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JEE Mains Previous Paper 1 (Held On: 10 Apr 2019 Shift 2)

Option 3 : \(2\sqrt 3 \)

JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)

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Given that A, B, C, are in A.P. ⇒ 2B = A + C

Now, A + B + C = π

⇒ B + 2B = π

⇒ 3B = π

\(B = \frac{\pi }{3}\)

\({\rm{Area\; = }}\frac{1}{2}\left( {4x} \right){\rm{sin\;}}60^\circ = \sqrt 3 x\) ----(i)

Now \({\rm{cos\;}}60^\circ = \frac{{16 + {x^2} - 3{x^2}}}{{8x}}\)

\( \Rightarrow \frac{1}{2} = \frac{{16 + {x^2} - 3{x^2}}}{{8x}}\)

\( \Rightarrow \frac{{8x}}{2} = 16 - 2{x^2}\)

⇒ 4x = 16 - 2x^{2}

⇒ 2x^{2 }+ 4x - 16 = 0

⇒ 2(x^{2 }+ 2x - 8) = 0

⇒ x^{2 }+ 2x - 8 = 0

⇒ x = 2

Substitute x = 2 in equation (i)

∴ Area \(\; = 2\sqrt 3 {\rm{\;sq}}{\rm{.\;cm}}\)