Geometry: Common Core (15th Edition)

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

Chapter 6 - Polygons and Quadrilaterals - 6-4 Properties of Rhombuses, Rectangles, and Squares - Practice and Problem-Solving Exercises - Page 382: 63

Answer

$PQ = 16$

Work Step by Step

The diagram indicates that $\overline{ST}$ bisects both $\overline{PR}$ and $\overline{QR}$. According to the triangle midsegment theorem, if a line segment joins two sides of a triangle at their midpoints, then that line segment is parallel to the third side of that triangle and is half as long as that third side. Knowing this information, we can deduce that $\overline{PQ}$, which is the third side, is parallel to $\overline{ST}$, which is the line segment, and that $PQ$ is two times the length of $ST$. We can now find $PQ$: $PQ = 2(ST)$ Let's plug in what we know: $PQ = 2(8)$ Multiply to solve for $PQ$: $PQ = 16$
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