Question 1054136
Both are Right Triangles.


The one using x:
One leg is x and hypotenuse is x+3;
Other leg found using Pythagorean Theorem Formula.
This other leg, r.


{{{r^2+x^2=(x+3)^2}}}
{{{r^2=(x+3)^2-x^2}}}
{{{r^2=x^2+6x+9-x^2}}}
{{{r^2=6x+9}}}
{{{r=sqrt(6x+9)}}}, but r is still unknown.


You can solve for x, but x will be in terms of the other leg, r.
{{{6x+9=r^2}}}
{{{6x=r^2-9}}}
{{{x=r^2/6-3/2}}}