Geometry: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281159
ISBN 13: 978-0-13328-115-6

Chapter 1 - Tools of Geometry - Common Core Cumulative Standards Review - Selected Response - Page 77: 9

Answer

B

Work Step by Step

use the distance formula to determine the distance from D(2,4) to C(-1,0) $d_{DC}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ substitute given values $d_{DC}=\sqrt{(-1-2))^2+(0-4))^2}$ simplify in parentheses $d_{DC}=\sqrt{(-3)^2+(-4)^2}$ simplify $d_{DC}=\sqrt{9+16}$ $d_{DC}=\sqrt{25}$ $d_{DC}=5$ use the distance formula to determine the distance from C(-1,0) to A(-2,1) $d_{CA}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ substitute given values $d_{CA}=\sqrt{(-2-(-1)^2+(1-0)^2}$ simplify in parentheses $d_{CA}=\sqrt{(-1)^2+(1)^2}$ simplify $d_{CA}=\sqrt{1+1}$ $d_{CA}=\sqrt{2}$ $d_{CA}=1.4$ use the distance formula to determine the distance from A(-2,1) to B(3,1) $d_{AB}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ substitute given values $d_{AB}=\sqrt{(3-(-2)^2+(1-1)^2}$ simplify in parentheses $d_{AB}=\sqrt{(5)^2+(0)^2}$ simplify $d_{AB}=\sqrt{25+0}$ $d_{AB}=\sqrt{25}$ $d_{AB}=5$ use the distance formula to determine the distance from B(3,1) to F(5,2) $d_{BF}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ substitute given values $d_{BF}=\sqrt{(5-3)^2+(2-1)^2}$ simplify in parentheses $d_{BF}=\sqrt{(2)^2+(1)^2}$ simplify $d_{BF}=\sqrt{4+1}$ $d_{BF}=\sqrt{5}$ $d_{BF}=2.2$ use the distance formula to determine the distance from F(5,2) to E(0,3) $d_{AB}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ substitute given values $d_{AB}=\sqrt{(0-5)^2+(3-2)^2}$ simplify in parentheses $d_{AB}=\sqrt{(-5)^2+(1)^2}$ simplify $d_{AB}=\sqrt{25+1}$ $d_{AB}=\sqrt{26}$ $d_{AB}=5.1$ Sum the individual distances to find the total distance. $d=5+1.4+5+2.2+5.1=18.7$
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