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Sides of a triangle given an area and an equation.

  1. Jun 7, 2014 #1

    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

    (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. jcsd
  3. Jun 7, 2014 #2


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    We can't use pythagoras theorem because we don't know whether it's a right angle triangle or not.
  4. Jun 7, 2014 #3
    In that case, I don't know what the third equation is. Does anyone know? :(
  5. Jun 7, 2014 #4


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    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.
  6. Jun 7, 2014 #5
    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)
  7. Jun 7, 2014 #6


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    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.
  8. Jun 7, 2014 #7


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    ^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
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