 ###### Brian McCall

Univ. of Wisconsin
J.D. Univ. of Wisconsin Law school

Brian was a geometry teacher through the Teach for America program and started the geometry program at his school

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# Triangle Side Inequalities - Problem 2

Brian McCall ###### Brian McCall

Univ. of Wisconsin
J.D. Univ. of Wisconsin Law school

Brian was a geometry teacher through the Teach for America program and started the geometry program at his school

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The sum of two side lengths of a triangle must be less than the length of the third side. So, given two side lengths, it is possible to find a range of values for the third side so that a triangle can be formed.

Let this third unknown side be of length x. If x is the greatest length, then the sum of the first two sides must be less than x. If x is one of the smaller lengths, the sum of x and the smaller length must be less than the third, largest length. With these inequalities, a range of values for x so that a triangle can be formed with the given lengths.

We can apply the triangle inequality theorem to some pretty complex problems. If we look at this problem right here, it says we have a triangle with two sides, one of length 4cm and one of length 12cm, describe the possible length of the third side x.

So if we’re to draw in some possibilities here, we could have a triangle with a side of 12 and a side of 4, and it will look something like that. Or we could have the 12 and the 4 a little bit further apart, like that, so you have 12 and 4, and we’d see that we have a big obtuse triangle. Or we can have the 12 and the four really close, something like that, and we would have another obtuse triangle.

So if we look at this side length here it could be pretty short, it could be pretty long or it could be something kind of in the middle. So to figure this out, let’s write three inequalities.

My first inequality is going to say, one of my sides plus 4, so x plus 4 is greater than 12. So I’m just taking any two sides, adding them up and saying that must be more than the third side. Our second inequality is going to say x plus 12 is greater than your other side, so we have X plus 12 is greater than 4. And your last inequality is going to say four plus 12 must be greater than x.

So we have three inequalities, pretty simple to solve. X plus 4 is greater than 12, we’ll subtract four on both sides and we see that X is greater than 8. So that’s one possibility. If we look at this one, we’re going to subtract 12 and we’ll see that x must be greater than -8. I can see that this is a negative number probably not going to use a negative number in geometry so we can just write a big x to that.

And last 4 plus 12 is 16, so we know that X must be less than 16. So we have two inequalities one says x must be greater than eight, the other says X must be less than 16. If you wanted to write it as a compound inequality, you’d say that x is greater than 8 and less than 16.