QJEEM17Mabbd0: JEE Main 2017 Mathematics - Class 11 JEEM17Mabbd0 Solution: The two adjacent sides of a cyclic quadrilateral are 2 and 5…

Here; cos $\theta$  =  ${{{a^2} + {b^2} - {c^2}} \over {2ab}}$

and   $\theta$ = 60^o

$\Rightarrow$    cos 60^o = ${{4 + 25 - {c^2}} \over {2.2.5}}$

$\Rightarrow$     10 = 29 $-$ c²

$\Rightarrow$    c² = 19

$\Rightarrow$    c = $\sqrt {19}$

also;  cos $\theta$ = ${{{a^2} + {b^2} - {c^2}} \over {2ab}}$

and $\theta$ = 120^o

$\Rightarrow$     $-$ ${1 \over 2}$ = ${{{a^2} + {b^2} - 19} \over {2ab}}$

$\Rightarrow$    a² + b² $-$ 19 = $-$ ab

$\Rightarrow$    a² + b² + ab = 19

$\therefore$    Area = ${1 \over 2} \times 2 \times 5$ sin 60 + ${1 \over 2}$ ab sin 120^o = 4 $\sqrt 3$

$\Rightarrow$     ${{5\sqrt 3 } \over 2} + {{ab\sqrt 3 } \over 4}$ = $4\sqrt 3$

$\Rightarrow$     ${{ab} \over 4}$ = 4 $-$ ${5 \over 2}$ = ${3 \over 2}$

$\Rightarrow$     ab = 6

$\therefore$    a² + b² = 13

$\Rightarrow$    a = 2, b = 3

Perimeter = Sum of all sides

= 2 + 5 + 2 + 3 = 12