Let's try to coordinatize the problem (you don't need to do it this way, but it will help me). The center of the semicircle will be the origin (0,0). That means the border of the semicircle has the formula $y = \sqrt{25-x^2}$.
Here are the coordinates for points A,B,C,D,E:
$A (-5,0)$
$B (-5,10)$
$C (5,10)$
$D (5,0)$
$E (0,5)$
So, we just need to find point $F$. We know that the slope of the line $CF$ is equal to the slope of the tangent line at $F$. Suppose $F (a,b)$. Then we have:
$y'(a) = \dfrac{10-y(a)}{5-a}$
Let's plug in and try to solve for $a$:
$y'(a) = -\dfrac{a}{\sqrt{25-a^2}} = \dfrac{10-\sqrt{25-a^2}}{5-a}$
Cross multiplying gives:
$a(a-5) = 10\sqrt{25-a^2}-(25-a^2)$
$a^2-5a = 10\sqrt{25-a^2}-25+a^2$
$25-5a = 10\sqrt{25-a^2}$
Squaring both sides gives:
$625-250a+25a^2 = 100(25-a^2)$
$125a^2-250a-1875 = 0$
$a^2-2a-15 = 0$
$(a-5)(a+3) = 0$
$a=5$ or $a=-3$
$x = a=-3$ is the only solution that makes sense. Now $y = \sqrt{25-a^2} = \sqrt{25-9} = 4$ gives us the $y$ coordinate. So, $F$ is the point $(-3,4)$.
Finally, we plug into the area formula (given coordinates):
Area of $DEF$:
$\left|\dfrac{D_x(E_y-F_y)+E_x(F_y-D_y)+F_x(D_y-E_y)}{2}\right| = \left|\dfrac{5(5-4)+0(4-0)-3(0-5)}{2}\right| = 10$