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Triangle inequality theorem

triangle inequalities

Triangle inequality theorem states that any side in a triangle is shorter than the sum of the last two. In other words you can not construct a triangle whose any side is lesser than the sum of the other two.

For example, let’s try to construct a triangle whose side c = 6, b = 2, and a = 3

triangle inequality theorem

And we got that it can’t be constructed. Let’s reflect on our theorem.
a + c = 9 which is greater than b which checks out.
c + b = 8 which is greater than a which also checks out.
a + b = 5 which is lesser than c, so here is our error.

If a + b would be equal to c, at best a and b would overlap with c which means that a + b must also be greater than c.

For example, for a triangle whose sides are 4, 5, 6.
4 + 5 = 9 > 6
5 + 6 = 11 > 4
4 + 6 = 10 > 5

And this triangle can be constructed:

construction of angle

Example 2.:Let’s say that we are given length of three line segments, and we are interested to find out if we can make a triangle out of these three lines. These lines are equal 4cm, 5cm and 12cm.

For every side, lengths of other two added together must be greater than it.

Let’s see:
4 + 12 = 16 which is greater than 5.
5 + 12 = 17 which is greater than 4.
4 + 5 = 9 which is lesser than 12, which can’t be if we want to construct a triangle.

This means that we can’t construct a triangle with these length of sides.

Example 3.: Now, you don’t want to guess every time the length of sides and then check it by this procedure. You want to set two sides and simply calculate all possible lengths of third side.

This is how you’d do it:
You have side a, and side b: a = 5, b = 7.
you know that c < a + b -> c < 12.
But, also c + a < b -> c > b - a -> c > 2,
and c + b < a -> c > a - b -> c > - 2

Now we look at the intersection of these intervals. c < 12, c > 2 -> c ɛ < 2, 12 >

In a triangle the longest side is across the greatest angle, and reversed, the greatest angle is across the longest side. This will be accurate for every triangle.

Try to draw any random triangle and measure its sides and angles.

inequality triangle

Observe this triangle for instance. Largest side here is obviously a with length of 12.96, across of a is α with measure 111.05^{\circ} which is the greatest measure in this triangle so this is true. Continue.
Second greatest side is c with length of 9.06, across c is \gamma whose measure equals 40.71^{\circ} which is the second greatest number. And of course across the smallest side is the angle with smallest measure.

Another important thing to remember is that triangles don’t have only one set of angles. There are also exterior angles. Those are angles that are supplementary to inner angles (they add up to 180^{\circ}).

inner and exterior angles in triangle

\alpha, \beta, \gamma – interior angles, \alpha ', \beta ', \gamma ' – exterior angles

What is interesting about these angles is that every exterior angle’s measure is equal to the sum of the other two interior angles (those who are not supplementary to that exterior angle).

That means that:
\gamma ' = \beta + \gamma
\beta ' = \alpha + \gamma
\gamma '= \alpha + \beta

From this you can conclude that every exterior angle is greater than any of the two other interior angles.

 

Triangle inequality theorem worksheets

  Angle - Integers (541.1 KiB, 203 hits)

  Angle - Decimals (554.1 KiB, 181 hits)

  Angle - Fractions/Mixed numbers (691.2 KiB, 206 hits)

  Side length (309.6 KiB, 256 hits)

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