Video Transcript
In this video, we will learn how to
define the triangle inequality and identify if the given side lengths are valid for
constructing a triangle. We will discover that a triangle
with side lengths π, π, and π only exists if the two shorter side lengths have a
sum that is greater than the third side length.
Letβs begin by considering how we
might construct a triangle using a ruler and a pair of compasses. First, we will use a ruler to
sketch the base of our triangle, naming it segment π΅πΆ and giving it a length of
π. Next, we will construct a circle
with radius π centered at point π΅. Then, we construct a second circle
with radius π centered at point πΆ. Now, if we sketch a point π΄ at the
intersection of circle π΅ and πΆ, we will be able to form triangle π΄π΅πΆ.
Now, itβs important to notice that
when we constructed these two circles at the end points of segment π΅πΆ, the circles
had two points of intersection. And we couldβve formed a triangle
with vertex π΄ at either of these two points of intersection. At this point, it will be helpful
for us to consider what we know about the lengths π, π, and π for this first
construction.
We recall that the shortest
distance between two points is a straight line, so that means that the shortest
distance between point π΅ and πΆ is length π. Therefore, any other path from
point π΅ to point πΆ would be longer than the length π. So, letβs say that we made a detour
from point π΅ to point π΄ before going to point πΆ. That distance is represented by the
sum of lengths π plus π, which is certainly longer than length π. This conclusion leads us to wonder
what if π plus π was not greater than π but instead π plus π equaled π.
To explore this idea, we will erase
our original construction and create a new one, where π plus π equals π. We will begin again with the
segment π΅πΆ of length π. But now, weβre working with a
restriction that says π plus π must equal π. So after using our compass to
construct circle π΅ with radius π and circle πΆ with radius π, we see that these
two circles meet at a point of tangency, and weβll name that point π΄. In our first construction, point π΄
became the third vertex of a triangle. But now that π plus π equals π,
we donβt have a triangle at all; we actually have a line segment.
Letβs stop to consider what we have
learned so far from these first two constructions. We have learned that to form a
triangle from lengths π, π, and π, π plus π can be greater than π, but π plus
π cannot equal π.
Now, thereβs only one more
possibility to be considered: can π plus π be less than π? Letβs do a new construction to
examine this possibility. As shown in this third
construction, if the sum of the two radii is less than π, then these two circles
will never intersect at all and they certainly will not form a triangle. Therefore, we conclude that to form
a triangle with side lengths π, π, and π, it is perfectly fine for π plus π to
be greater than π, as we saw in our first construction. However, it is not possible to form
a triangle when π plus π equals π or when π plus π is less than π.
Now, we will clear the screen in
order to make space to write out the formal expression of this idea, called the
triangle inequality theorem. The triangle inequality theorem
states that the sum of the lengths of any two sides of a triangle must be greater
than the length of the third side. This means that for any triangle
π΄π΅πΆ, π΄π΅ plus π΄πΆ is greater than π΅πΆ, π΄π΅ plus π΅πΆ is greater than π΄πΆ,
and π΄πΆ plus π΅πΆ is greater than π΄π΅. We can use the triangle inequality
theorem to determine if it is possible to construct a triangle with three given side
lengths. Letβs put this into practice in our
first example.
Is it possible to form a triangle
with side lengths six meters, seven meters, and 18 meters?
To answer this question, we will
need to recall the triangle inequality theorem, which tells us that the sum of the
lengths of any two sides of a triangle must be greater than the length of the third
side. This means that for any triangle
π΄π΅πΆ, π΄π΅ plus π΄πΆ is greater than π΅πΆ, π΄π΅ plus π΅πΆ is greater than π΄πΆ,
and π΄πΆ plus π΅πΆ is greater than π΄π΅.
We are being asked if itβs possible
to form a triangle with the lengths of six meters, seven meters, and 18 meters. For this to be the case, these
three values must make all three triangle inequalities true. So, we will be checking to see if
18 plus seven is greater than six, if 18 plus six is greater than seven, and if six
plus seven is greater than 18.
The first inequality is true
because 25 is greater than six. The second inequality is also true
because 24 is greater than seven. But when we add the two shorter
side lengths, we get 13, which is not greater than the third side length of 18. This means that we cannot form a
triangle with side lengths six meters, seven meters, and 18 meters.
As we see in this problem, it is
most important to check the sum of the two shorter side lengths. That is because this will be the
inequality that is most likely to fail. So, it is helpful to note that we
could check this inequality first in the future. And if that inequality fails, then
we wouldnβt even need to go on to check the other two inequalities.
