Question

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

पर्याय

• 1

• `(sqrt(6) + sqrt(3))/2`

• `(3 + sqrt(3))/4`

• `(sqrt(6) + 2sqrt(3))/4`

Answer 1

Let a triangle ABC be inscribed in the circle x² `-sqrt(2)(x + y) + y^2` = 0 such that ∠BAC = `π/2`. If the length of side AB is `sqrt(2)`, then the area of the ΔABC is equal to 1.

Explanation:

x² `-sqrt(2)(x + y) + y^2` = 0

∴ Coordinates of centre of circle is `(1/sqrt(2) 1/sqrt(2))`

r = `sqrt(1/2 + 1/2 - 0)`

r = 1

BC = 2

Apply Pythogoras theorem in ΔABC, we get

AC² + AB² = BC²

⇒ AC² = 4 - 2 = 2

⇒ AC = `sqrt2`

∴ Area of ΔABC = `1/2xx"AB"xx"AC"`

`1/2 xx sqrt(2) xx sqrt(2) = \cancel2/\cancel2` = 1 square unit.