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

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

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Answer 1 · 2015-12-26T00:02:02

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

Explanation:

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

$a + b > c$ $a+b > c$

$b + c > a$ $b+c > a$

$c + a > b$ $c+a > b$

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

If you try to apply Heron's formula to lengths $a = 2$ $a=2$ , $b = 9$ $b=9$ , $c = 2$ $c=2$ , 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 $s p$ $sp$ is given by:

$s p = \frac{a + b + c}{2} = \frac{2 + 9 + 2}{2} = \frac{13}{2}$ $sp = (a+b+c)/2 = (2+9+2)/2 = 13/2$

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

$A = \sqrt{s p (s p - a) (s p - b) (s p - c)}$ $A = sqrt(sp(sp-a)(sp-b)(sp-c))$

$= \sqrt{\frac{13}{2} (\frac{13}{2} - 2) (\frac{13}{2} - 9) (\frac{13}{2} - 2)}$ $=sqrt(13/2(13/2-2)(13/2-9)(13/2-2))$

$= \sqrt{(\frac{13}{2}) (\frac{9}{2}) (- \frac{5}{2}) (\frac{9}{2})}$ $=sqrt((13/2)(9/2)(-5/2)(9/2))$

$= \sqrt{- \frac{5265}{16}}$ $=sqrt(-5265/16)$

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

1 Answer

Explanation:

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