Question

The two adjacent sides of a cyclic quadrilateral are $2$ and $5$ and the angle between them is ${60}^0.$ If the area of the quadrilateral is $4\sqrt{3}$.sq.unit, then the remaining two sides are

• A. $2,3$
• B. $1,2$
• C. $3,4$
• D. $2,2$

Answer 1

The correct option is A $2,3$

$\because$ Area of a quadrilateral$=4\sqrt{3}$.sq.units (given)
$\Rightarrow \frac{1}{2}\times 2\times 5\sin {60}^0+\frac{1}{2}ab\sin {120}^0=4\sqrt{3}$
$\Rightarrow 5\times \frac{\sqrt{3}}{2}+\frac{1}{2}ab.\sin (180-60)=4\sqrt{3}$
$\Rightarrow \frac{5\sqrt{3}}{2}+\frac{ab\sqrt{3}}{2}\sin 60=4\sqrt{3}$
$\Rightarrow \frac{5\sqrt{3}}{2}+\frac{ab\sqrt{3}}{4}=4\sqrt{3}$
$\Rightarrow \frac{5}{2}+\frac{ab}{4}=4$
$\Rightarrow \frac{ab}{4}=4-\frac{5}{2}=\frac{3}{2}$
$\therefore \frac{ab}{4}=\frac{3}{2}$
Hence $ab=\frac{3}{2}\times 4=6$
and $\cos A=\frac{2^2+5^2-a^2-b^2}{2(10+ab)}=\frac{1}{2}$
$\Rightarrow -a^2-b^2-ab=10-29$
$\Rightarrow a^2+b^2+ab=19$
Since $ab=6$ we have
$\Rightarrow a^2+b^2=19-6=13$
$\therefore a+b=\sqrt{a^2+b^2+2ab}=\sqrt{13+2\times 6}=5$
Hence, $a+b=5,ab=6$
$a,b$ are the roots of $t^2-5t+6=0$
$\therefore t=2,3$
$\Rightarrow a=2,b=3$ or $a=3,b=2$