High School Basic geometry problem with triangle

[Alex126]

Given a right triangle, where the hypotenuse measures 5, and the area measures 6, calculate the measurements of the legs.

This is a problem I thought of, and I was wondering how to mathematically solve it with an equation.

I tried calling one leg x.
So the other leg, because of Pitagora's theorem, is: √(5² - x²)

The area is equal to the product of the legs divided by two, so:

6 = (x * √(5² - x²))/2
12 = x * √(5² - x²)

Problem is, I don't know how to solve the equation beyond the first basic step lol

Btw, I know it can be solved with Euclide's second theorem, but I was wondering how to solve it with Pitagora, if it's possible to solve that equation I wrote. So basically I'm asking how to solve that equation.

 

[mjc123]

Well, you know a² + b² = 25 and ab/2 = 6. So you can work out a² + 2ab + b² and a² - 2ab + b²; hence a + b and a - b, hence a and b.

 

[mfb]

Btw, I know it can be solved with Euclide's second theorem, but I was wondering how to solve it with Pitagora, if it's possible to solve that equation I wrote. So basically I'm asking how to solve that equation.

You can solve the last equation by squaring it, afterwards it is a quadratic equation in x².
The approach described by mjc123 is easier, however.

 

[Alex126]

Well, you know a2 + b2 = 25 and ab/2 = 6. So you can work out a2 + 2ab + b2 and a2 - 2ab + b2; hence a + b and a - b, hence a and b.

That's actually nice lol Thanks for the input.

You can solve the last equation by squaring it, afterwards it is a quadratic equation in x2.

I think that's what I was looking forward to doing, but I don't know how that would work. So, could you elaborate on the passages required? I only know about the property where you can sum/subtract, multiply/divide something left and right in an equation, but I don't know how to proceed with "squaring".

 

[mfb]

If c=d, then c²=d². If you know that both c and d are positive (which is the case here), the reverse direction is true as well.

 

[Alex126]

Wouldn't I get 144 = x² * (25-x²) from 12 = x * √(5² - x²) then?
It would continue as 144 = 25 x² - x⁴, which would be a grade-four equation.

 

[mfb]

It is a quartic equation in x, but a quadratic equation in x².

144 = 25 (x²) - (x²)²

You can solve for x² with the usual formula for quadratic equations. Afterwards you can take the square root to find x.

 

[Alex126]

Aaaah I see. Neat, thanks a lot :D