Question 1204789

 the area of a right triangle is:

{{{A=(1/2)b*a}}}  where {{{a}}} and {{{b }}}are legs



if hypotenuse is {{{c=2}}}, then


{{{a=sqrt(2^2-b^2)}}}

{{{a=sqrt(4-b^2)}}}......eq.1


 if area is {{{1}}}, then


{{{(1/2)b*a =1}}}...substitute{{{ a}}}

{{{(1/2)b*sqrt(4-b^2) =1}}}

{{{b*sqrt(4-b^2) =2}}}....square both sides

{{{(b*sqrt(4-b^2))^2 =2^2}}}

{{{b^2*(4-b^2) =4}}}

{{{4b^2-b^4 -4=0}}}

{{{-b^4 +4b^2-4=0}}}

{{{-(b^4 -4b^2+4)=0}}}

{{{-(b^4 -2b^2-2b^2+4)=0}}}
{{{-((b^4 -2b^2)-(2b^2-4))=0}}}

{{{-(b^2(b^2 -2)-2(b^2-2))=0}}}

{{{-((b^2 -2)(b^2-2))=0}}}


solution

{{{b^2-2=0}}}

{{{b^2=2}}}

{{{b=sqrt(2)}}}


then 


{{{a=sqrt(4-sqrt(2)^2)}}}......eq.1

{{{a=sqrt(4-2)}}}

{{{a=sqrt(2)}}}



perimeter:

{{{P=a+b+c}}}

{{{P=sqrt(2)+sqrt(2)+2}}}

{{{P=2sqrt(2)+2}}}