Geometry: Common Core (15th Edition)

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

Chapter 5 - Relationships Within Triangles - 5-4 Medians and Altitudes - Practice and Problem-Solving Exercises - Page 312: 9


$ZU = \frac{27}{2}$ $ZY = \frac{9}{2}$

Work Step by Step

According to the concurrency of medians theorem, the medians of a triangle are concurrent at a point that is two-thirds of the way between each vertex and the midpoint of the side opposite to the vertex. $\overline{ZY}$ is one-third of the way from $U$ to $Z$. Let's set up an equation incorporating what we know: $YU = \frac{2}{3}(ZU)$ Let's plug in what we know: $9 = \frac{2}{3}(ZU)$ Divide each side by $\frac{2}{3}$ to solve for $TW$. To divide by $\frac{2}{3}$ means to multiply by its reciprocal, which is $\frac{3}{2}$: $ZU = 9(\frac{3}{2})$ Multiply to solve: $ZU = \frac{27}{2}$ If $ZU$ is the sum of $YU$ and $ZY$, we can subtract $YU$ from $ZU$ to get $ZY$: $ZY = ZU - YU$ Let's plug in what we know: $ZY = \frac{27}{2} - 9$ Rewrite $9$ as an equivalent fraction with $2$ as its denominator: $ZY = \frac{27}{2} - \frac{18}{2}$ Subtract to solve: $ZY = \frac{9}{2}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.