Question

Let a triangle ABC be inscribed in the circle
\(x² - \sqrt2(x+y)+y² = 0\)
such that ∠BAC= π/2. If the length of side AB is √2, then the area of the ΔABC is equal to :

• A. \(\frac{(\sqrt2+\sqrt6)}{3}\)
• B. \(\frac{(\sqrt6+\sqrt3)}{2}\)
• C. \(\frac{(3+\sqrt3)}{4}\)
• D. \(\frac{(\sqrt6+2\sqrt3)}{4}\)

Answer 1

The Correct Option is A

The correct answer is 1 , not there in the options
\(x² -\sqrt2(x+y)+y²=0\)
∴ Coordinates of centre of circle is \(( \frac{1}{\sqrt2} \frac{1}{\sqrt2} )\)
\(r = \sqrt{\frac{1}{2} + \frac{1}{2} - 0}\)
r = 1

Fig.

BC = 2
Apply Pythagoras theorem in ΔABC, we get
AC² + AB² = BC²
⇒ AC² = 4-2 = 2
\(⇒ AC = \sqrt2\)
\(∴\) Area of ΔABC = \(\frac{1}{2}\) × AB × AC
\(\frac{1}{2} × \sqrt2 × \sqrt2 = \frac{2}{2} = 1 \) sq. unit