Finding the area of a triangle

**lanierms** · February 26th 2011, 05:02 AM

$a, b, c$ are the lengths of the sides of a triangle.

When they satisfy the equations : $a^2+b^2+c^2=6, a^4+b^4+c^4=14$ , find the area of the triangle.

I got to finding the value of $a^2b^2+b^2c^2+c^2a^2$ , but I can't progress any further.

**Unknown008** · February 26th 2011, 06:39 AM

Another version of Heron's formula is:

$Area = \dfrac14 \sqrt{(a^2 + b^2 + c^2)^2 - 2 (a^4 + b^4 + c^4)}$

**lanierms** · February 26th 2011, 07:00 PM

Isn't there another way of solving it? I haven't learned Heron's formula yet.

**Unknown008** · February 26th 2011, 11:51 PM

Oh? I thought you did, since the $a^2b^2 + b^2c^2 + a^2c^2$ is in one of the Heron's formula for the area of a triangle..

Heron's formula - Wikipedia, the free encyclopedia

$A = \sqrt{\dfrac{2(a^2b^2 + b^2c^2 + a^2c^2) - (a^4 + b^4 + c^4)}{16}}$

**lanierms** · February 27th 2011, 01:12 AM

Can't we just make it end up as a right triangle or a regular triangle (without using Heron's formula) so it would actually be possible to get the area?

Something like $a=b=c$ , or $a^2+b^2=c^2$ .

Anyway, very useful. I didn't know Heron's formula could change like that.

**Unknown008** · February 27th 2011, 01:36 AM

Sorry, but form what is given, I can't find of any way...

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