Trignometry - Geometry

Since a regular hexagon can be considered to be made up of 6 equilateral triangles, a line joining the farthest vertices of a hexagon can be considered to be made up using the sides of two opposite equilateral triangle forming the hexagon. Hence, its length should be twice the side of the hexagon, in this case, 4 cm.

Now, AD divided the hexagon into two symmetrical halves. Hence, AD bisects angle D, and hence, angle ADC is $$60^{\circ\ }$$.

We can find out the value of AT using cosine formula:

$$AT^2=\ 4^2+1^2-2\times\ 1\times\ 4\cos60$$

$$AT^2=\ 17-4=13$$

$$AT=\sqrt{\ 13}$$

Applying cosine rule in triangle ACD, 

$$100+X^2-2\times\ 10\times\ X\cos30=400$$

$$X^2-10X\sqrt{\ 3}-300=0$$

Solving, we get X = $$\left(\frac{10\sqrt{\ 3}+10\sqrt{\ 15}}{2}\right)$$

Hence, area = 10Xsin 30 = $$\frac{\left(\frac{10\sqrt{\ 3}+10\sqrt{\ 15}}{2}\right)10}{2}$$

= $$25(\sqrt{3}+\sqrt{15})$$

It is given, Angle BAE = 45 degrees

This implies AE = BE

Let AE = BE = x 

In right-angled triangle ABD, it is given $$\angle ABC=\theta\ $$

$$\sin\theta=\dfrac{AD}{AB}\ $$

$$\sin\theta=\dfrac{AD}{x\sqrt{\ 2}}\ $$

$$\sqrt{\ 2}\sin\theta=\dfrac{AD}{BE}\ $$

The answer is option D.