## Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole

# Chapter 4 - Section 4.4 - The Trapezoid - Exercises - Page 202: 42

#### Answer

(a) $m\angle P = 63^{\circ}$ (b) $NP = 13~cm$

#### Work Step by Step

(a) Since $\angle M \cong \angle Q$, the measure of the angle $m\angle M = 90^{\circ}$ We can see that $m\angle NRQ = 90^{\circ}$ since it is supplementary to $\angle PRN$ Then $m\angle MNR = 90^{\circ}$, since the sum of the four angles in $MNRQ$ is $360^{\circ}$ We can find $m\angle P$: $m\angle P + m\angle PRN + m\angle RNP = 180^{\circ}$ $m\angle P + m\angle PRN + m\angle RNP = 180^{\circ}$ $m\angle P + m\angle PRN + (m\angle MNP - m\angle MNR) = 180^{\circ}$ $m\angle P + m\angle PRN + (m\angle P+54^{\circ} - m\angle MNR) = 180^{\circ}$ $2m\angle P + m\angle PRN + 54^{\circ} - m\angle MNR = 180^{\circ}$ $2m\angle P + 90^{\circ} + 54^{\circ} - 90^{\circ} = 180^{\circ}$ $2m\angle P + 54^{\circ} = 180^{\circ}$ $2m\angle P = 126^{\circ}$ $m\angle P = \frac{126^{\circ}}{2}$ $m\angle P = 63^{\circ}$ (b) We can find $PR$: $PR = PQ- QR$ $PR = PQ-MN$ $PR = 20~cm-15~cm$ $PR = 5~cm$ Note that $RN = MQ = 12~cm$ We can find the length of $\overline{NP}$: $NP = \sqrt{(PR)^2+(RN)^2}$ $NP = \sqrt{(5~cm)^2+(12~cm)^2}$ $NP = \sqrt{25~cm^2+144~cm^2}$ $NP = \sqrt{169~cm^2}$ $NP = 13~cm$

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