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$8$A) REGULARITY Determine whether $\triangle D E F \cong \triangle P Q R$.$\newline$$$D(-6,1), E(1,2), F(-1,-4), P(0,5), Q(7,6), R(5,0)$$$\newline$Find the side lengths of each triangle.$\newline$$D E=5 \sqrt{2}, P Q=5 \sqrt{2}, E F=\sqrt{10}, Q R=\sqrt{10}, D F=5 \sqrt{2}, P R=5 \sqrt{2}$$\newline$$D E=\sqrt{2}, P Q=\sqrt{2}, E F=\sqrt{10}, Q R=\sqrt{10}, D F=\sqrt{2}, P R=\sqrt{2}$$\newline$$D E=5 \sqrt{2}, P Q=5 \sqrt{2}, E F=2 \sqrt{10}, Q R=2 \sqrt{10}, D F=5 \sqrt{2}, P R=5 \sqrt{2}$$\newline$$D E=5 \sqrt{2}, P Q=5 \sqrt{10}, E F=2 \sqrt{10}, Q R=2 \sqrt{10}, D F=5 \sqrt{10}, P R=5 \sqrt{2}$$\newline$$8$B) Fill in the blanks using the available answer choices.$\newline$Is $\qquad$ $D E F \cong \triangle P Q R$ ? Explain.$\newline$$\triangle$$\newline$$D E F$ $\qquad$ to $\qquad$ $D E=5 \sqrt{2}, P Q=5 \sqrt{2}, E F=\sqrt{10}, Q R=\sqrt{10}, D F=5 \sqrt{2}, P R=5 \sqrt{2}$$1$ by $D E=5 \sqrt{2}, P Q=5 \sqrt{2}, E F=\sqrt{10}, Q R=\sqrt{10}, D F=5 \sqrt{2}, P R=5 \sqrt{2}$$2$ because corresponding sides have $\qquad$ and are $\qquad$ (Blank I)$\newline$(Blank $2$)$\newline$(Blank: $3$)$\newline$Blank $1$ options$\newline$Blank $2$ options$\newline$Blank $3$ options$\newline$- is congruent$\newline$- the same measure$\newline$- is not congruent$\newline$- different measures$\newline$- congruent$\newline$- not congruent