Question

the graph shows triangles cde and rst. is cde similar to rst? justify your answer. yes, because a rotation 180° around the origin followed by a translation up 8 units maps cde onto rst. yes, because a rotation 180° around the origin followed by a reflection over the x - axis maps cde onto rst. no, because ∠e and ∠t do not have the same measure.

Explanation:

Step1: Calculate the side lengths of triangles CDE and RST

• For triangle CDE:
• \(CD=\sqrt{(5 - 3)^{2}+(1 - 2)^{2}}=\sqrt{4 + 1}=\sqrt{5}\)
• \(CE=\sqrt{(5 - 3)^{2}+(6 - 2)^{2}}=\sqrt{4+16}=\sqrt{20} = 2\sqrt{5}\)
• \(DE=\sqrt{(5 - 5)^{2}+(6 - 1)^{2}}=5\)
• For triangle RST:
• \(RS=\sqrt{(-2+5)^{2}+(6 - 7)^{2}}=\sqrt{9 + 1}=\sqrt{10}\)
• \(ST=\sqrt{(-2+5)^{2}+(6 - 1)^{2}}=\sqrt{9+25}=\sqrt{34}\)
• \(RT=\sqrt{(-2+2)^{2}+(6 - 1)^{2}}=5\)

Step2: Check the similarity ratios

• If two triangles are similar, the ratios of their corresponding side lengths are equal.
• \(\frac{CD}{RS}=\frac{\sqrt{5}}{\sqrt{10}}=\frac{1}{\sqrt{2}}\), \(\frac{CE}{ST}=\frac{2\sqrt{5}}{\sqrt{34}}=\frac{\sqrt{10}}{\sqrt{17}}\), \(\frac{DE}{RT}=\frac{5}{5} = 1\)
• Since the ratios of the side lengths are not equal, we can also check angles.
• \(\angle E\) in \(\triangle CDE\): \(\tan\angle E=\frac{CD}{CE/2}=\frac{\sqrt{5}}{ \sqrt{5}} = 1\), so \(\angle E = 45^{\circ}\)
• \(\angle T\) in \(\triangle RST\): \(\tan\angle T=\frac{RS/2}{ST/2}

eq1\) (from side - length calculations, and using the right - triangle - like property for angle calculation in coordinate - based triangles). So \(\angle E
eq\angle T\)

Answer:

No, because \(\angle E\) and \(\angle T\) do not have the same measure.