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 \(TU\)

For \(QR\), using the distance formula \(d=\vert y_2 - y_1\vert\) (since \(x\) - coordinate of \(Q(-2,0)\) and \(R(-2,4)\) is same). \(QR=\vert4 - 0\vert=4\).
For \(TU\), using the distance formula \(d=\vert y_2 - y_1\vert\) (since \(x\) - coordinate of \(T(-1,-2)\) and \(U(-1,-4)\) is same). \(TU=\vert-2+ 4\vert=2\). So, \(\frac{QR}{TU}=\frac{4}{2}=2\).

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

For \(PR\), using the distance formula \(d=\vert x_2 - x_1\vert\) (since \(y\) - coordinate of \(P(4,4)\) and \(R(-2,4)\) is same). \(PR=\vert4 + 2\vert=6\).
For \(SU\), using the distance formula \(d=\vert x_2 - x_1\vert\) (since \(y\) - coordinate of \(S(2,-4)\) and \(U(-1,-4)\) is same). \(SU=\vert2 + 1\vert=3\). So, \(\frac{PR}{SU}=\frac{6}{3}=2\).

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

We know that \(\sqrt{52}=\sqrt{4\times13}=2\sqrt{13}\). Then \(\frac{\sqrt{52}}{\sqrt{13}}=\frac{2\sqrt{13}}{\sqrt{13}} = 2\).

Step4: Determine the similarity theorem

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

Answer:

\(\frac{QR}{TU}=2\), \(\frac{PR}{SU}=2\), \(\frac{\sqrt{52}}{\sqrt{13}} = 2\), SSS (Side - Side - Side) Similarity