Question

How do you use Heron's formula to determine the area of a triangle with sides of that are 25, 28, and 22 units in length?

Answer 1

Area of triangle is `262.72` units

Explanation:

Heron's formula gives the area of a triangle with sides `a`, `b` and `c` as `sqrt(s(s-a)(s-b)(s-c))` where `s=1/2(a+b+c)`.

Hence to determine the area of a triangle with sides of `25, 28` and `22` units, first we find `s`, which is given by `s=1/2(25+28+22)=37.5`.

Hence area of triangle is

`sqrt(37.5xx(37.5-25)xx(37.5-28)xx(37.5-22))` or

`sqrt(37.5xx12.5xx9.5xx15.5)` or `sqrt(69,023.4375)` or `262.72` units

Answer 2

`16A^2=(25+28+22)(-25+28+22)(25-28+22)(25+28-22)` `= (75)(25)(19)(31)`

`A = \sqrt{{ 3(25)(25)(19)(31)}/16} =25/4 sqrt{1767}`

Explanation:

Heron's formula is usually among the poorest choices. Here are some modern forms:

For a triangle with sides `a,b,c` and area `A`,

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

`= 16 s(s-a)(s-b)(s-c)` where `s=1/2 (a+b+c)`

The last is of course Heron.

Usually given integer sides the factored form is fast and leads to the least messy exact answer:

`16A^2=(25+28+22)(-25+28+22)(25-28+22)(25+28-22)`

`= (75)(25)(19)(31)`

`A = \sqrt{{ 3(25)(25)(19)(31)}/16} = 25/4 \sqrt{3(19)(31)} = 25/4 sqrt{1767}`

If we're given 2d coordinates, the Shoelace Theorem is the quickest way to the area. If we're given 3 or more dimensional coordinates, the first two forms, which rely only on squared lengths, are best.

The first form is great for deriving the special cases as well:

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

Equilateral triangle : `a=b=c`

`16A^2 = 4a^4 - a^4 = 3a^4 quad` or `quad A = \sqrt{3}/4 a^2`

Isosceles triangle: `a=c,` base `b.`

`16A^2 = 4a^2b^2 - b^4 =b^2(4a^2-b^2) or A=b/4 \sqrt{4a^2-b^2}`

Right Triangle `c^2=a^2+b^2`

`16A^2 = 4a^2b^2 - (c^2 - a^2-b^2)^2 = 4a^2b^2 - 0` or `A=1/2 ab`

Cool. I'm not sure why they don't teach it in school.