How do you use Heron's formula to determine the area of a triangle with sides of that are 12, 16, and 19 units in length?

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Answer 1 · 2018-07-16T11:16:01

#95.503\ \text{unit}^2#

The semi-perimeter #s# of the triangle having sides #a=12#, #b=16# & #c=19# is given as

#s=\frac{a+b+c}{2}#

#=\frac{12+16+19}{2}#

#=23.5#

Now, using Hero's formula, the area #\Delta# of triangle is given as follows

#\Delta=\sqrt{s(s-a)(s-b)(s-c)}#

#=\sqrt{23.5(23.5-12)(23.5-16)(23.5-19)}#

#=95.503\ \text{unit}^2#

Answer 2 · 2018-07-16T11:21:38

#"The area of the triangle is " Delta~~95.5036 sq.units#

We have,

#"Heron's formula : the area of the triangle is"#

#Delta=sqrt(s(s-a)(s-b)(s-c)) " , where s is semi perimeter"#

Let ,

#a=12, b=16 ,and c=19#

#=>s=(a+b+c)/2=(12+16+19)/2=23.5#

#=>s-a=23.5-12=11.5#

#s-b=23.5-16=7.5#

#s-c=23.5-19=4.5#
So ,
#Delta=sqrt(23.5(11.5)(7.5)(4.5))~~95.5036# square units