Is there a way to find the area of a triangle by only knowing the lengths of each side? I have been trying to find a way (and am almost done) I would just like to know if this has been solved before...

Printable View

- Sep 18th 2006, 02:43 PMQuickArea of Any Triangle
Is there a way to find the area of a triangle by only knowing the lengths of each side? I have been trying to find a way (and am almost done) I would just like to know if this has been solved before...

- Sep 18th 2006, 02:52 PMPlato
Is this what you mean?

Heron's Formula -- from Wolfram MathWorld - Sep 18th 2006, 02:55 PMQuick
- Sep 18th 2006, 04:47 PMSoroban
Hello, Quick!

Quote:

Is there a way to find the area of a triangle by only knowing the lengths of each side?

I have been trying to find a way (and am almost done).

I would just like to know if this has been solved before.

Yes . . . a**long**time ago . . .

__Heron's Formula__

If*a, b, c*are the sides of the triangle and s = (a + b + c)/2, the semiperimeter,

. . . . . . . . . . . . . . . . . . . . . . . . . _________________

then the area is given by: . A . = . √s(s - a)(s - b)(s - c)

- Sep 18th 2006, 05:19 PMThePerfectHacker
I can post a proof! (The one I knew from 8th grade, still remember)

(Once LaTeX is installed) - Sep 19th 2006, 12:39 AMGlaysher
- Sep 19th 2006, 03:13 AMQuick
- Sep 19th 2006, 04:51 AMtopsquark
- Sep 19th 2006, 02:54 PMQuick
- Sep 19th 2006, 04:03 PMtopsquark
Given a triangle with sides a, b, and c and angles A, B, and C across from that side (respectively) (ie. angle A is across from side a, etc.)

Law of Sines:

a/sin(A) = b/sin(B) = c/sin(C)

Law of Cosines:

a^2 = b^2 + c^2 - 2bc*cos(A)

b^2 = a^2 + c^2 - 2ac*cos(B)

c^2 = a^2 + b^2 - 2ab*cos(C)

(Notice the pattern.)

(Note also that the Law of Cosines is a generalization of the Pythagorean Theorem to a triangle that is not a right triangle.)

-Dan - Sep 19th 2006, 06:10 PMThePerfectHacker