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 quadrirateral is $4\sqrt{3}$, find the remaining two sides.

Answer 1

Let AB = 2 and BC = 5. $\angle ABC={60}^0$ (given).
Since the quadrilateral ABCD= Area of $△$ ABC+ Area of $△$ACD
=$\frac{1}{2}AB.BCsin{60}^0+\frac{1}{2}CD.DAsin{120}^0$
=$\frac{1}{2}.2.5.\frac{\sqrt{3}}{2}+\frac{1}{2}c.d.\frac{\sqrt{3}}{2}=4\sqrt{3}$ (given)
$\therefore \frac{\sqrt{3}}{4}cd=4\sqrt{3}-\frac{-5\sqrt{3}}{2}=\frac{3\sqrt{3}}{2}2$
or cd= 6 ....(1)
Also $A{B}^2+B{C}^2-2AB.BCcos{60}^0=A{C}^2$
=$C{D}^2+D{A}^2-2CD.DAcos{120}^0$ by cosine rule.
or $4+25-\mathrm{2.2.5.}\frac{1}{2}=c^2+d^2+cd$
or $c^2+d^2+cd=19$ or $c^2+d^2=13$, By (1) ......(2)
Now from (1) and (2), we have
$c^2+d^2=13,c^2{d}^2=36$
$\therefore c^2$ and $d^2$ are the roots of
$t^2-13t+36=0\therefore t=9,4.$
or $c^2=9,d^2=4$ or $c^2=4,d^2=9$
$\therefore c=3,d=2orc=2,d=3.$
Hence the other two sides are 2 and 3.