Question

the graph shows quadrilaterals opqr and defg. is opqr similar to defg? justify your answer. yes, because a dilation by a scale factor of 1/2 centered at point q maps opq onto defg.

Explanation:

Step1: Calculate side lengths

Use the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\) or count grid units. For \(OPQR\), \(OQ\) (vertical segment from \((8,6)\) to \((8,-2)\)) has length \(6-(- 2)=8\). For \(DEFG\), \(DG\) (vertical segment from \((4,3)\) to \((4,-2)\)) has length \(3 - (-2)=5\). Wait, no, better use vectors or count. For \(OPQR\): \(OQ\) (from \(O(8,6)\) to \(Q(2,4)\)): horizontal change \(8 - 2=6\), vertical change \(6 - 4 = 2\). For \(DEFG\): \(DG\) (from \(D(4,3)\) to \(G(1,3)\)) horizontal change \(4 - 1=3\), vertical change \(0\). Wait, wrong approach. Let's use similarity criteria. Two polygons are similar if their corresponding angles are equal and corresponding sides are in proportion.
Another way: Check the ratio of corresponding sides.
For \(OPQR\): \(O(8,6)\), \(Q(2,4)\), \(R(5,6)\), \(P(8,-2)\). For \(DEFG\): \(D(4,3)\), \(G(1,3)\), \(F(0,2)\), \(E(4,-2)\).
The ratio of \(OQ\) (length from \(O\) to \(Q\): \(\sqrt{(8 - 2)^2+(6 - 4)^2}=\sqrt{36 + 4}=\sqrt{40}=2\sqrt{10}\)) and \(DG\) (length from \(D\) to \(G\): \(\sqrt{(4 - 1)^2+(3 - 3)^2}=3\)). Wait, no. Let's count grid - units for horizontal and vertical segments.
In \(OPQR\), the horizontal segment \(OR\) (from \(O(8,6)\) to \(R(5,6)\)) has length \(8 - 5=3\). In \(DEFG\), the horizontal segment \(DG\) (from \(D(4,3)\) to \(G(1,3)\)) has length \(4 - 1 = 3\). The vertical segment \(OP\) (from \(O(8,6)\) to \(P(8,-2)\)) has length \(6-(-2)=8\). The vertical segment \(DE\) (from \(D(4,3)\) to \(E(4,-2)\)) has length \(3-(-2)=5\). The ratio of vertical sides \(OP:DE = 8:5\), and for horizontal sides \(OR:DG=3:3 = 1:1\). Since the ratios of corresponding sides are not equal (\(\frac{8}{5}
eq1\)), the quadrilaterals are not similar.

Answer:

No, because the ratios of corresponding sides are not equal.