Question

Let a triangle $A B C$ be inscribed in the circle $x ^{2}-$ $\sqrt{2}(x+y)+y^{2}=0$ such that $\angle B A C=\frac{\pi}{2}$. If the length of side $A B$ is $\sqrt{2}$, then the area of the $\triangle ABC$ is equal to:

• A. $(\sqrt{2}+\sqrt{6}) / 3$
• B. $(\sqrt{6}+\sqrt{3}) / 2$
• C. $(3+\sqrt{3}) / 4$
• D. None of the above

Answer 1

Answer: (D)

Radius of given circle is $1 .$
$ BC =$ diameter $=2, AB =\sqrt{2}$
$ AC =\sqrt{ BC ^{2}- AB ^{2}}=\sqrt{2} $
$\triangle ABC =\frac{1}{2} AB \cdot AC =1$