Question

A triangle has sides with lengths: 16, 14, and 19. How do you find the area of the triangle using Heron's formula?

Answer 1

≈ 109.665 square units

Explanation:

This is a two step process.

step 1 : Calculate half the perimeter (s) of the triangle

let a = 16 , b = 14 and c = 19

`s = (a+b+c)/2 = (16+14+19)/2 = 49/2 = 24.5`

step 2 : Calculate area

area `= sqrt(s(s-a)(s-b)(s-c)`

`= sqrt(24.5(24.5-16)(24.5-14)(24.5-19)`

`= sqrt(24.5xx8.5xx10.5xx5.5) ≈ 109.665" square units "`

Answer 2

`16("area")^2 = (16+14+19)(-16+14+19)(16-14+19)(16+14-19)`

`"area" = 7/4 sqrt{(17)(21)(11)} = 7 /4 sqrt(3927)`

Explanation:

Jim's answer seems to be the magic Heron's Formula answer when I asked the question

How do you use Heron's formula to find the area of a triangle with sides of lengths 19, 14, and 14?

using the app camera. There was a question with those exact numbers but it brought me the other answer to this question, with slightly different numbers. Hmmm.

If this is going to live forever, here's how to do it saving the fraction to the end.

`a=16, b=14, c=19,` area `S`

It's a cool formula, remember the pattern of the minus sign:

`16S^2 = (a+b+c)(-a+b+c)(a-b+c)(a+b-c)`

`16S^2 = (16+14+19)(-16+14+19)(16-14+19)(16+14-19)=(49)(17)(21)(11)`

`S = 7/4 sqrt{(17)(21)(11)} = 7 /4 sqrt(3927)`

While we're here, did you know

`16S^2 = 4a^2 b^2 - (c^2-a^2-b^2)^2 = (a^2+b^2+c^2)-2(a^4+b^4+c^4)`

All the lengths are squared in these, so they work great given coordinates, especially more than two dimensional coordinates.