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: 10


$VY = 6$ $YX = 3$

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. $\overline{YX}$ is one-third of the way from $V$ to $X$. Let's set up an equation incorporating what we know: $VY = \frac{2}{3}(VX)$ Let's plug in what we know: $VY = \frac{2}{3}(9)$ Multiply to solve for $VY$: $VY = \frac{18}{3}$ Divide both the numerator and denominator by their greatest common factor, which is $3$: $VY = 6$ If $VX$ is the sum of $VY$ and $YX$, we can subtract $VY$ from $VX$ to get $YX$: $YX = VX - VY$ Let's plug in what we know: $YX = 9 - 6$ Subtract to solve: $YX = 3$
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