Right triangle inscribed in a square. Find the square area? I hope it's valid to ask for "a more neat solution" of a problem on this network, despite the fact that I don't have a strict definition of the word "neat".
Here is the square and the right triangle inscribed in it.

I did the following:
$$AC = Ah + hB$$
$$4\sin\theta = 4\cos\theta + 3\sin\theta$$
So
$$\tan\theta = 4$$
But
$$\sin\theta = \frac{\tan\theta}{\sqrt{1+\tan^2\theta}}$$
Therefore
$$AC = 4\sin\theta = \frac{16}{\sqrt {17}}$$
$$\text{Area} = \left(\frac{16}{\sqrt {17}}\right)^2$$
 A: Here it's another way. 
The triangles AhC and Bhf are similar. If you put $Bh=x$ and $Bf=y$ you get the relations
$$x+\frac{4}{3}y=\frac{4} {3} x$$
$$9=x^2+y^2$$
From which you can obtain the length of the side and then the area
A: Here is another way of doing it.
After discovering that $\tan \theta = 4$, all the line segments can then be expressed in terms of k with Ah = 1k as a start. 

Find the value of $k^2$ from 
$$16k^2 – \triangle yellow – \triangle green – \triangle blue = \triangle red = \dfrac {3 \times 4}{2}$$
Required area follows.
A: I wanted to make this a comment but my reputation is not high enough. I believe your way is the best and it's actually rather clever.  Your formula has a typo (forgot to square the tangent in the denominator. $$\sin(\theta)=\frac{\tan(\theta)}{\sqrt{1+\tan^2(\theta)}}$$
This formula is easily derived by considering a right triangle with legs $\tan(\theta)$ and $1$ (and therefore hypotenuse $\sqrt{1+\tan^2(\theta)}$.  That will be true for all right triangles by the definition of tangent.
A: Without using trigonometry:

enjoy...
By coloring triangles the same I am trying to emphasize the similar triangles, which are not necessarily equal (although red and blue triangles are). I don't mean that the triangles with the same color has the same area, be careful. 
A: I think your way is good, but we don't need to find $\tan\theta$.
From $4\sin\theta=4\cos\theta+3\sin\theta$, we have
$$\sin\theta=4\cos\theta$$
Squaring the both sides gives
$$\sin^2\theta=16(1-\sin^2\theta)$$
from which we can have
$$\sin^2\theta=\frac{16}{17}\quad\Rightarrow\quad \text{(area)}=16\sin^2\theta=\frac{16^2}{17}$$
A: Denote: $AB=a, BF=x, BH=y$. Then:
$$\begin{cases}a^2+(a-x)^2=25 \ (1)\\
(a-y)^2+a^2=16 \ (2)\\
x^2+y^2=9 \ (3)\end{cases}$$
$(1)-(2)$:
$$(a-x)^2-(a-y)^2=x^2+y^2 \Rightarrow 2a(y-x)=2y^2 \Rightarrow a=\frac{y^2}{y-x} \ \ \ \ \ \ \ \ \ \ \ (4)$$
$(4)\to (2)$:
$$\begin{align}\left(\frac{y^2}{y-x}-y\right)^2+\left(\frac{y^2}{y-x}\right)^2&=16 \Rightarrow \\
\frac{y^2(x^2+y^2)}{(y-x)^2}&=16 \stackrel{(3)}\Rightarrow \\
9y^2&=16(9-2xy) \stackrel{(3)}\Rightarrow \\
144-9y^2&=32y\sqrt{9-y^2} \Rightarrow \\
1104y^4-11808y^2+20736&=0 \Rightarrow \\
y_1&=\frac{12}{\sqrt{17}}; y_2=\frac{12}{\sqrt{65}} \Rightarrow \\
x_1&=\frac3{\sqrt{17}}; x_2=\frac{21}{\sqrt{65}}.\end{align}$$
Hence, from $(4)$:
$$a=\frac{y^2}{y-x}=\frac{\frac{144}{17}}{\frac9{\sqrt{17}}}=\frac{16}{\sqrt{17}}\\
y-x<0 \Rightarrow a\in \emptyset$$ 
Reference: WA answer.
A: Noting CAh and hBf are similar triangles in the linear ratio 4:3, redraw the original figure in a 4x4 square. The square of its hypotenuse Ch² is therefore 17 (see figure) - which is too large, it should be 4² or 16. Since the area of any similar figure is proportional to the square of any of its linear dimensions, simply scale the 4x4 square by 16/17 to find the area of the original figure: 16²/17
redrawn figure
A: let's say AC=x Ah=y and hB=x-y. hAC and fBh are similar triangles so:
Divide[x,4]=Divide[x-y,3]
y=Divide[x,4]
Divide[Power[x,2],16]+Power[x,2]=16
Power[x,2]=Divide[256,17]
Link to the solution
