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

# Question Video: Finding the Range of All Possible Values of a Triangleβs Side Length Given the Lengths of the Other Sides Mathematics • Second Year of Preparatory School

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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, π, and π, this theorem tells us that if we add side lengths π and π, it must be larger than side π. Also, if we add side lengths π and π, they must be larger than side length π. And if we add side lengths π and π, they must be larger than side length π.

Our three sides are π₯ plus six, two, and 25. We can let π₯ plus six be side π, two be side π, and 25 be side π. That means we need to say that π₯ plus six plus two must be greater than 25. Thatβ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 weβ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 itβs also true that π₯ has to be greater than 17. Since both of these must be true, weβre saying π₯ is greater than 17. And 21 is greater than π₯.

Before we do anything else, letβs check our third inequality. This one is side π plus side π has to be greater than side π. But in this case, side length π is already greater than side length π. 25 is greater than two. And that means any positive value for π will also make this statement true. Remember, weβ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 thatβ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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