Finding the shaded area in a triangle Here is the diagram:

I only know that the middle segment is a median of the big triangle. But nothing else.
 A: Let $h$ be the height of the triangles.  Then the area of the large triangle is $$\Delta_{\text{ large}} = \frac{1}{2}\times 6\times \sqrt{h^2+36} \times \sin a$$ and the area of the white triangle is $$\Delta_{\text{ white}} =\frac{1}{2} \times\sqrt{h^2+1}\times\sqrt{h^2+9}\times\sin a$$
But since the large triangle has the same height as the white triangle, but three times its base, we have $\Delta_{\text{ large}} = 3 \Delta_{\text{white}}$.  So
$$ 2\sqrt{h^2+36} = \sqrt{h^2+1}\sqrt{h^2+9}$$
Squaring both sides and simplifying gives
$$h^4+10h^2+9=4h^2+144$$
$$\Rightarrow (h^2+15)(h^2-9) = 0$$
So $h=3$, and the shaded area is $6$.
A: $\hspace1in$ 
Let $a, b, c$ be the 3 angles on the base (from left to right) and $h$ be the height of triangle.
We have 
$$h = 6 \tan a = 3 \tan b = \tan c$$
Since $c$ is an external angle for the white triangle in the middle,
$c = a + b$ and hence
$$\tan a = \tan(c-b) = \frac{\tan c - \tan b}{1 + \tan c \tan b}
= \frac{4\tan a}{1 + 12\tan^2 a} \implies \tan a = \frac12 \implies h = 3
$$
So the area of the shaded area is $\frac12(3+1)(3) = 6$.
A: Proof without words:
$\hspace{3cm}$
$$\frac gh=\frac{2}{\sqrt{h^2+9}}=\frac{\sqrt{h^2+1}}{\sqrt{h^2+36}} \Rightarrow h^4+6h^2-135=0 \Rightarrow h=3.$$
$$S=\frac12(3+1)3=6.$$
