Properties of Hexagon and Octagon

Area of a regular hexagon = $$\frac{3\sqrt{3}}{2}x^2$$

Area of an equilateral triangle = $$\frac{\sqrt{3}}{4}\left(a\right)^2$$ ; where a = side of the triangle

Since the area of the two figures are equal, we can equate them as folllows: $$\frac{3\sqrt{3}}{2}x^2=\frac{\sqrt{3}}{4}\left(12\right)^2$$

On simplifying: $$x^2=24\ $$

$$\therefore\ x=2\sqrt{6}$$

Let the side of the hexagon be a. 

The area of the whole hexagon is going to be $$\frac{3\sqrt{3}a^2}{2}$$

We know that the longest diagonal of a hexagon is 2 times the side of the hexagon.

PQ = $$\frac{\left(BC+AD\right)}{2}= \frac{\left(a+2a\right)}{2} =1.5a$$

The area of the trapezium =  Average of parllel sides * Height = $$\frac{\left(BC+PQ\right)}{2}\cdot h=\frac{5}{4}a\cdot h$$

The distance between BC and AD will be $$\sqrt{3}a$$, that is, the length of diagonal BF.

The distance between BC and AD will be $$\frac{\sqrt{3}a}{2}$$ 

And the height of the trapezium BCQP will be  $$\frac{\sqrt{3}a}{4}$$

Area of trapezium = $$\frac{5}{4}a\cdot \frac{\sqrt{3}a}{4}$$ = $$\frac{5\sqrt{3}a^2}{16}$$ 

Area of hexagon ABCDEF = $$\frac{3\sqrt{3}}{2}a^2$$

We have to find the ratio  = $$\frac{5\sqrt{3}a^2}{16}$$ : $$\frac{3\sqrt{3}}{2}a^2$$ = 5:24

The area of the hexagon is 3$$\frac{\sqrt{3}}{2}*a^{2}$$.

Let x be the length of BF

Each interior angle=$$\frac{(n-2)*180}{n}$$

=$$120^{0}$$

From cosine rule 

cos 120=$$\frac{(a^{2}+a^{2}-x^{2})}{2*a*a}$$

x=$$\sqrt{3}$$a

The length of BF and CE equal a$$\sqrt{3}$$. 

Ratio of the areas of the hexagon and the rectangle BCEF =  3$$\frac{\sqrt{3}}{2}*a^{2}$$ : a$$\sqrt{3}$$.*a

So, the required ratio is 3:2