Let a triangle A B C be inscribed in the circle x 2- √2(x+y)+y2=0 such that angle B A C=(π/2). If the length of side A B is √2, then the area of the triangle ABC is equal to:

Question

Let a triangle A B C be inscribed in the circle x 2- √2(x+y)+y2=0 such that angle B A C=(π/2). If the length of side A B is √2, then the area of the triangle ABC is equal to: (A) (√2+√6) / 3 (B) (√6+√3) / 2 (C) (3+√3) / 4 (D) None of the above

Tardigrade Admin · Accepted Answer

Radius of given circle is 1 . BC = diameter =2, AB =√2 AC =√ BC 2- AB 2=√2 triangle ABC =(1/2) AB ⋅ AC =1 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/ba8c99d0108afffa4a3eb7110b8364e2-.png" />

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

$(2 + 6) /3$

$(6 + 3) /2$

$(3 + 3) /4$

None of the above

Radius of given circle is $1.$
$BC =$ diameter $= 2, A B = 2$
$A C = B C^{2} - A B^{2} = 2$
$△ A BC = \frac{1}{2} A B \cdot A C = 1$

Q. Let a triangle ABC be inscribed in the circle x2− 2​(x+y)+y2=0 such that ∠BAC=2π​. If the length of side AB is 2​, then the area of the △ABC is equal to:

(2​+6​)/3

(6​+3​)/2

(3+3​)/4