, 2014

The perimeter of the region includes the diameter and the semi-circle.

Since the radius of the region is 10, then the length of its diameter is 20.

Since the radius of the region is 10, then the circumference of an entire circle with this radius is $2\pi(10) = 20\pi$, so the arc length of the semi-circle is one-half of $20\pi$, or $10\pi$.

Therefore, the perimeter of the region is $10\pi + 20$.
The $x$-intercepts of the parabola with equation $y=10(x+2)(x-5)$ are $-2$ and $5$.

Since the line segment, $PQ$, joining these points is horizontal, then its length is the difference in the intercepts, or $5-(-2)=7$.
The slope of the line joining the points $C(0,60)$ and $D(30,0)$ is $\dfrac{60-0}{0-30}=\dfrac{60}{-30}=-2$.

Since this line passes through $C(0,60)$, then the $y$-intercept of the line is 60, and so an equation of the line is $y=-2x+60$.

We thus want to find the point of intersection, $E$, between the lines with equations $y=-2x+60$ and $y=2x$.

Equating $y$-coordinates, we obtain $-2x+60=2x$ or $4x=60$, and so $x=15$.

Substituting $x=15$ into the equation $y=2x$, we obtain $y=2(15)=30$.

Therefore, the coordinates of $E$ are $(15,30)$.