[Math] Let $ABC$ be a right angled triangle at $C$.If the inscribed circle touches the side $AB$ at $D$ and $(AD)(BD)=11$

Question

Let $ABC$ be a right angled triangle at $C$.If the inscribed circle touches the side $AB$ at $D$ and $(AD)(BD)=11$,then find the area of the triangle $ABC.$

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Area of triangle $ABC=\frac{1}{2}AC\times BC$

Let the inscribed circle touches the side $AB$ at $D$,$AC$ at $E$ and $BC$ at $F.$Let $CE=CF=x$
Area of triangle $ABC=\frac{1}{2}AC\times BC=\frac{1}{2}(AE+EC)(BF+CF)=\frac{1}{2}(AD+x)(BD+x)$
I am not able to solve further.

Answer 1

Let $CE=CF=x,BD=BF=y,AD=AE=z$.

We have $yz=11$. Since the triangle is a right triangle, you can have $$(z+y)^2=(z+x)^2+(x+y)^2,$$ i.e. $$x^2+xz+xy=11.$$

So, the area is $$\frac{1}{2}(x+z)(x+y)=\frac{1}{2}(x^2+xy+xz+yz)=\frac{11+11}{2}$$