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What is the number of non-congruent scalene triangles whose sides all have integral length, and the longest side has length 11?

What does it mean by non-congruent scalene triangles? Do we have to use Pythagoras' Theorem and use trial-and-error for each integer until 11?

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[..]the method of hit-and-trial or the famous trial-and-error is being referred here. – user21436 Mar 20 '12 at 15:14
@kannappan ya.i am sorry.its trial and error. – vikiiii Mar 20 '12 at 15:15

A triangle is scalene if all sides have different lengths.

You will not need the Pythagorean Theorem. You will need the fact that three positive numbers are the sides of a triangle if and only if the sum of any two is greater than the third. This is called the Triangle Inequality.

You will need the fact that two triangles are congruent if and only if they have the same side lengths.

There is general theory, but just try to make a complete list, by being very systematic. The longest side is $11$. There is no other side of length $11$, since the triangle is scalene, and in particular not isosceles. So the biggest candidate for second longest side is $10$.

How many scalene triangles have second longest side equal to $10$? To list and count, what are the possibilities for the smallest side? Can you see it is anything from $2$ to $9$, inclusive?

How many of our triangles have second longest side equal to $9$? Continue. It is easy to make a mistake, so make full explicit lists. Shortcuts come later!

Remark: If you work through this example systematically, you can learn enough to tackle similar problems without much trouble.

Let's introduce some notation. List the sides of the triangle as $(a,b,c)$, where $a>b>c$ (we can't have equality anywhere, since we are looking for scalene triangles.) In our case we have $a=11$, so we only care about the pair $(b,c)$.

A useful fact is that automatically we have $a+b>c$ and $a+c>b$. So for the Triangle Inequality, we only need to make sure that $b+c>a$.

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Can you please give an example by taking largest side=10.I am not able to get your answer totally , thats why needs some more explaination. – vikiiii Mar 27 '12 at 2:53
@vikiiii: You mean second largest side $10$, the largest is $11$. So for second largest side $10$, the smallest side is any of $9$, $8$, $7$, $6$, $5$, $4$, $3$, $2$ (eight possibilities). For second largest side $9$, smallest is any of $8$, $7$, $6$, $5$, $4$, $3$ (six possibilities). For second largest side $8$, smallest is any of $7$, $6$, $5$, $4$. For second largest side $7$, smallest is any of $6$, $5$. For second largest $6$ or below, there are no triangles. Add up. – André Nicolas Mar 27 '12 at 3:08

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