Question

A triangle has sides with lengths: 2, 9, 1. How do you find the area of the triangle using Heron's formula?

Answer 1

There is no such triangle, but...

Explanation:

As noted by other contributors, it is not possible to form a triangle with sides `2`, `9` and `1`. If the sides of a triangle have lengths `a`, `b` and `c` then all of the following must hold:

`{(a+b > c), (b+c > a), (c+a > b):}`

and these conditions are sufficient.

For fun, let us try to apply Heron's formula and see what happens.

The semi-perimeter `s` is given by the formula:

`s = 1/2(a+b+c) = 1/2(2+9+1) = 6`

Then Heron's formula tells us that the area is:

`sqrt(s(s-a)(s-b)(s-c)) = sqrt(6(6-2)(6-9)(6-1))`

`color(white)(sqrt(s(s-a)(s-b)(s-c))) = sqrt(6(4)(-3)(5))`

`color(white)(sqrt(s(s-a)(s-b)(s-c))) = sqrt(-360)`

i.e. not a real area.

Notice that the three conditions we wrote down above hold if and only if all of `(s-a) > 0`, `(s-b) > 0` and `(s-c) > 0`.

Simply put, Heron's formula applies and gives a positive real value if and only if the conditions for sides of length `a`, `b` and `c` to be able to form a triangle hold.