Elementary Geometry for College Students (5th Edition)

Published by Brooks Cole
ISBN 10: 1439047901
ISBN 13: 978-1-43904-790-3

Chapter 10 - Section 10.1 - The Rectangular Coordinate System - Exercises - Page 456: 25b

Answer

equilateral triangle

Work Step by Step

D(0,0) E(4,0) and F(2,2$\sqrt 3$) Using distance formula x1=0,y1=0,x2=4,y2=0 then, $\overline{DE}$ = $\sqrt (x2-x1)^{2} + (y2-y1)^{2}$ = $\sqrt (4-0)^{2} + (0-0)^{2}$ =4 units Similarly, $\overline{EF}$ = $\sqrt (2-4)^{2} + (2\sqrt 3-0)^{2}$ =$\sqrt 2^{2} + 2\sqrt 3^{2}$ = $\sqrt 4+12$ = $\sqrt 16$ =4 units $\overline{FD}$ = $\sqrt (2-0)^{2} + (2\sqrt 3-0)^{2}$ =$\sqrt 2^{2} + 2\sqrt 3^{2}$ = $\sqrt 4+12$ = $\sqrt 16$ =4 units Therefore $\overline{DE}$ = $\overline{EF}$ =$\overline{FD}$ in triangle DEF therefore triangle DEF is equilateral triangle.
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