} (the hypotenuse of the right triangle) measures {{{highlight(15cm)}}} .