Question

Let the two adjacent sides of a cyclic quadrilateral be 2, 5 and the angle between them is $\frac{\pi }{3}$. If the area of quadrilateral is $4\sqrt{3}$ square units, then the remaining sides can be

• A. 2
• B. 4
• C. 3
• D. 6

Answer 1

Answer: (A), (C)

2
3

In the figure, the area of upper triangle is $\frac{1}{2}\times 2\times 5\times \sin {60}^o=\frac{5\sqrt{3}}{2}$

Using cosine rule in the upper triangle, we get
$\cos {60}^o=\frac{4+25-d^2}{20}$

$\Rightarrow d^2=29-10=19$

Area of the lower triangle= area of the quadrilateral - area of the upper triangle$=4\sqrt{3}-\frac{5\sqrt{3}}{2}=\frac{3\sqrt{3}}{2}$

Area of the lower triangle is $\frac{1}{2}ab\sin {120}^o=\frac{3\sqrt{3}}{2}$
$\Rightarrow ab=6$....(1)

Now, using cosine rule in the lower triangle, we get
$\cos {120}^o=\frac{a^2+b^2-d^2}{2ab}$

Using (1) in the above equation, we get $a^2+b^2=13$....(2)

Using both (1) and (2), we get $a=3$ and $b=2$ (or vice versa).