Question Video: Bounding the Perimeter of Triangles Using the Triangle Inequality Mathematics

What is the range of possible perimeters, 𝑃, of the given figure 𝐴𝐡𝐢𝐷𝐸 if 𝐴𝐢 = 5 and 𝐷𝐢 = 9?

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Video Transcript

What is the range of possible perimeters, 𝑃, of the given figure 𝐴𝐡𝐢𝐷𝐸 if 𝐴𝐢 equals five and 𝐷𝐢 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 𝐴𝐢 is five and 𝐷𝐢 is nine. We can also see in the figure that side 𝐡𝐢 has the same length as side 𝐷𝐢 and side 𝐴𝐢 has the same length as side 𝐢𝐸. 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 𝐴𝐡 plus 𝐡𝐢 plus 𝐢𝐸 plus 𝐸𝐷 plus 𝐷𝐢 plus 𝐢𝐴. We can then substitute in the known side lengths from the figure to obtain that 𝑃 equals 𝐴𝐡 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 𝐴𝐡 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 𝐴𝐡𝐢 in the figure and using the known side lengths gives us three inequalities. We have nine plus five is greater than 𝐴𝐡, nine plus 𝐴𝐡 is greater than five, and 𝐴𝐡 plus five is greater than nine. We can rewrite each of these inequalities to isolate 𝐴𝐡. In the first inequality, we have 14 is greater than 𝐴𝐡. We can reorder this inequality to get that 𝐴𝐡 is less than 14. In the second inequality, we can subtract nine from both sides to get that 𝐴𝐡 is greater than negative four, though we already knew this since 𝐴𝐡 is a length. We can subtract five from both sides of the third inequality to obtain that 𝐴𝐡 is greater than four.

We could apply the same process with triangle 𝐢𝐷𝐸. However, we can note that the two known side lengths in this triangle are the same length as those in triangle 𝐴𝐡𝐢. 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 𝐴𝐡 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 𝐴𝐡 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.

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