Area of a right triangle with little data

In summary, the student is trying to find x and y from x^2+y^2=9, but is having trouble because (x+y)^2=x^2+2xy+y^2 and (x+y)^2-9 cannot be solved for x and y separately. They use the law of sines to solve for x and y in terms of sin(\pi/4+\theta) and are then able to find x and y from x^2+y^2=9.
  • #1
mafagafo
188
12

Homework Statement


9ld4qx.png


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.
 
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  • #2
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?
 
  • #3
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?
 
  • #4
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##.
 
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  • #5
Pranav-Arora said:
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:

[tex]\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}[/tex]
[tex]\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)[/tex]
[tex]y=2x[/tex]
[tex]y^2+x^2=5x^2=9[/tex]
[tex]x=\sqrt{\frac{9}{5}}\Rightarrow A=\frac{x\cdot y}{2}=\frac{9}{5}[/tex]

Can anyone, please, confirm this?
 
  • #6
Looks right to me! :)
 
  • #7
Is the law of sines derived from the Pythagorean Theorem? If not, do you think it is doable without the law of sines?
 
  • #8
mafagafo said:
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)?
 
  • #9
[tex]\frac{x}{y}=\frac{1}{2}[/tex]
Holy ****. I went all over the mountain instead of taking the tunnel!
 

What is the formula for finding the area of a right triangle with little data?

The formula for finding the area of a right triangle is A = (1/2)bh, where A is the area, b is the length of the base, and h is the height of the triangle.

What if I only have the length of one side and the measure of one angle?

In order to find the area of a right triangle with limited data, you will need to use the formula A = (1/2)ab*sinC, where a and b are the known sides of the triangle and C is the measure of the known angle.

Can I use the Pythagorean Theorem to find the area of a right triangle with little data?

No, the Pythagorean Theorem can only be used to find unknown side lengths in a right triangle. It cannot be used to find the area.

What is the unit of measurement for area in a right triangle?

The unit of measurement for area in a right triangle will be the square of the unit of measurement for the sides. For example, if the sides are measured in feet, the area will be measured in square feet.

Is there an easy way to remember the formula for finding the area of a right triangle?

One easy way to remember the formula is to think of the triangle as half of a rectangle. The base and height of the triangle are equivalent to the base and height of the rectangle, and the area of a rectangle is simply base times height.

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