Note that $\Delta = r*s =1*s = s$. So, we have to minimize $s$. WLOG, we may assume $a \ge b$ and $c^2=a^2 + b^2$. Let $a=b+t$, for some $t \ge 0$. $s= \frac{(a+b+ \sqrt{a^2+b^2})}{2}$. This is minimised if $t=0$ since it reduces the terms of $t$ after expanding $a$ in terms of $b,t$. Thus, $a=b$. So, $\Delta = s \Rightarrow \frac{1}{2}*a*a = \frac{2*a + \sqrt{2}*a}{2}$. So, $a=2+\sqrt{2}$ and $\Delta_{min} = 3+2\sqrt{2}$.

Suppose that the right angle is at point $B$. Then it is easy to see that the altitude from $B$ to $AC$ is not more than $1+\sqrt{2}$. Let $AB=1+x$, $BC=1+y$. Then $AC=x+y$ and $(1+x)(1+y)\ge (x+y)(1+\sqrt{2})$ So $(x-\sqrt{2})(y-\sqrt{2})\ge 1$. Now we want to minimise $(1+x)(1+y)$. Replace $x-\sqrt{2}$ by $x_1$ and $y-\sqrt{2}$ by $y_1$. Then we have $x_1y_1\ge 1$ and we want to minimise $(1+\sqrt{2}+x_1)(1+\sqrt{2}+y_1)=\text{Some Constant}+ (1+\sqrt{2})(x_1+y_1)$. Obviously $x_1+y_1$ is minimised with the condition $x_1y_1\ge 1$ when $x_1=y_1=1$. Hence we get that the minimum area is $\frac{1}{2}(2+\sqrt{2})(2+\sqrt{2})=3+2\sqrt{2}$ Sniped.

$r= \frac{P+B-H}{2}$ $\implies 1=\frac{P+B-H}{2}$ Let $P=a, B=b$ $\implies 2= a+b - \sqrt {a^2+b^2}$ And we have to minimize $ab$ Setting $a+b=u$ and $ab=v$ $2=u-\sqrt {u^2-2v}$ $\iff u^2-2v= u^2-4u+4$ $ 4=4u-2v \ge 8 \sqrt v -2v$ $\implies \sqrt v \ge 2+ \sqrt 2$ $ab \ge (2+ \sqrt 2 )^2$ Min. Area is $ \frac{1}{2}(2+\sqrt{2})(2+\sqrt{2})=3+2\sqrt{2} $

Oh lol. I did not expect you can use Pythagorean theorem in this question. Hence my above solution.

is my solution correct = let $a,b,c$ be the sides of triangle and $ c^2 = a^2 + b^2 $ by the given condition we have $ ab = a+b+c = a+b+(a^2+b^2){1/2} $ now , $a+b\geq 2(ab)^{1/2} $ and $ (a^2+b^2)^{1/2}\geq (a+b)/(2)^{1/2}\geq (2ab)^{1/2}$ now let $ab^{1/2} =x$ so that $ x^2\geq x(2+2^{1/2} $ and thus $ x\geq 2+2^{1/2} $ and thus $(x^{2})/2 = ab/2 \geq 1/2(2+2^{1/2})^{2} = 3+2\sqrt{2}$

Yes. You have the same idea as $ARCH999$