Geometry: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281159
ISBN 13: 978-0-13328-115-6

Chapter 6 - Polygons and Quadrilaterals - 6-5 Conditions for Rhombuses, Rectangles, and Squares - Practice and Problem-Solving Exercises - Page 388: 33



Work Step by Step

Using the Triangle Inequality Theorem, we need to see if the sum of each of the combinations of two sides is greater than the other side. Let's set $x$ as the length of the third side. Now, let's look at the possible combinations of sides to see if the lengths of two sides is greater than the length of a third side: 1st combination: $x + 7 > 11$ Solve for $x$ by subtracting $7$ from each side of the inequality: $x > 4$ 2nd combination: $x + 11 > 7$ Subtract $11$ from each side of the equation: $x > -4$ 3rd combination: $7 + 11 > x$ Switch the inequality around and add to simplify: $x < 18$ $x$ cannot be a negative number because it is a length, so let's eliminate the inequality that includes a negative number. We know $x$ has to be greater than $4$ and $x$ has to be less than $18$, so we have $x$ in the following range: $4 < x < 18$ Let's look at the answer options to see which one cannot be the length of the third side of the triangle. The lengths $13$, $7$, and $5$ lie within this interval; therefore, we are not looking for options $F$, $G$, and $H$. Option $I$ is correct because we are looking for the length that is impossible for this triangle.
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