Join CE, OC and OE

Since, CD is parallel to AB, $x = 30$ (Alternate Interior angles are equal)
i.e. $∠CBA = 60$
Thus $∠COE = 120$ (Central angle is twice the angle subtended between two chords)

Arc Length CE $= \frac{120}{360}2π(5) = \frac{10π}{3}$

Now, CBE is an equilateral triangle inscribed in a circle with O as the centroid of the triangle
i.e. $OB = radius = \frac{2}{3}H$

$5 = \frac{2}{3}H$

$H = \frac{15}{2}$

We know, height of an equilateral triangle is given by $\frac{\sqrt{3}}{2}a$ where $a$ is the side of equilateral triangle
Thus, $\frac{\sqrt{3}}{2}a = \frac{15}{2}$

$a = \frac{15}{\sqrt{3}} = 5\sqrt{3}$

Perimeter of Shaded region =$\frac{10π}{3} + 5\sqrt{3} + 5\sqrt{3} = \frac{10π}{3} + 10\sqrt{3}$

Hence, option D