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

1 Answer
Dec 23, 2015

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)#