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

1 Answer
George C.
Dec 26, 2015

There is no such triangle, since #2+2 < 9#

Explanation:

If a triangle has sides of length #a#, #b# and #c# then all of these conditions hold:

#a+b > c#

#b+c > a#

#c+a > b#

...unless you count empty triangles, in which case change the #>#'s into #>=#'s.

If you try to apply Heron's formula to lengths #a=2#, #b=9#, #c=2#, then you will find that you end up attempting to take the square root of a negative number, hence no Real area:

The semi-perimeter #sp# is given by:

#sp = (a+b+c)/2 = (2+9+2)/2 = 13/2#

Then Heron's formula for the area #A# is:

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

#=sqrt(13/2(13/2-2)(13/2-9)(13/2-2))#

#=sqrt((13/2)(9/2)(-5/2)(9/2))#

#=sqrt(-5265/16)#

It is possible to simplify this further, but there's no real point since it's clearly the square root of a negative quantity.

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