# A triangle has sides with lengths: 2, 8, and 3. How do you find the area of the triangle using Heron's formula?

Jan 19, 2016

There is no such triangle since $2 + 3 < 8$

#### Explanation:

It is not possible to form a triangle with such lengths of sides.

If the lengths of the sides of a triangle are $a$, $b$ and $c$ then all of the following must hold:

$a + b > c$

$b + c > a$

$c + a > b$

Jan 19, 2016

$s = \frac{1}{2} \left(8 + 2 + 3\right) = \frac{13}{2}$
sqrt(13/2 (6.5-8)(6.5-2)(6.4-3)
Oops the this is not going to work, why do you think it is?

#### Explanation:

heron's formula is sqrt(s(s-a)(s-b)(s-c)
Where s = the semi-perimeter
And a,b,c are the sides.

Obviously this is not going to work, but the problem is
Not in the Heron's formula but your arbitrary apparent triangle.
But on the triangle inequality the a+b>c, i.e you must have
The sum of any two sides of triangle alway bigger than any single side. So the sum of sides 2 and 3 must add to more than 8 such that the semi perimeter is bigger than 8...