Question

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

• A. 12.5
• B. 13.2
• C. 12
• D. 13

Answer 1

The correct option is C 12

$\Rightarrow$ $\angle B+\angle D={180}^o$ [ Sum of opposite angles of cyclic quadrilateral is ${180}^o$ ]

$\Rightarrow$ ${60}^o+\angle D={180}^o$

$\therefore$ $\angle D={120}^o$

Area of cyclic quadrilateral $ABCD=$ Area of $△ABC+$ Area of $△ACD$

$\Rightarrow$ $4\sqrt{3}=\frac{1}{2}\times AB\times BC\sin {60}^o+\frac{1}{2}\times CD\times DA\sin {120}^o$

$\Rightarrow$ $4\sqrt{3}=\frac{1}{2}\times 2\times 5\times \frac{\sqrt{3}}{2}+\frac{1}{2}xy\times \frac{\sqrt{3}}{2}$

$\Rightarrow$ $4\sqrt{3}=\frac{\sqrt{3}}{4}(10+xy)$

$\Rightarrow$ $16=10+xy$

$\therefore$ $xy=6$ ----- ( 1 )

$\Rightarrow$ $(AC{)}^2=(AB{)}^2+(BC{)}^2-2AB\times BC\cos {60}^o$

$=(2{)}^2+(5{)}^2-2\times 2\times 5\times \frac{1}{2}$

$=4+25-10$

$=19$ ----- ( 2 )

$\Rightarrow$ $(AC{)}^2=(CD{)}^2+(DA{)}^2-2CD\times DA\cos {120}^o$

$=x^2+y^2+xy$ ----- ( 3 )

Equating ( 2 ) and ( 3 )

$\Rightarrow$ $x^2+y^2+xy=19$

$\Rightarrow$ $x^2+y^2+\frac{6}{y}y=19$ [ From ( 1 ) ]

$\Rightarrow$ $x^2+y^2=13$

Now, substitute $y=\frac{6}{x}$ we get,

$\Rightarrow$ $x^2+\frac{36}{x^2}=13$

$\Rightarrow$ $x^4-13x^2+36=0$

$\Rightarrow$ $x^4-9x^2-4x^2+36=0$

$\Rightarrow$ $x^2(x^2-9)-4(x^2-9)=0$

$\Rightarrow$ $(x^2-9)(x^2-4)=0$

$\therefore$ $x=3,2$

After substituting $x$ in $( 1 ) we get,

$y=2,3$

Let $x=3$ and $y=2$

$\Rightarrow$ Perimeter of cyclic quadrilateral $ABCD=2+5+3+2=12$