Question

the graph shows quadrilaterals cdef and lmno
is cdef similar to lmno? justify your answer.
yes, because a dilation by a scale factor of \\( \frac { 1 } { 2 } \\) centered at the origin followed by a rotation \\( 180 ^ { \circ } \\) around the origin maps cdef onto lmno.
yes, because a dilation by a scale factor of \\( \frac { 1 } { 2 } \\) centered at the origin followed by a reflection over the y - axis maps cdef onto lmno.
no, because \\( \frac { l m } { c d } \\) is not equal to \\( \frac { m n } { d e } \\)

Explanation:

Step1: Find the coordinates of vertices

Assume \(C(-7,2)\), \(D(-3,2)\), \(E(-3, - 4)\), \(F(-7,-8)\), \(L(3,-1)\), \(M(2,-1)\), \(N(2,2)\), \(O(3,4)\)

Step2: Calculate the lengths of sides

For \(CD\): Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), \(CD=\sqrt{(-3+7)^2+(2 - 2)^2}=4\)
For \(DE\): \(DE=\sqrt{(-3 + 3)^2+(-4 - 2)^2}=6\)
For \(LM\): \(LM=\sqrt{(3 - 2)^2+(-1+1)^2}=1\)
For \(MN\): \(MN=\sqrt{(2 - 2)^2+(2 + 1)^2}=3\)

Step3: Calculate the ratios

\(\frac{LM}{CD}=\frac{1}{4}\), \(\frac{MN}{DE}=\frac{3}{6}=\frac{1}{2}\)

Answer:

No, because \(\frac{LM}{CD}\) is not equal to \(\frac{MN}{DE}\)