Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Chapter 4 - Section 4.4 - The Trapezoid - Exercises - Page 213: 43

Answer

(a) $m\angle P = 59^{\circ}$ (b) $NR = 4~in$

Work Step by Step

(a) Since $\overline{MN} \parallel \overline{QP}$, the measure of the angle $m\angle M = 90^{\circ}$ We can find $m\angle P$: $m\angle P + m\angle Q + m\angle M + m\angle MNP = 360^{\circ}$ $m\angle P + m\angle Q + m\angle M + (m\angle M+31^{\circ}) = 360^{\circ}$ $m\angle P + 90^{\circ} + 90^{\circ}+ (90^{\circ}+31^{\circ}) = 360^{\circ}$ $m\angle P + 301^{\circ}) = 360^{\circ}$ $m\angle P = 59^{\circ}$ (b) We can find $PR$: $PR = PQ- QR$ $PR = PQ-MN$ $PR = 9~in-6~in$ $PR = 3~in$ We can find the length of $\overline{NR}$: $NR = \sqrt{(NP)^2-(PR)^2}$ $NR = \sqrt{(5~in)^2-(3~in)^2}$ $NR = \sqrt{25~in^2-9~in^2}$ $NR = \sqrt{16~in^2}$ $NR = 4~in$
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