Question

A triangle has sides with lengths: 1, 5, and 3. How do you find the area of the triangle using Heron's formula?

Answer 1

There is no such triangle, since `1 + 3 < 5`.

Explanation:

Interestingly, if you attempt to apply Heron's formula with such lengths then you will find yourself trying to take the square root of a negative number, hence no Real area...

Let `a=1`, `b=5`, `c=3`.

Then the semiperimeter is defined by the formula:

`sp = (a+b+c)/2 = (1+5+3)/2 = 9/2`

And the area is given by the formula:

`A = sqrt(sp(sp-a)(sp-b)(sp-c))`

`=sqrt(9/2(9/2-1)(9/2-5)(9/2-3))`

`=sqrt(9/2*7/2*-1/2*3/2)`

`=sqrt(-189/16)`