In the January Metroplex Math Circle presentation, the speaker also showed the equation to calculate the area (A) of a triangle from the length of its sides a, b, c. It is simple, clean and frankly quite elegant:

A = sqrt(s (s-a) (s-b) (s-c)), where s= 0.5 (a+b+c) half perimter of the triangle.

The derivation of the formula is not hard at all. Continuing from my proof that the altitudes of a triangle is concurent (orthocenetr):

m+n=a

n^2+h^2=b^2

m^2+ h^2 = c^2

so h^2 = 0.5[ c^2+ b^2 - 0.5h (a^2 +((c^2-b^2)/a)^2)]

A = 0.5 h a =0.5 a sqrt{0.5[ c^2+ b^2 - 0.5h (a^2 +((c^2-b^2)/a)^2)]}

using the relation (a+b+c)^2 = a^2+b^2+c^2+2ab + 2bc + 2ca, the above equation is simplified as

A =sqrt(s (s-a) (s-b) (s-c))

or h= 2 A/a = 2 sqrt(s (s-a) (s-b) (s-c))/a

**Radii of circumcircle, 9 point circle and incircle**

I showed in part 1, that radius of circumcircle R = b c /(2h), so

R= abc/(4 sqrt(s (s-a) (s-b) (s-c))) = abc/(4A).

The radius of 9 point circle R9 = 0.5 R

R9= 0.5 R = abc/(8 sqrt(s (s-a) (s-b) (s-c))) = abc/(8A)

The radius of inccirlce r, and it is easy to show that A = 0.5 s

r = 2A/s = 2 sqrt(s (s-a) (s-b) (s-c)))/s

All quatities are uniquely determined by the sides of the triangle, and the formulas are all clean and simple! There are many more interesting relations exist in a triangle - as a saying goes - **the more you know, the more you don't know!**

**Exploration is fun.**

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