Find area of triangle.

[@iMíccoli]

I've another solution
ABOC is cyclic because angle BAC +angle BOC=180° so using Ptolomeu's theorem we have BO×AC + AB×OC= AO×BC.
Notice that AO=2 because it's the radius and BC=√5 by using Pythagoras in ∆BOC.
So 1×AC + AB×2=2√5 which means AC=2(√5-AB) but remember that ∆ABC is right triangle so:
AB²+AC²=BC²
AB²+[2(√5-AB)]²=(√5)²
AB²+4(5-2√5AB+AB²)=5
AB²+20-8√5AB+4AB²=5
5AB²-8√5AB+15=0 and now let AB=x so then we have 5x²-8√5x+15=0 and by solving the quadratic equation you'll get AB=x=√5 or AB=x=(3√5)/5 but remember AC=2(√5-AB) so if AB=√5 then AC=0 which is an absurd because AC is the side length of a triangle so it must be a positive real number therefore AB=(3√5)/5 and now:
AC=2[√5-(3√5)/5]
AC=2[(2√5)/5]
AC=(4√5)/5 and finally Area(∆ABC)=(AC×AB)/2
Area(∆ABC)=[(4√5)/5 × (3√5)/5]/2
Area(∆ABC)=6/5cm².
❤

[@relishmath5632]

Impressive . . . Thanks for sharing it.