Sides of a triangle given an area and an equation.

  • Context: Undergrad 
  • Thread starter Thread starter maistral
  • Start date Start date
  • Tags Tags
    Area Triangle
Click For Summary

Discussion Overview

The discussion revolves around finding the sides of a triangle given the conditions that the sum of the sides equals 10 and the area equals 10. Participants explore various mathematical approaches and the implications of the problem's constraints.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant attempts to use the Pythagorean theorem, but questions its applicability since the triangle may not be a right triangle.
  • Another participant suggests that an equilateral triangle would maximize the area for a given perimeter, noting that the area of such a triangle with a perimeter of 10 is less than 10.
  • A participant introduces Heron's formula for the area of a triangle but expresses difficulty in finding a solution that satisfies the given conditions.
  • There is a suggestion to use the semi-perimeter in conjunction with Heron's formula, with the area set to 10 and the semi-perimeter calculated from the sum of the sides.
  • One participant points out that the problem may be ill-posed, suggesting a possible typo in the problem statement.

Areas of Agreement / Disagreement

Participants express uncertainty about the solvability of the problem, with some suggesting that it may be ill-posed. There is no consensus on a definitive method to find the sides of the triangle under the given constraints.

Contextual Notes

Limitations include the ambiguity of the triangle type (right or otherwise) and the potential for a typo in the problem statement, which affects the ability to find a valid solution.

maistral
Messages
235
Reaction score
17
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?
 
Mathematics news on Phys.org
maistral said:
So I tried using the Pythagorean theorem;

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

In that case, I don't know what the third equation is. Does anyone know? :(
 
maistral said:
Am I missing something here; or is this problem really unsolvable?

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

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.
 
^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
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
4
Views
2K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 9 ·
Replies
9
Views
3K