Now that we have covered the points of concurrency within a triangle, let’s talk about another relationship within a triangle: amongst its sides and angles.
TRIANGLE INEQUALITY (sides only)
Would you be able to create a triangle given any three side lengths? Use the following Desmos activity to help you gather your ideas and thoughts about this question:
TRIANGLE INEQUALITY (sides only)
Would you be able to create a triangle given any three side lengths? Use the following Desmos activity to help you gather your ideas and thoughts about this question:
Play around with each of the side lengths. Which combination of lengths worked? Which combination of lengths did not work? Let’s look at a specific example where \(AC = 25\) units, \(AB = 7.9\) units, and \(BC = 16.8\) units. Notice that these side lengths do NOT make a triangle. Why is that? Since the length of \(\overline{AC}\) is \( 25\) units, the other two sides are not long enough to intersect. So, in order to make a triangle, we would need either \(\overline{AB}\) or \(\overline{BC}\) to be longer.
So to answer the question, a triangle cannot be formed with any three side lengths. Here’s the rule that needs to occur in order to make a triangle:
So to answer the question, a triangle cannot be formed with any three side lengths. Here’s the rule that needs to occur in order to make a triangle:
Triangle Inequality Theorem
The sum of the lengths of the two smaller sides must be greater than the length of the third side.
The sum of the lengths of the two smaller sides must be greater than the length of the third side.
Let’s take a look at some examples.
Example 1:
Would the following lengths create a triangle?
a) \(3, 6, 4\)
b) \(12, 2, 10\)
c) \(1, 2, 5\)
Solution:
a) The longest side is \(6\) units. So we need to determine if the sum of \(3\) and \(4\) is greater than \(6\):
\(3 + 4 > 6\)?
\(7 > 6\)? Since \(7\) is greater than \(6\), that means YES, these side lengths would indeed create a triangle.
b) The longest side is \(12\) units. So we need to determine if the sum of \(2\) and \(10\) is greater than \(12\):
\(2 + 10 > 12\)?
\(12 > 12\)? Since \(12\) isn’t greater itself, that means NO, these side lengths would NOT create a triangle.
c) The longest side is \(5\) units. So we need to determine if the sum of \(1\) and \(2\) is greater than \(5\):
\(1 + 2 > 5\)?
\(3 > 5\)? Since \(3\) isn’t greater than \(5\), that means NO, these side lengths would NOT create a triangle.
Example 2:
Given two sides of a triangle have lengths \(15\) cm and \(10\) cm, what values could represent the value of the third side length?
Solution:
In this situation, we are unaware of what the third side length could be. Let’s label it \(x\) cm:
Example 1:
Would the following lengths create a triangle?
a) \(3, 6, 4\)
b) \(12, 2, 10\)
c) \(1, 2, 5\)
Solution:
a) The longest side is \(6\) units. So we need to determine if the sum of \(3\) and \(4\) is greater than \(6\):
\(3 + 4 > 6\)?
\(7 > 6\)? Since \(7\) is greater than \(6\), that means YES, these side lengths would indeed create a triangle.
b) The longest side is \(12\) units. So we need to determine if the sum of \(2\) and \(10\) is greater than \(12\):
\(2 + 10 > 12\)?
\(12 > 12\)? Since \(12\) isn’t greater itself, that means NO, these side lengths would NOT create a triangle.
c) The longest side is \(5\) units. So we need to determine if the sum of \(1\) and \(2\) is greater than \(5\):
\(1 + 2 > 5\)?
\(3 > 5\)? Since \(3\) isn’t greater than \(5\), that means NO, these side lengths would NOT create a triangle.
Example 2:
Given two sides of a triangle have lengths \(15\) cm and \(10\) cm, what values could represent the value of the third side length?
Solution:
In this situation, we are unaware of what the third side length could be. Let’s label it \(x\) cm:
Since we don’t know what \(x\) is, there are two options to weigh:
Option 1:
If \(x\) is the longest side, then the sum of the other two sides must be greater than \(x\). In other words:
\(10 + 15 > x\)
\(25 > x\)
We can re-write the inequality as \(x < 25\)
Option 2:
If \(15\) is the longest side, then the sum of the other two sides must be greater than \(15\). In other words:
\(x + 10 > 15\)
\(x > 5\)
So now we know that \(x > 5\) and \(x < 25\). This can be re-written as compound inequaltity: \(5 < x < 25\).
TRIANGLE INEQUALITY (sides and angles)
Another triangle inequality involves relating the side lengths and angles within a triangle. Compare the angle measures with the sides while moving around the vertices to see if you can come up with our next triangle inequality!
Option 1:
If \(x\) is the longest side, then the sum of the other two sides must be greater than \(x\). In other words:
\(10 + 15 > x\)
\(25 > x\)
We can re-write the inequality as \(x < 25\)
Option 2:
If \(15\) is the longest side, then the sum of the other two sides must be greater than \(15\). In other words:
\(x + 10 > 15\)
\(x > 5\)
So now we know that \(x > 5\) and \(x < 25\). This can be re-written as compound inequaltity: \(5 < x < 25\).
TRIANGLE INEQUALITY (sides and angles)
Another triangle inequality involves relating the side lengths and angles within a triangle. Compare the angle measures with the sides while moving around the vertices to see if you can come up with our next triangle inequality!
Let’s take a snapshot of the triangle to discuss our upcoming theorem:
Based on the angle measures, we know that is the smallest angle measure \(\angle C\). Now, imagine if you are standing at \(\angle C\). Then the side opposite of you would be the side you are facing, \(\overline{AB}\). Notice that \(\overline{AB}\) is the shortest side length, since it is opposite of the smallest angle measure.
Similarly, if we choose the angle with the largest measure, \(\angle A\), you will notice that the longest side, \(\overline{BC}\), is opposite of \(\angle A\). That brings us to our newest theorem:
Similarly, if we choose the angle with the largest measure, \(\angle A\), you will notice that the longest side, \(\overline{BC}\), is opposite of \(\angle A\). That brings us to our newest theorem:
Side & Angle Relationships in Triangles
In a triangle the longest side is opposite the largest angle, the medium side is opposite the middle angle, and the shortest side is opposite the smallest angle.
In a triangle the longest side is opposite the largest angle, the medium side is opposite the middle angle, and the shortest side is opposite the smallest angle.
Example 4:
Given the following triangle, write an inequality, from smallest to largest, that would represent the measures of the angles. Solution: Let’s first list the side lengths from shortest to longest: \(EM < MJ < EJ\). Let’s start with the shortest side: \(\overline{EM}\). If you were standing on \(\overline{EM}\), which angle would be opposite of you? This is the smallest angle, which is \(\angle J\). Now the medium side is \(\overline{MJ}\). Opposite that side is \(\angle E\). The longest side is \(\overline{EJ}\). Opposite this side is \(\angle M\), the largest angle. Therefore, the angles from smallest to largest would be written as: \(m\angle J < m\angle E < m\angle M\). |
Quick Check
1) Would the following lengths create a triangle?
a) \( 5, 5, 5\)
b) \(3, 12, 11\)
c) \(25, 12, 10\)
2) If two sides of a triangle have lengths \(12\) in and \(9\) in, what values could represent the value of the third side length?
3) Write an inequality to represent the measures of the angles and the lengths of sides from smallest to largest.
1) Would the following lengths create a triangle?
a) \( 5, 5, 5\)
b) \(3, 12, 11\)
c) \(25, 12, 10\)
2) If two sides of a triangle have lengths \(12\) in and \(9\) in, what values could represent the value of the third side length?
3) Write an inequality to represent the measures of the angles and the lengths of sides from smallest to largest.