Geometry: Common Core (15th Edition)

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

Chapter 5 - Relationships Within Triangles - 5-3 Bisectors in Triangles - Practice and Problem-Solving Exercises - Page 305: 18


$x = 4$

Work Step by Step

$R$ is the incenter of the triangle because it marks the point of concurrency of a triangle's angle bisectors. Therefore, $R$ is the center of the circle that is inscribed inside the triangle, so any point on that circle is equidistant from $R$. In this diagram, $S$ and $T$ are located on the inscribed circle; therefore, $RS$ is equal to $RT$. We can now set these distances equal to one another to solve for $x$: $4(x - 3) + 6 = 5(2x - 6)$ Let's distribute first to simplify: $4x - 12 + 6 = 10x - 30$ Add the constants on the left side of the equation: $4x - 6 = 10x - 30$ Subtract $4x$ from each side of the equation to isolate the variable on one side of the equation: $-6 = 6x - 30$ Add $30$ to each side of the equation to isolate constants on one side of the equation: $6x = 24$ Divide each side of the equation by $6$ to solve for $x$: $x = 4$
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