A semicircular arch for a tunnel under a river has a diameter of 80 feet (see figure) . Determine the height of the arch 4 feet from the edge of the tunnel.

A) 36 feet
B) $12\sqrt{11}$ feet
C) $4\sqrt{19}$ feet
D) $4\sqrt{399}$ feet
E) $6$ feet

Using the diameter of the arch, we can find the radius to be 40 feet.
Let's call the center of the semicircle point O, and let's call the point where the arch meets the ground (four feet from the edge of the tunnel) point A.
Then we can draw a right triangle OAB, where OB is the radius of the semicircle (40 ft), OA is the height we want to find (let's call it h), and AB is half of the length of the tunnel (40 ft).
Using the Pythagorean theorem, we get:
$$A{B}^2+O{A}^2=O{B}^2$$
$$(40^2)+(h^2)=(80{)}^2$$
Simplifying and solving for h, we get:
$$h=\sqrt{80^2-40^2}=\sqrt{(40)(120)}=\sqrt{(4\cdot 10)(4\cdot 30)}=4\sqrt{10\cdot 30}=4\sqrt{300}=4\sqrt{3\cdot 100}=4\sqrt{3}\cdot \sqrt{100}=4\sqrt{3}\cdot 10=40\sqrt{3}$$
Then, using similar triangles, we can find that the height 4 feet from the edge of the tunnel is $\frac{4}{40}=\frac{1}{10}$ of this value:
$$\frac{1}{10}(40\sqrt{3})=4\sqrt{3}\cdot \frac{\sqrt{10}}{10}=4\sqrt{\frac{30}{100}}=4\sqrt{0.3}=4\sqrt{3}\cdot \sqrt{0.1}=2\sqrt{3}\cdot \sqrt{10}\approx \boxed{\text{(C)}4\sqrt{19}\text{ feet}}$$