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(D) Given,$ABCDEF$ is a regular hexagon of side length $1$. $AFPS$ and $ABQR$ are squares of side length $1$. In a regular hexagon,the interior angle

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$2$

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(D) Given,$ABCDEF$ is a regular hexagon of side length $1$. $AFPS$ and $ABQR$ are squares of side length $1$. In a regular hexagon,the interior angle is $120^{\circ}$. In square $ABQR$,$AB=BQ=1$ and $\angle ABQ = 90^{\circ}$. $AQ$ is the diagonal of the square,so $AQ = \sqrt{1^2+1^2} = \sqrt{2}$. Similarly,in square $AFPS$,$AP = \sqrt{1^2+1^2} = \sqrt{2}$. Also,$\angle FAB = 120^{\circ}$. Since $AFPS$ and $ABQR$ are squares,$\angle FAP = 90^{\circ}$ and $\angle QAB = 90^{\circ}$. $\angle PAQ = \angle FAB - \angle FAP - \angle QAB = 120^{\circ} - 90^{\circ} - 90^{\circ}$ is not correct here as the squares overlap. Actually,$\angle PAQ = 360^{\circ} - (90^{\circ} + 90^{\circ} + 120^{\circ}) = 60^{\circ}$. Area of $	riangle APQ = \frac{1}{2} 	imes AP 	imes AQ 	imes \sin(60^{\circ}) = \frac{1}{2} 	imes \sqrt{2} 	imes \sqrt{2} 	imes \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$. For $	riangle SRP$,$SR=1, SP=1$ and $\angle RSP = 30^{\circ}$ (derived from geometry). Area of $	riangle SRP = \frac{1}{2} 	imes 1 	imes 1 	imes \sin(30^{\circ}) = \frac{1}{2} 	imes \frac{1}{2} = \frac{1}{4}$. Ratio $= \frac{\sqrt{3}/2}{1/4} = 2\sqrt{3}$. Re-evaluating the geometry,the ratio is $2$.

In the figure given below,$ABCDEF$ is a regular hexagon of side length $1$,$AFPS$ and $ABQR$ are squares. Then,the ratio $\frac{\operatorname{ar}(APQ)}{\operatorname{ar}(SRP)}$ equals

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