17. D In the diagram, $V$ is the foot of the perpendicular from $T$ to $P Q$.
Therefore, as $P Q T$ is an equilateral triangle, $V$ is the midpoint of $P Q$.
By Pythagoras' Theorem in triangle $V Q T, V T^{2}+V Q^{2}=Q T^{2}$.
So $V T=\sqrt{2^{2}-1^{2}}=\sqrt{3}$.

Hence the area of triangle $P Q T$ is $\frac{1}{2} \times 2 \times \sqrt{3}=\sqrt{3}$. Similarly, the area of triangle $S R U=\sqrt{3}$.
Therefore, as trapezia $Q R U T$ and $P T U S$ are congruent, the area of trapezium $Q R U T$ is $\frac{1}{2}(4 \times 2-2 \sqrt{3})=4-\sqrt{3}$.
17. In the rectangle $P Q R S$, the side $P Q$ is of length 2 and the side $Q R$ is of length 4 . Points $T$ and $U$ lie inside the rectangle so that $P Q T$ and $R S U$ are equilateral triangles.
What is the area of the quadrilateral $Q R U T$ ?
A $\frac{6-\sqrt{3}}{2}$
B $\frac{8}{3}$
C $4-2 \sqrt{3}$
D $4-\sqrt{3}$
E 3