Question

How do you use Heron's formula to find the area of a triangle with sides of lengths `15`, `16`, and `14`?

Answer 1

`"Area"~~96.56`

Explanation:

Heron's formula says that:
`"Area"=sqrt(S(S-A)(S-B)(S-C))` given `S=(A+B+C)/2`

For this triangle, `S=(14+15+16)/2=22.5`

`"Area"=sqrt(22.5(22.25-14)(22.5-15)(22.5-16))=sqrt(149175/16)=(15sqrt(663))/4~~96.56`

Answer 2

Know more than the other students! Archimedes' Theorem is a modern form of Heron's Formula. We set `a=15, b=16, c=14` in

`16A^2 = 4a^2b^2 - (c^2-a^2-b^2)^2 = 149175`

`A = (15 sqrt(663))/4`

Explanation:

Archimedes' Theorem is usually superior to Heron's Formula.

Given a triangle with sides `a,b,c` and area `A`, we have

`16A^2 = 4a^2b^2 - (a^2-a^2-b^2)^2`

`= (a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4)`

`= (a+b+c)(-a+b+c)(a-b+c)(a+b-c)`

No semiperimeter, no fractions till the end, no multiple square roots when given coordinates. We're free to choose whichever form is convenient and assign `a,b,c` however we see fit.