Finding the shaded area in a triangle

Solution 1:

$\hspace1in$

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 $\frac12(3+1)(3) = 6$.

Solution 2:

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 $6$.

Solution 3:

Proof without words:

$\hspace{3cm}$

$$\frac gh=\frac{2}{\sqrt{h^2+9}}=\frac{\sqrt{h^2+1}}{\sqrt{h^2+36}} \Rightarrow h^4+6h^2-135=0 \Rightarrow h=3.$$ $$S=\frac12(3+1)3=6.$$