Exercise 4

Chapter 5, Section 5-1, Page 204
enVision Geometry 1st Edition by Al Cuoco
ISBN: 9780328931583

Solution

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Step 1
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(a)\textbf{(a)}

From the figure, WY=2n+7WY=2n+7 so we must solve for nn first.

From the figure, WZXY\overline{WZ}\bot\overline{XY} and XZYZ\overline{XZ}\cong\overline{YZ} so by the Converse of the Perpendicular Bisector Theorem, XWWY\overline{XW}\cong\overline{WY} and we can write:

XW=WYXW=WY

5n2=2n+75n-2=2n+7

5n=2n+95n=2n+9

3n=93n=9

n=3n=3

Hence,

WY=2(3)+7WY=2(3)+7

WY=13\color{#c34632}{WY=13}

(b)\textbf{(b)}

From the figure, KNJL\overline{KN}\bot\overline{JL} and JKKL\overline{JK}\cong\overline{KL} so by the Converse of Perpendicular Bisector Theorem, JNNL\overline{JN}\cong\overline{NL}. So, JN=NL=14JN=NL=14.

OMNJL\overline{OM}\bot\overline{NJL} and NMLM\overline{NM}\cong\overline{LM} so by the Converse of Perpendicular Bisector Theorem, NOOL\overline{NO}\cong\overline{OL}. So, NO=OLNO=OL. By Segment Addition Postulate,

NL=NO+OLNL=NO+OL

14=OL+OL14=OL+OL

14=2OL14=2OL

7=OL7=OL

OL=7\color{#c34632}{OL=7}

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