Sides of a triangle given an area and an equation.

1. Jun 7, 2014

maistral

WARNING: THIS IS NOT HOMEWORK~!!

Okay, so the problem goes like this:
"Find a,b,c of a triangle; If a+b+c = 10 ; Area = 10"

I know it sounds totally vague (I think so too). So I tried using the Pythagorean theorem;

c2 = a2+b2

then the given equation;
10 - a - b = c;

then the formula for a right triangle;
10 = 1/2 * a * b; b = 20/a

finally;
(10 - a - 20/a)2 = a2+(20/a)2

And solving for a, it does not converge to a solution. I tried Heron's formula as well for the area; it doesn't converge properly as well (negative value for b). Am I missing something here; or is this problem really unsolvable?

2. Jun 7, 2014

We can't use pythagoras theorem because we don't know whether it's a right angle triangle or not.

3. Jun 7, 2014

maistral

In that case, I don't know what the third equation is. Does anyone know? :(

4. Jun 7, 2014

jbriggs444

It seems clear that an equilateral triangle would maximize area for a given perimeter. [You could demonstrate this by fixing one side and reasoning that an isosceles triangle maximizes the area of a triangle with that fixed side. Apply the same reasoning to the remaining sides and you end up with an equilateral triangle]

The area of an equilateral triangle with perimeter 10 is way less than 10.

5. Jun 7, 2014

bhillyard

Try the formula for the area of a triangle sides a, b, c.
s is the semi perimeter, s=(a+b+c)/2

Area = √s(s-a)(s-b)(s-c)

6. Jun 7, 2014

mathman

There is no third equation.

Use Area = √{s (s - a)(s - b)(s - c)}, where area = 10, and s = a+b+c = 10.

Try fixing one side and see if you can come up with a sensible solution.

7. Jun 7, 2014

lurflurf

^that should be
Area = √{s (s - a)(s - b)(s - c)}, where area = 10, and 2s = a+b+c = 10
as mentioned above the maximal area is (25/9) √3~4.811<10
The problem is ill-posed
maybe a typo