Video Transcript
What is the range of possible perimeters, 𝑃, of the given figure 𝐴𝐵𝐶𝐷𝐸 if 𝐴𝐶
equals five and 𝐷𝐶 equals nine?
In this question, we are asked to find the range of possible perimeters of a shape
using the lengths of two sides and a given figure. We can start by adding the two lengths we are given. We are told that 𝐴𝐶 is five and 𝐷𝐶 is nine. We can also see in the figure that side 𝐵𝐶 has the same length as side 𝐷𝐶 and
side 𝐴𝐶 has the same length as side 𝐶𝐸. So, we can add these lengths onto the figure.
We can then recall that the perimeter of a polygon is the sum of its side
lengths. So the perimeter 𝑃 of the given polygon is equal to 𝐴𝐵 plus 𝐵𝐶 plus 𝐶𝐸 plus
𝐸𝐷 plus 𝐷𝐶 plus 𝐶𝐴. We can then substitute in the known side lengths from the figure to obtain that 𝑃
equals 𝐴𝐵 plus nine plus five plus 𝐸𝐷 plus nine plus five. We can then simplify the right-hand side of the equation to get that 𝑃 equals 𝐴𝐵
plus 𝐸𝐷 plus 28.
We now need to find upper and lower bounds for the two unknown side lengths. We can do this by noting that both sides are in triangles with two known side
lengths, so we can apply the triangle inequality. This tells us that in any triangle, the sum of any two of its sides’ lengths must be
greater than the length of the final side. Applying this to each pair of sides gives us three inequalities that must be
satisfied for any triangle.
Applying this result to triangle 𝐴𝐵𝐶 in the figure and using the known side
lengths gives us three inequalities. We have nine plus five is greater than 𝐴𝐵, nine plus 𝐴𝐵 is greater than five, and
𝐴𝐵 plus five is greater than nine. We can rewrite each of these inequalities to isolate 𝐴𝐵. In the first inequality, we have 14 is greater than 𝐴𝐵. We can reorder this inequality to get that 𝐴𝐵 is less than 14. In the second inequality, we can subtract nine from both sides to get that 𝐴𝐵 is
greater than negative four, though we already knew this since 𝐴𝐵 is a length. We can subtract five from both sides of the third inequality to obtain that 𝐴𝐵 is
greater than four.
We could apply the same process with triangle 𝐶𝐷𝐸. However, we can note that the two known side lengths in this triangle are the same
length as those in triangle 𝐴𝐵𝐶. So we would obtain the same inequalities. In particular, we want an upper and lower bound for 𝐷𝐸. So we want the first inequality that tells us 𝐷𝐸 is less than 14 and the third
inequality that tells us 𝐷𝐸 is greater than four.
We can now note that 𝐴𝐵 and 𝐷𝐸 must both be less than 14. So we substitute 14 for these lengths into the perimeter formula to find an upper
bound for the perimeter. Similarly, we can substitute 𝐴𝐵 and 𝐷𝐸 equals four to find a lower bound for the
perimeter. Substituting 14 into the perimeter equation and writing the inequality gives us 𝑃 is
less than 14 plus 14 plus 28, which simplifies to give us that 𝑃 is less than
56. Substituting four for the lengths in the perimeter formula and writing the inequality
gives us that 𝑃 is greater than four plus four plus 28, which simplifies to give us
𝑃 is greater than 36. We can combine these inequalities into the single compound inequality 36 is less than
𝑃 is less than 56.