The following theorem answers the
question βif three side lengths satisfy the triangle inequality, will they always
form a triangle?β The answer to this question is yes,
because according to the following theorem, if π, π, and π are positive real
numbers that satisfy the triangle inequality, then there exists a triangle with side
lengths π, π, and π. This means that we can verify that
positive real numbers, such as two, five, and six, do form a triangle, whereas the
positive real numbers three, five, and eight do not form a triangle because the two
shorter sides three and five do not add to a sum greater than eight.
In our next example, we will find
the range of all possible values for an unknown side length of a triangle using the
triangle inequality.
Find the range of all possible
values of π₯ if π₯ plus six centimeters, two centimeters, and 25 centimeters
represent the lengths of the sides of a triangle.
To find the range of all possible
values of π₯, we must first recall that the side lengths in a triangle must be
positive and satisfy the triangle inequality. If these three positive real
numbers satisfy the triangle inequality, then these values can represent the side
lengths of a triangle. The triangle inequality states that
the sum of any two sides of a triangle must be greater than the length of the third
side. We will use this fact to construct
the three inequalities that the side lengths must satisfy in order to form a
triangle.
First, we write out our three
inequalities using every combination of π₯ plus six, two, and 25. Then, we will solve each inequality
for π₯ to see what new information it will tell us about the possible range of
values. By simplifying the left side of the
first inequality, we have π₯ plus eight is greater than 25. Subtracting eight from both sides
yields the strict inequality π₯ is greater than 17. This is certainly an important
piece of information that tells us that the range of all possible values of π₯ must
be above 17.
Solving the second inequality
doesnβt give us any new information since we know that side lengths of a triangle
must be positive and we already have that π₯ is greater than 17. Subtracting six from both sides of
the last inequality gives us 21 is greater than π₯. If we turn this inequality around,
it reads that π₯ is strictly less than 21. Since we need π₯ to be greater than
17 and less than 21, we can write our final answer as a compound inequality: 17 less
than π₯ less than 21. We found this answer by
constructing three inequalities according to the triangle inequality theorem with
side lengths π₯ plus six, two, and 25.
In our last example, we will
combine the triangle inequality theorem with our knowledge of isosceles triangles to
determine the missing side length in an isosceles triangle.
If π΄π΅πΆ is an isosceles triangle
with π΄π΅ equal to two centimeters and π΅πΆ equal to five centimeters, find
π΄πΆ.
To begin, we will recall the
definition of an isosceles triangle. An isosceles triangle is a triangle
that has two congruent sides. The congruent sides are called the
legs of the triangle, and the third side is called the base. The following triangle diagrams are
not drawn to scale, but they can help us think through the possible arrangement of
sides in this triangle. For triangle π΄π΅πΆ to be
isosceles, the missing side π΄πΆ must be congruent to either side π΄π΅ or side
π΅πΆ. If π΄π΅ is congruent to π΄πΆ, then
π΄πΆ equals two. If π΅πΆ is congruent to π΄πΆ, then
π΄πΆ has to equal five.
To determine whether the third side
length is two centimeters or five centimeters, we will need to recall the triangle
inequality theorem. The triangle inequality tells us
that the sum of the lengths of any two sides of a triangle must be greater than the
length of the third side. And if this holds true, then the
triangle can be constructed. Therefore, we can check each of the
triplets two, two, five and two, five, five to determine if they can represent the
side lengths of a triangle.
When considering the two, two, five
triplet, we will need to check if the sum of the lengths of the two shorter sides is
greater than the length of the third side. However, two plus two is less than
five. This means that this triplet does
not satisfy the triangle inequality. For the second triplet, we see that
two plus five is greater than five, five plus two is greater than five, and five
plus five is greater than two. Since all three sums are larger
than the length of the third side, this means that the triplet satisfies the
triangle inequality.
We have shown that the only
possibilities for the length of π΄πΆ were two centimeters or five centimeters. Then, we found out that a triangle
with side lengths two centimeters, two centimeters, and five centimeters cannot
exist because the sum of the two shorter side lengths is not greater than the third
side length, whereas the lengths two centimeters, five centimeters, and five
centimeters fulfills the requirements of all three triangle inequalities. Therefore, the third side of the
isosceles triangle π΄πΆ has a length of five centimeters.
We can now summarize some key
points of this video. The central idea of this video was
the triangle inequality, which tells us that in any triangle π΄π΅πΆ the sum of the
lengths of any two sides must be greater than the length of the third side. That is, π΄π΅ plus π΄πΆ is greater
than π΅πΆ, π΄π΅ plus π΅πΆ is greater than π΄πΆ, and π΄πΆ plus π΅πΆ is greater than
π΄π΅. We also saw that if π, π, and π
are positive real numbers that satisfy the triangle inequality, then there exists a
triangle with side lengths π, π, and π.