Video: Finding the Range of All Possible Values of a Triangle’s Side Length Given the Lengths of the Other Sides

Find the range of all possible values of π‘₯ if (π‘₯ + 6) cm, 2 cm, and 25 cm represent the lengths of the sides of a triangle.

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Video Transcript

Find the range of all possible values of π‘₯ if π‘₯ plus six centimetres, two centimetres, and 25 centimetres represent the lengths of the sides of a triangle.

We should know a rule for the lengths of triangle sides. The triangle inequality theorem states, the lengths of any two sides of a triangle must sum to more than the length of the third side. If we imagine a triangle with side lengths π‘Ž, 𝑏, and 𝑐, this theorem tells us that if we add side lengths π‘Ž and 𝑏, it must be larger than side 𝑐. Also, if we add side lengths 𝑏 and 𝑐, they must be larger than side length π‘Ž. And if we add side lengths π‘Ž and 𝑐, they must be larger than side length 𝑏.

Our three sides are π‘₯ plus six, two, and 25. We can let π‘₯ plus six be side π‘Ž, two be side 𝑏, and 25 be side 𝑐. That means we need to say that π‘₯ plus six plus two must be greater than 25. That’s our first inequality. Two plus 25 must be greater than π‘₯ plus six, which is our second inequality. And π‘₯ plus six plus 25 must be greater than two, which is our third inequality.

Back to our first inequality, we can remove the parentheses since we’re only dealing with addition. And then, we combine like terms. Six plus two is eight. π‘₯ plus eight must be greater than 25. And so we subtract eight from either side to get π‘₯ by itself. And we see that π‘₯ is greater than 25 minus eight, which is 17. Our first inequality tells us that π‘₯ must be greater than 17.

Moving on to our second inequality, we can combine like terms on the left side. And we get 27 is greater than π‘₯ plus six. To get π‘₯ by itself, we subtract six from both sides. 27 minus six is 21. 21 is greater than π‘₯. Our second inequality tells us that 21 has to be greater than π‘₯. But it’s also true that π‘₯ has to be greater than 17. Since both of these must be true, we’re saying π‘₯ is greater than 17. And 21 is greater than π‘₯.

Before we do anything else, let’s check our third inequality. This one is side π‘Ž plus side 𝑏 has to be greater than side 𝑐. But in this case, side length 𝑏 is already greater than side length 𝑐. 25 is greater than two. And that means any positive value for π‘Ž will also make this statement true. Remember, we’ve already said that π‘₯ must be greater than 17. If π‘₯ is greater than 17, it will make the third inequality true.

We now need to consider the best way to write π‘₯ is greater than 17 and 21 is greater than π‘₯. We know that 21 is greater than π‘₯ and that π‘₯ is greater than 17. We can combine those statements to make something that looks like this. 21 is greater than π‘₯ and π‘₯ is greater than 17. Another way to write that would be 17 is less than π‘₯, which is less than 21. This is the more common way to write an inequality like this. The lower bound for the value of π‘₯ is on the left. And the upper bound for the value of π‘₯ is on the right, which is probably because that’s the way most number lines operate.

The range for π‘₯-values is between 17 and 21. And so we say that 17 is greater than π‘₯, which is greater than 21.

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