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Mathematics Question on Area under Simple Curves

Let a triangle ABC be inscribed in the circle
x2−2(x+y)+y2=0x² - \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

(2+6)3\frac{(\sqrt2+\sqrt6)}{3}

B

(6+3)2\frac{(\sqrt6+\sqrt3)}{2}

C

(3+3)4\frac{(3+\sqrt3)}{4}

D

(6+23)4\frac{(\sqrt6+2\sqrt3)}{4}

Answer

(2+6)3\frac{(\sqrt2+\sqrt6)}{3}

Explanation

Solution

The correct answer is 1 , not there in the options
x2−2(x+y)+y2=0x² -\sqrt2(x+y)+y²=0
∴ Coordinates of centre of circle is (1212)( \frac{1}{\sqrt2} \frac{1}{\sqrt2} )
r=12+12−0r = \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=2⇒ AC = \sqrt2
∴∴ Area of ΔABC = 12\frac{1}{2} × AB × AC
12×2×2=22=1\frac{1}{2} × \sqrt2 × \sqrt2 = \frac{2}{2} = 1 sq. unit