Question

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

• A. $12$
• B. $12.5$
• C. $13$
• D. $13.2$

Answer 1

Answer: (A)

$12$

Area $(\Delta$ ABCD$)=4\sqrt{3}$ -given
$=A\left(\Delta ABC\right)+A\left(\Delta ADC\right)$
In $\Delta ABC$,
Area $=\frac{1}{2}x$AB$\Delta$BC $\sin$B
$=\frac{1}{2}\times 2\times 5\times \sin {60}^o$
$=5\frac{\sqrt{3}}{2}$
$\therefore$Area$(\Delta$ADC$=4\sqrt{2}-\frac{5}{2}\sqrt{3}=\frac{3}{2}\sqrt{3}$ .........(I)
In $\Delta$ADC,
Area $=\frac{1}{2}\times AD\times CD\times \sin {120}^o$
$\frac{3}{2}\sqrt{x}=\frac{1}{x}\times c\times d\times \frac{\sqrt{3}}{2}$
from (i) cd$=6$ ......(ii)
In $\Delta$ABC,
$\cos B=\frac{2^2+5^2-A{C}^2}{2\times 2\times 5}$ ......cosive rule
$\frac{1}{2}=\frac{4+25-A{C}^2}{20}$

$(AC{)}^2=19$

..........(iii)
In $\Delta$ADC,
$\cos {120}^o=\frac{c^2+d^2-(AC{)}^2}{2cd}$ .......cosive rule
$-\frac{1}{2}=\frac{c^2{d}^2-19}{2\times 6}$ $\left(iii\right)-(AC{)}^2$ $\left(ii\right)-cd$
$c^2+d^2=13$ ......(iv)
$(iv+2(ii)=c^2+d^2+2cd$
$=13+2\times 6=13+12=25$
$(c+d{)}^2=25$
$\Rightarrow$ Perimeter $\square$ABCD$=2+5+c+d=7+5=12$