Area of a right triangle with little data

[mafagafo]

Homework Statement

Homework Equations

x^2 + y^2 = 9
A = 0.5xy
x ≠ y

The Attempt at a Solution

x^2 + y^2 = 9
A = xy/2

(x + y)^2 = x^2 + 2xy + y^2 = 9 + 2xy = 9 + 4A
A = ((x+y)^2 - 9)/4

Then I am lost. I need to find the area.

 

[BruceW]

it looks like you're trying to use some combination of x and y to find the answer. But why not try to find x and y individually, then find the answer?

 

[mafagafo]

Because I simply couldn't? The pdf that contained the problem is dealing with the Pythagorean Theorem. I do not think that I need more trigonometry for it.

If we call the point on the hypotenuse (that divides it in 1 and 2) P:
As x != y, the two acute angles are different from each other and, subsequently, I do not have right angles at the intersection with the hypotenuse.

To where?

 

[Saitama]

Name the triangle as $ABC$ right-angled at $B$ with $AB=x$. Let D be the point on AC such that AD=1.

Also, let $\angle BAC=\theta$, then you have $\angle ADB=\frac{3\pi}{4}-\theta$ and $\angle BDC=\frac{\pi}{4}+\theta$. Apply law of sines separately in triangles ADB and BDC to obtain x and y in terms of $\sin(\pi/4+\theta)$, you should then be able to find x and y from $x^2+y^2=9$.

 

[mafagafo]

Name the triangle as $ABC$ right-angled at $B$ with $AB=x$. Let D be the point on BC such that AD=1.

Also, let $\angle BAC=\theta$, then you have $\angle ADB=\frac{3\pi}{4}-\theta$ and $\angle BDC=\frac{\pi}{4}+\theta$. Apply law of sines separately in triangles ADB and BDC to obtain x and y in terms of $\sin(\pi/4+\theta)$, you should then be able to find x and y from $x^2+y^2=9$.

Thank you SO much. I remade the image, made the problem clearer with your info.
Right to the point:

\frac{x}{\sin{\left(\frac{3\pi}{4}-\theta\right)}}=\frac{2}{\frac{\sqrt{2}}{2}}\Rightarrow x=\frac{2}{\sqrt{2}}\sin{\left(\frac{3\pi}{4}-\theta\right)}=\left(\frac{2}{\sqrt{2}}\right)\left(\frac{\sqrt{2}}{2}\cos{\theta}+\frac{\sqrt{2}}{2}\sin{\theta}\right)\Rightarrow x=\cos{\theta}+\sin{\theta}
\frac{y}{\sin{\left(\frac{\pi}{4}+\theta\right)}}=\frac{2}{\frac{\sqrt{2}}{2}}\Rightarrow y=\frac{4}{\sqrt{2}}\sin{\left(\frac{\pi}{4}+\theta\right)}=\left(\frac{4}{\sqrt{2}}\right)\left(\frac{\sqrt{2}}{2}\cos{\theta}+\frac{\sqrt{2}}{2}\sin{\theta}\right)\Rightarrow y=2\left(\cos{\theta}+\sin{\theta}\right)
y=2x
y^2+x^2=5x^2=9
x=\sqrt{\frac{9}{5}}\Rightarrow A=\frac{x\cdot y}{2}=\frac{9}{5}

Can anyone, please, confirm this?

 

[Saitama]

Looks right to me! :)

 

[mafagafo]

Is the law of sines derived from the Pythagorean Theorem? If not, do you think it is doable without the law of sines?

 

[Saitama]

Is the law of sines derived from the Pythagorean Theorem? If not, do you think it is doable without the law of sines?

Yes, there is an alternative way, I missed it before.

Do you see that AD is an angle bisector? Can you recall some property related to it (hint, it is related to ratio of sides)?

 

[mafagafo]

\frac{x}{y}=\frac{1}{2}
Holy ****. I went all over the mountain instead of taking the tunnel!