The lengths of two adjacent sides of a cyclic quadrilateral are 2 units and 5 units and the angle between…

1. $12.5$ units
2. $13$ units
3. $13.2$ units
4. $12$ units

Solution

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

$\Rightarrow 10=29-c^2$

$\Rightarrow c^2=19$

$\Rightarrow c=\sqrt{19}$

Now,

$\cos 120°=\frac{a^2+b^2-c^2}{2ab}$

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

$\Rightarrow a^2+b^2-19=-ab$

$\Rightarrow a^2+b^2+ab=19$

Area of quadrilateral  $=\frac{1}{2}\times 2\times 5\sin \mathrm{60°}+\frac{1}{2}ab\sin \mathrm{120°}=4\sqrt{3}$

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

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

$\Rightarrow ab=6$

$\Rightarrow a^2+b^2=13$

$\Rightarrow a=2, b=3$

Perimeter

$=2+5+2+3$

$=12$ units

Asked in: JEE Main 2017 (09 Apr Online)

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