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

Correct answer: (A) 1. Given, circle x^2-√2x+y+y^2=0 With centre (1√2,1√2 and radius 1.Since ∠BAC=π2, so BC will act as diameter,BC=diameter=2 (given)So, AC=√(BC^2-AB^2)=√2 and AB=√2Now, △ABC=1)/2AB·AC=1

Q40 / 90 · JEE Main 2022 (29 Jun Shift 2) · Mathematics

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

Correct answer
A
$1$
B
$\frac{(\sqrt{6} + \sqrt{3})}{2}$
C
$\frac{(\sqrt{3} + \sqrt{3})}{2}$
D
$\frac{(\sqrt{6} + 2 \sqrt{3})}{4}$

Solution

Given,

circle $x^{2} - \sqrt{2} (x + y) + y^{2} = 0$

With centre $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ and radius $1$ .

Since $∠ B A C = \frac{π}{2}$ , so $B C$ will act as diameter,

$B C =$ diameter $= 2$ (given)

So, $A C = \sqrt{B C^{2} - A B^{2}} = \sqrt{2}$ and $A B = \sqrt{2}$

Now, $△ A B C = \frac{1}{2} A B \cdot A C = 1$

Hence, the correct answer is option: 'A'.

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Let a triangle ABC be inscribed in the circle x2-2x+y+y2=0 such that ∠BAC=π2. If the length of side AB is 2, then the area of the △ABC is equal to:

Solution

Relevant Questions

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