Question

the graph shows quadrilaterals efgh and uvwx. is efgh congruent to uvwx? justify your answer. yes, because a reflection across the x - axis maps efgh onto uvwx. yes, because a translation up 10 units maps efgh onto uvwx. no, because \\( \overline { g h } \\) and \\( \overline { w x } \\) do not have the same length. no, because \\( \angle f \\) and \\( \angle v \\) do not have the same measure.

Explanation:

Step1: Calculate the length of \( GH \)

Use the distance formula \( d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2} \). For \( G(7,-8) \) and \( H(5,-5) \), \( d_{GH}=\sqrt{(7 - 5)^2+(-8+5)^2}=\sqrt{4 + 9}=\sqrt{13} \).

Step2: Calculate the length of \( WX \)

For \( W(6,8) \) and \( X(5,5) \), \( d_{WX}=\sqrt{(6 - 5)^2+(8 - 5)^2}=\sqrt{1+9}=\sqrt{10} \).
Since \( \sqrt{13}
eq\sqrt{10} \), \( GH\) and \( WX \) are not equal. Congruent figures have all corresponding sides equal.

Answer:

No, because \( \overline{GH} \) and \( \overline{WX} \) do not have the same length.