How do you use Heron's formula to find the area of a triangle with sides of lengths #1 #, #1 #, and #2 #?

1 Answer
Jan 25, 2016

Heron's formula for finding area of the triangle is given by
#Area=sqrt(s(s-a)(s-b)(s-c))#

Where #s# is the semi perimeter and is defined as
#s=(a+b+c)/2#

and #a, b, c# are the lengths of the three sides of the triangle.

Here let #a=1, b=1# and #c=2#

#implies s=(1+1+2)/2=4/2=2#

#implies s=2#

#implies s-a=2-1=1, s-b=2-1=1 and s-c=2-2=0#
#implies s-a=1, s-b=1 and s-c=0#

#implies Area=sqrt(2*1*1*0)=sqrt0=0# square units

#implies Area=0# square units

Why is are 0?

The area is 0,because there exists no triangle with the given measurements the given measurements represent a line and a line has no area.

In any triangle the sum of any two sides must be greater than the third side.

If #a,b and c# are three sides then
#a+b>c#
#b+c>a#
#c+a>b#

Here #a=1, b=1# and #c=2#

#implies b+c=1+2=3>a# (Verified)
#implies c+a=2+1=3>b# (Verified)
#implies a+b=1+1=2cancel>c# (Not Verified)

Since, the property of triangle is not verified therefore, there exists no such triangle.