JEE Main 2022 · 29 Jun · Shift 2 · Mathematics Q11

← JEE Main 2022 · 29 Jun · Shift 2

MathematicsSingle correct+4 / -1Question 11 of 90

Let a triangle ABC be inscribed in the circle $x^2-\sqrt2(x+y)+y^2=0$ such that $\angle BAC=\dfrac\pi2$ . If the length of side AB is $\sqrt2$ , then the area of the $\Delta ABC$ is equal to :

\left(\sqrt2+\sqrt6\right)/3

\left(\sqrt6+\sqrt3\right)/2

\left(3+\sqrt3\right)/4

\left(\sqrt6+2\sqrt3\right)/4

Answer and worked solution

Official answer: Dropped: full marks to everyone

This question was dropped in NTA's final key. The working shows why no option is right.

Circle. Centre $\left(\frac1{\sqrt2},\frac1{\sqrt2}\right)$ , radius 1. A right angle at A makes BC a diameter, so $BC=2$ .

Area. $AC=\sqrt{4-2}=\sqrt2$ , so the area is $\frac12\cdot\sqrt2\cdot\sqrt2=1$ , which is none of the options.

Topics:Equation of a circle

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JEE Main 2022 · 29 Jun · Shift 2, Mathematics question 11