An immediate alternative to a trigonometry problem (high school) I have a right triangle, $AH\perp BC$, where $\cos \beta=\sqrt 5/5$ and $\overline{AH}+\overline{CH}+\overline{HB}=7$.

I have to found the area.
My steps (or solution):
$$\mathcal A(\triangle ABC)=\frac 12\overline{AB}\cdot \overline{BC} \sin \beta$$
with $0<\beta<\pi$. Hence $\sin \beta=\sqrt{1-5/25}=2\sqrt5/5$. But $\overline{AB}=\overline{BC}\cos \beta$ and the area
$$\mathcal A(\triangle ABC)=\frac12(\overline{BC}\cos \beta)\cdot (\overline{BC}\sin \beta)=\frac12\overline{BC}^2 \tag 1$$
We know that:
$$\overline{AH}+\overline{CH}+\overline{HB}=7, \quad \text{with}\quad  \overline{CH}+\overline{HB}=\overline{BC}$$
Hence
$$\overline{AH}+\overline{BC}=7 \iff \overline{BC}=7-\overline{AH}$$
and $\overline{AH}=\overline{AB}\cos\gamma=\overline{BC}\cos\gamma\cos\beta$.
Putting this last identity to the condition $\overline{BC}=7-\overline{AH}$ I will have:
$$\overline{BC}=7-\overline{BC}\cos\gamma\cos\beta \iff \overline{BC}(1+\cos\gamma\cos\beta)=7$$
After,
$$\overline{BC}=\frac{7}{1+\cos\gamma\cos\beta}$$
with $\cos \gamma=\cos(\pi/2-\cos\beta)=\sin\beta=2\sqrt5/5$.
Definitively
$$\overline{BC}^2=\frac{49}{\left(1+\frac 25\right)^2}=\frac{49}{\frac{49}{25}}$$
and
$$\mathcal{A}(\triangle ABC)=\frac15\overline{BC}^2=5 \tag 2$$
I'm very tired and often don't find immediate solutions to problems. I had thought of using Euclid's second theorem by placing
$\overline{AH}=x$ and $\overline{BC}=y$, and $x+y=7$. After I will have $x^2=\overline{BH}\cdot \overline{CH}$. But I have left this path.

Can this possible alternative be used or is there another more immediate way?

 A: The picture should really look like this:

We have $\sin\beta=2/\sqrt5$ and $\tan\beta=CA/AB=2$. Then $HB,HA,HC$ are in geometric progression with ratio $2$; since their sum is $7$ it can easily be worked out that the hypotenuse $BC$ is $5$. Then $AB/BC=\cos\beta$ gives $AB=\sqrt5$, $CA=2\sqrt5$ and $A(\triangle ABC)=5$.
A: $$\cos \beta = \frac{|AB|}{|BC|} = \frac{\sqrt 5}{5},\quad  \sin \beta = \frac{|AH|}{|AB|} = \frac{2\sqrt 5}{5}$$
Given $|AB|: |BC| = \sqrt5:5$, we assume $$|AB| = \sqrt5k, |BC| = 5k $$ where $k$ is a constant. Now using $$\frac{|AH|}{|AB|} = \frac{2\sqrt 5}{5}, \quad |AH| = 2k$$
$$|AH| + |BH| + |HC| = |AH| + |BC| = 7k = 7\implies k = 1.$$
So, $$|AH| = 2, |BC| = 5 \implies \mathcal{A}(\triangle ABC) = \frac{1}{2} \cdot |AH| \cdot |BC| = 5.$$
A: HINT
Let $a = \overline{AH}$, $b = \overline{CH}$ and $c = \overline{HB}$. Then we have that
\begin{align*}
\begin{cases}
a = \tan(\beta)c\\\\
a = \tan(\pi/2 - \beta)b
\end{cases} \Rightarrow a(1 + \tan(\beta) + \cot(\beta)) = 7
\end{align*}
Can you take it from here?
A: $\cos\beta=\dfrac{\sqrt5}{5}$ and $\sin\beta=\dfrac{2\sqrt5}{5}.$ Let $\overline{AH}=x,$ then you can obtain values $$\overline{AB}= \dfrac{\sqrt{5}x}{2},\qquad \overline{BC}= \dfrac{5x}{2},\qquad \overline{AC}=\sqrt{5}x$$ and $$\overline{AH}=x,\qquad \overline{BH}=\dfrac{x}{2},\qquad \overline{CH}=2x$$ using simple trigonometry. Now, compute the value of $x$ via $\overline{AH}+\overline{BH}+\overline{CH}=7$ and use it to compute the area.
A: HINT

Start with length of $ HB =1$.
Chase the sides using Pythagoras thm. A property of right triangles can be used ( square of altitude equals product of base segments ; $HC=HA^2/HB$).
$$\text{ScaleFactor=}\frac{7}{4+2+1}=1$$
$$\text{Area = ScaleFactor}^2 \cdot \frac12 \cdot (4+1) \cdot 2 =5. $$
