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

##### 1 Answer

There is no such triangle, but...

#### Explanation:

As noted by other contributors, it is not possible to form a triangle with sides

#{(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 = 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

Simply put, Heron's formula applies and gives a positive real value if and only if the conditions for sides of length