Question 1096436
The area of a right triangle can be calculated as
{{{area=leg[1]*leg[2]/2}}} ,
Because if we call one leg the base of the triangle,
the other leg is the altitude.
 
In this case, leg lengths in meters are {{{x}}} and {{{x+4}}} ,
so {{{x(x+4)/2=30}}} ,
{{{x(x+4)=30*2}}} ,
{{{x^2+4x=60}}} ,
{{{x^2+4x-60=0}}} , and
{{{(x-6)*(x+10)=0}}} .
The solution is {{{highlight(x=6)}}} from {{{x-6=0}}} .
The other solution to the equation, {{{x=-10}}} from {{{x+10=0}}}
is not a solution to the problem, 
because {{{x}}} is the length of a triangle side in meters,
and it must be a positive number.
So, the length of the shorter leg is {{{6meters}}} ,
and the length of the longer leg is {{{6meters+4 meters=highlight(10meters)}}} .