Question

complete the statements to verify that the triangles are similar.
\\( \frac { q r } { t u } = \\)
\\( \frac { p r } { s u } = \\)
\\( \frac { p q } { s t } = \frac { \sqrt { 5 2 } } { \sqrt { 1 3 } } = \\)
therefore, \\( \triangle p q r \sim \triangle s t u \\) by the theorem.

Explanation:

Step1: Calculate \(QR\) and \(TU\)

\(QR\) is the vertical distance from \(Q(-2,0)\) to \(R(-2,4)\). Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), here \(x_1=x_2=-2\), so \(QR=\vert4 - 0\vert = 4\).
\(TU\) is the vertical distance from \(T(-1,-2)\) to \(U(-1,-4)\). Here \(x_1=x_2=-1\), so \(TU=\vert- 2-(-4)\vert=2\). Then \(\frac{QR}{TU}=\frac{4}{2}=2\).

Step2: Calculate \(PR\) and \(SU\)

\(PR\) is the horizontal distance from \(P(4,4)\) to \(R(-2,4)\). Using the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), here \(y_1=y_2 = 4\), so \(PR=\vert4-(-2)\vert=6\).
\(SU\) is the horizontal distance from \(S(2,-4)\) to \(U(-1,-4)\). Here \(y_1=y_2=-4\), so \(SU=\vert2-(-1)\vert = 3\). Then \(\frac{PR}{SU}=\frac{6}{3}=2\).

Step3: Simplify \(\frac{\sqrt{52}}{\sqrt{13}}\)

Using the property \(\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\) (\(a = 52\), \(b = 13\)), \(\frac{\sqrt{52}}{\sqrt{13}}=\sqrt{\frac{52}{13}}=\sqrt{4}=2\).

Step4: Determine the similarity theorem

Since \(\frac{QR}{TU}=\frac{PR}{SU}=\frac{PQ}{ST}=2\), by the Side - Side - Side (SSS) similarity theorem, \(\triangle PQR\sim\triangle STU\).

Answer:

\(\frac{QR}{TU}=2\), \(\frac{PR}{SU}=2\), \(\frac{\sqrt{52}}{\sqrt{13}} = 2\), and \(\triangle PQR\sim\triangle STU\) by the SSS (Side - Side - Side) similarity theorem.