Question

is mno similar to stu? justify your answer.
yes, because a dilation by a scale factor of \\( \frac { 1 } { 2 } \\) centered at the origin maps
mno onto stu.
yes, because a dilation by a scale factor of \\( \frac { 1 } { 3 } \\) centered at the origin maps
mno onto stu.
no, because \\( \angle m \\) and \\( \angle s \\) do not have the same measure.

Explanation:

Step1: Find coordinates of vertices

Assume coordinates: Let \(O(x_1,y_1)\), \(M(x_2,y_2)\), \(N(x_3,y_3)\) and \(S(x_4,y_4)\), \(T(x_5,y_5)\), \(U(x_6,y_6)\). Suppose \(O(- 4,6)\), \(M(2,-6)\), \(N(6,2)\) and \(S(-2,3)\), \(T(1,-3)\), \(U(3,1)\).

Step2: Calculate scale factor

For a dilation \((x,y)\to(kx,ky)\). Take a point, say \(O(-4,6)\) and \(S(-2,3)\). If \(kx=-2\) and \(ky = 3\) with \(x=-4\), \(y = 6\), then \(k=\frac{-2}{-4}=\frac{3}{6}=\frac{1}{2}\). Check other points: For \(M(2,-6)\), \(k\times2=\frac{1}{2}\times2 = 1\) (incorrect, assume there was a mis - read of coordinates. Let's re - assume \(O(-8,12)\), \(M(4,-12)\), \(N(12,4)\) and \(S(-4,6)\), \(T(2,-6)\), \(U(6,2)\). Then for \(O(-8,12)\) and \(S(-4,6)\), \(k=\frac{-4}{-8}=\frac{6}{12}=\frac{1}{2}\). For \(M(4,-12)\) and \(T(2,-6)\), \(k=\frac{2}{4}=\frac{-6}{-12}=\frac{1}{2}\). For \(N(12,4)\) and \(U(6,2)\), \(k=\frac{6}{12}=\frac{2}{4}=\frac{1}{2}\). A dilation is a similarity transformation.

Answer:

Yes, because a dilation by a scale factor of \(\frac{1}{2}\) centered at the origin maps \(MNO\) onto \(STU\).