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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 .

Blue
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Vinoth Kumar C M
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    Such triangle doesn't exist because $20+30\le60$, see wikipedia. – Yai0Phah Jul 01 '12 at 12:45
  • @FrankScience. Thank you. I completely missed that point. – Vinoth Kumar C M Jul 01 '12 at 12:46
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    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 01 '12 at 12:53
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    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 01 '12 at 13:02

1 Answers1

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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 $s(s-a)(s-b)(s-c)\ge0$.

robjohn
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  • The triangle exists in a complex environment. – Ardent Coder Jun 15 '20 at 15:03
  • @ArdentCoder: even in $\mathbb{C}^n$, distances are real and positive. and the triangle inequality holds: $|z_1+z_2|\le|z_1|+|z_2|$. – robjohn Jun 15 '20 at 15:31
  • I simply felt so, I'm not a mathematician lol. But I still feel that the triangle somehow exists because using the same Heron's formula, we get a Complex area. Using the area and 0.5 * b * h, we get a Complex height as well. So, seemingly that traingle has a Real base and a Complex height. I'm afraid my question might get closed if I ask this, so I'm talking to you in comments. It's not very important, just that someone asked me a simple doubt but I got more involved in it and finally reached here haha – Ardent Coder Jun 15 '20 at 15:36
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    If you give context to your question, such as what you mention in this comment, then I don't think it should be closed because of context. I cannot promise that it will be well-received; that is a matter of public opinion. There are other questions about imaginary lengths in triangles, e.g. Pythagorean Theorem for imaginary numbers. Look for those and read those before posting. – robjohn Jun 15 '20 at 17:14
  • That's so polite to new users! Thanks for your advice, and the link where I see attempts to explain the point but in a very mathematical way. No problem, I might ask that question (if I remember lol but this is not a big deal, I sometimes dig deep into simple questions; actually it started with finding a bug) after graduation where I have Maths as a complementary course so that I could understand advanced Maths language in a better way :) – Ardent Coder Jun 15 '20 at 17:38