Question

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

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

The correct option is C $5$

Let $AD=2$ and $AB=5$ and $\angle DAB={60}^{\circ }$
Since the quadrilateral is cyclic, $\angle BCD={120}^{\circ }$
Area of $\Delta ABD=\frac{1}{2}\mathrm{2.5.}\sin {60}^{\circ }=\frac{5\sqrt{3}}{2}\cdots \left(i\right)$
Area of $\Delta BCD=$Area of quadrialteral $ABCD-$ area of $\Delta ABD$
$\text{Area of}\Delta BCD=4\sqrt{3}-\frac{5\sqrt{3}}{2}=\frac{3\sqrt{3}}{2}\cdots \left(ii\right)$
Let $CD=x$ and $BC=y$
So, area of $\Delta BCD=\frac{1}{2}x.y.\sin {120}^{\circ }$
$\Rightarrow \frac{3\sqrt{3}}{2}=\frac{1}{2}x.y\frac{\sqrt{3}}{2}\left(\text{Using equation}\right(ii\left)\right)$
$\Rightarrow xy=6\cdots \left(iii\right)$
Now applying cosine rule in $\Delta BAD$, we get
$\cos {60}^{\circ }=\frac{A{D}^2+A{B}^2-B{D}^2}{2AD.AB}$
$\Rightarrow \frac{1}{2}=\frac{2^2+5^2-B{D}^2}{\mathrm{2.2.5}}\Rightarrow B{D}^2=19$
Again, applying cosine rule in $\Delta BCD$, we get
$\cos {120}^{\circ }=\frac{x^2+y^2-19}{2x.y}=-\frac{1}{2}$
$\Rightarrow x^2+y^2+xy=19\Rightarrow x^2+y^2=13(\because xy=6)$
Now, $x^2+y^2+2xy=13+12=25\Rightarrow (x+y{)}^2=25\Rightarrow x+y=5$