Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Chapter 8 - Section 8.2 - Perimeter and Area of Polygons - Exercises - Page 359: 3

Answer

4$\sqrt 29$m

Work Step by Step

In parallelogram if adjacent sides are equal then it will become rhombus In rhombus diagonals are perpendicular bisectors So if assume the point of intersection of both the diagonals as E, so EC = EA = 2m and BE = ED = 5m and $\angle$ CED = 90$^{\circ}$ By pythagoras theorem $CD^{2}$ =$CE^{2}$ + $ED^{2}$ = $5^{2}$ + $2^{2}$ = 25 + 4 = 29 CD = $\sqrt 29$ m As we know in rhombus all the sides are equal therefore AB=BC=CD=DA therefore perimeter = AB+BC+CD+DA = $\sqrt 29$+$\sqrt 29$+$\sqrt 29$+$\sqrt 29$ =4$\sqrt 29$m
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