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 :
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 :
\(\frac{(\sqrt2+\sqrt6)}{3}\)
\(\frac{(\sqrt6+\sqrt3)}{2}\)
\(\frac{(3+\sqrt3)}{4}\)
\(\frac{(\sqrt6+2\sqrt3)}{4}\)
The Correct Option is A
Solution and Explanation
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
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
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