Question 169672
ok ........so when she walks diagonally across the field that ends up being the hypothenuse of the right triange which is 15 meters. what we dont know is the other 2 legs...but we have a hint since the perimeter of the triangel is 36m. So now we know that both legs add up to 21----->(36-15)....so we will call one leg a and the other 21-a.
:
{{{a^2+(21-a)^2=15^2}}}
:
{{{a^2+441-42a+a^2=225}}}
:
{{{2a^2-42a+216=0}}}divide by 2
:
{{{a^2-21a+108=0}}}
:
 using the quadratic we find that {{{highlight(a=9)}}}or {{{highlight(a=12)}}}
each leg can be either value.  The dimensions of the field is 9m by 12m

*[invoke quadratic "a", 1, -21, 108]