Finding the ratio How to find the ratio between the area of the big  regular pentagon $ABCDE$ and the small  regular pentagon $PQRST$


The ratio of the areas is the same as the square of the ratios of the sides, so the ratio you want is
$$\frac{{\rm area}(ABCDE)}{{\rm area}(PQRST)} = \left( \frac{{\rm length}(AB)}{{\rm length}(PQ)}\right)^2$$
To calculate this, note that the angles $BAC$ and $CAD$ are all $\pi/5$, which you can get by noting that (a) the angle $BAE$ is $3\pi/5$ (sum of exterior angles is $2\pi$) and that $AEDB$ is a trapezium, so its interior angles sum to $2\pi$ and it two upper angles and two lower angles are equal.
Now you can use trigonometry. If we call the side length of the larger pentagon $a$, then dropping the perpendicular from point $A$ to intersect with $PQ$ at $S$ and $DC$ at $T$, leads to consideration of the triangles $ASE$ and $ATD$. The ratio of their heights is the same as the ratio of the length $AB:PQ$.
For the triangle $ASE$, you have
$$\sin(\pi/5) = {\rm length}(AS) / a$$ 
and for $ATD$ you have
$$\tan(2\pi/5) = 2\, {\rm length}(AT) / a$$
and combining these gives
$$\frac{{\rm length}(AT)}{{\rm length}(AS)} = \frac{\tan(2\pi/5)}{2\sin(\pi/5)}$$
and so the ratio of the areas is
$$\frac{\tan^2(2\pi/5)}{4\sin^2(\pi/5)} = \frac{1}{2}(7 + 3\sqrt{5}) \approx 6.85$$
