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: 8


$TW = 27$ $TY = 18$

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. $\overline{YW}$ is one-third of the way from $T$ to $W$. Let's set up an equation incorporating what we know: $YW = \frac{1}{3}(TW)$ Let's plug in what we know: $9 = \frac{1}{3}(TW)$ Divide each side by $\frac{1}{3}$ to solve for $TW$. To divide by $\frac{1}{3}$ means to multiply by its reciprocal, which is $3$: $TW = 9(3)$ Multiply to solve: $TW = 27$ If $TW$ is the sum of $YW$ and $TY$, we can subtract $YW$ from $TW$ to get $TY$: $TY = TW - YW$ Let's plug in what we know: $TY = 27 - 9$ Subtract to solve: $TY = 18$
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