7. Special segments

Lesson

We know that the sides of a triangle can all be the same, two can be the same and one different, or all of them can be different. But how different can they be?

Can you construct a triangle with sides of lengths $1$1, $2$2, and $10$10? No matter how you might try, this is not possible - there just isn't enough length in the two shorter sides to get from one end of the long side to the other.

What about a triangle with side lengths $14$14, $12$12, and $25$25? To make one, we start by drawing a segment of length $25$25, and then place two circles - one of radius $14$14, and the other of radius $12$12 - at either end:

By creating two new segments using both ends of the original segment and one of the intersection points, we produce a triangle of the correct dimensions:

This is a very thin triangle! It only just made it across.

Now we know that any three numbers don't always make a triangle - but how can we tell whether three numbers do, or don't, represent the lengths of the sides of a triangle?

The answer is called the triangle inequality, and it says the following:

The triangle inequality

In any triangle, the sum of the lengths of any two sides is always greater than the length of the third side.

As an extra challenge, read this proof of the triangle inequality and see if you can replicate it on your own.

In practice, the contrapositive of this statement is also very useful:

Triangle inequality (contrapositive)

If the sum of two numbers is not greater than a third, then there is no triangle with those three numbers as side lengths.

Can a triangle have side lengths of $3$3, $7$7, and $9$9?

**Think**: We should find all three pairwise sums of the three numbers, and see if any of them are smaller than the number we didn't include in the sum.

**Do**: $3+7=10>9$3+7=10>9, $3+9=12>7$3+9=12>7, and $7+9=16>3$7+9=16>3. Since all three sums satisfy the triangle inequality, we know there is such a triangle.

Can a triangle have side lengths of $12$12, $7$7, and $4$4?

**Think**: We should find the three pairwise sums as before.

**Do**: $12+7=19>4$12+7=19>4 and $12+4=16>7$12+4=16>7, but $7+4=11<12$7+4=11<12. Since the final sum does **not** satisfy the triangle inequality, we know there is no such triangle.

**Reflect**: Unlike in the beginning, we didn't have to draw these lengths as triangles to determine whether or not it is possible. Many of the most powerful results in geometry tell you not just what is true for a particular shape or figure - they tell you all the possibilities of what might be true, and save you a lot of time and effort when solving problems!

Given three side lengths $7$7, $12$12, and $17$17:

Complete the following statements, using $=$=, $>$>, or $<$<:

$7+12\editable{}17$7+1217

$7+17\editable{}12$7+1712

$17+12\editable{}7$17+127

Do these lengths form a valid triangle?

Yes

ANo

B

Given three sides $3.1$3.1, $7.7$7.7 and $11.8$11.8.

Complete the following statements, using $=$=, $>$>, or $<$<:

$3.1+7.7\editable{}11.8$3.1+7.711.8

$3.1+11.8\editable{}7.7$3.1+11.87.7

$11.8+7.7\editable{}3.1$11.8+7.73.1

Is this a valid triangle?

Yes

ANo

B