JEE 2022 — Mathematics PYQ

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Question

JEE 2022 — Mathematics PYQ

JEE | Mathematics | 2022

Let a triangle $ABC$ be inscribed in the circle $x^2 - \sqrt{2}(x + y) + y^2 = 0$ such that $\angle BAC = \frac{\pi}{2}$ . If the length of side $AB$ is $\sqrt{2}$ , then the area of the $\triangle ABC$ is equal to:

Choose the correct answer:

Explanation

Solution

1. Circle Parameters:

Equation: $x^2 + y^2 - \sqrt{2}x - \sqrt{2}y = 0$ .

Center: $( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} ) = ( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} )$
Radius ( $R$ ): $\sqrt{(\frac{1}{\sqrt{2}})^2 + (\frac{1}{\sqrt{2}})^2 - 0} = \sqrt{\frac{1}{2} + \frac{1}{2}} = 1$

2. Hypotenuse ( $BC$ ):

Kyonki $\angle BAC = 90^\circ$ , isliye $BC$ circle ka diameter hai.

$BC = 2R = 2(1) = 2$

3. Side $AC$ Calculation:

Pythagoras theorem se:

AB^2 + AC^2 = BC^2

(\sqrt{2})^2 + AC^2 = (2)^2

2 + AC^2 = 4 \implies AC^2 = 2 \implies AC = \sqrt{2}

4. Final Area:

\text{Area}(\triangle ABC) = \frac{1}{2} \times AB \times AC

\text{Area} = \frac{1}{2} \times \sqrt{2} \times \sqrt{2}

\text{Area} = \frac{1}{2} \times 2 = 1

Choose the correct answer:

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