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


By angle-chasing one can show that $EP=AP=AQ$.  
Without loss of generality we can assume that $PQ=1$.  Let $x=EA$. Then $EP=AP=AQ=x-1$. 
Note that triangles $EQA$ and $APQ$ are similar. It follows that $\dfrac{EA}{AQ}=\dfrac{AQ}{PQ}$. Thus 
$$\frac{x}{x-1}=\frac{x-1}{1}.$$
This simplifies to $x^2-3x+1=0$, which has roots $\frac{3\pm\sqrt{5}}{2}$. One of these is less than $1$. So $x=\dfrac{3+\sqrt{5}}{2}$. 
Now the ratio of the area of the big pentagon to the little one is $x^2$ to $1$. 
Compute. The number $x^2$ simplifies to $\dfrac{7+3\sqrt{5}}{2}$. 
Remark: Note that $\frac{3+\sqrt{5}}{2}$ is the square of the famous "Golden Ratio" $\frac{1+\sqrt{5}}{2}$. The Golden Ratio is often denoted by $\varphi$. So the ratio of the areas is $\varphi^4$. 
