CL_JEE MAIN_29-06-2022_(S2) - Competishun – Live JEE & NEET Courses and Exam Prep

JEE MAIN 2022

29-06-2022 S2

Question

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

Select the correct option:

$\frac{(\sqrt{2}+\sqrt{6})}{3}$

$\frac{(\sqrt{6}+\sqrt{3})}{2}$

$\frac{(3+\sqrt{3})}{4}$

$\frac{(\sqrt{6}+2 \sqrt{3})}{4}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Question Tags

JEE Main

Mathematics

Easy

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