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