# B Basic geometry problem with triangle

1. Mar 13, 2017

### Alex126

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.

Last edited: Mar 13, 2017
2. Mar 13, 2017

### 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.

3. Mar 13, 2017

### Staff: Mentor

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.

4. Mar 13, 2017

### Alex126

That's actually nice lol Thanks for the input.
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".

5. Mar 13, 2017

### Staff: Mentor

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.

6. Mar 13, 2017

### 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.

7. Mar 13, 2017

### Staff: Mentor

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.

8. Mar 13, 2017

### Alex126

Aaaah I see. Neat, thanks a lot :D