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if the sides of the triangle are given by 20 cm, 30 cm, and 60 cm find the area of the triangle.

I tried a long time. Apparently, Heron's formula does not seem to work

$\sqrt{s(s-a)(s-b)(s-c)}$

where $s = (a+b+c)/2$

In the above problem $s=55$ and thus we end up with a negative number inside square root. I am not sure if there is any other formula to be applied to this problem .

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Such triangle doesn't exist because $20+30\le60$, see wikipedia. –  Frank Science Jul 1 '12 at 12:45
    
@FrankScience. Thank you. I completely missed that point. –  sun123 Jul 1 '12 at 12:46
1  
In fact the boolean expression $s(s-a)(s-b)(s-c) > 0$ provides a perfect test if $a$, $b$ and $c$ are side lengths for a nondegenerate triangle. –  ncmathsadist Jul 1 '12 at 12:53
3  
BTW, failure to satisfy the Triangle Inequality is precisely the reason you end up with a negative inside the square root. My preferred way to write Heron's Formula is $\frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}$. A negative factor appears when a side is too long (and it's never the case ---for non-negative $a$,$b$,$c$--- that more than one factor goes negative). Thus, non-negativity under the Heron radical is a Litmus test for satisfaction of the Triangle Inequality. (Ah ... @ncmathsadist's comment appeared while I was composing mine. Same thing.) –  Blue Jul 1 '12 at 13:02

1 Answer 1

up vote 4 down vote accepted

Heron's formula works if the triangle exists. This triangle does not exist. By the triangle inequality applied to this triangle, we should have $60\le20+30$, which is false.

If the triangle inequality holds, then each of $s-a=\frac{b+c-a}{2}$, $s-b=\frac{a+c-b}{2}$, and $s-c=\frac{a+b-c}{2}$ is non-negative and so $s(s-a)(s-b)(s-c)\ge0$.

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