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Question
complete the statements to verify that the triangles are similar.\\(\frac{qr}{tu} = 2\\)\\(\frac{pr}{su} = 2\\)\\(\frac{pq}{st} = \frac{sqrt{52}}{sqrt{13}} = \\)\\(\text{therefore, } \triangle \\) ~ \\(\triangle stu\\) by the theorem.\\(\text{options for the last blank: } 1/2, 2, 4, 6\\)
Step1: Simplify the ratio of square roots
We have \(\frac{\sqrt{52}}{\sqrt{13}}\). Using the property \(\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\) (for \(a\geq0,b > 0\)), we get \(\sqrt{\frac{52}{13}}\).
Step2: Calculate the fraction inside the square root
Calculate \(\frac{52}{13}=4\). So now we have \(\sqrt{4}\).
Step3: Evaluate the square root
\(\sqrt{4} = 2\). So \(\frac{PQ}{ST}=2\).
Step4: Identify the similar triangle and the theorem
We know that \(\frac{QR}{TU} = 2\), \(\frac{PR}{SU}=2\) and \(\frac{PQ}{ST}=2\). So the corresponding sides of \(\triangle PQR\) and \(\triangle STU\) are in proportion. By the SSS (Side - Side - Side) Similarity Theorem, if the corresponding sides of two triangles are in proportion, then the triangles are similar. So \(\triangle PQR\sim\triangle STU\) (the first triangle should be \(\triangle PQR\) as we are taking the ratio of its sides to the sides of \(\triangle STU\)) and the ratio is \(2\), and the theorem is SSS Similarity.
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For \(\frac{PQ}{ST}=\frac{\sqrt{52}}{\sqrt{13}}=\boxed{2}\)
The triangle \(\triangle PQR\) (assuming the first triangle is \(\triangle PQR\) with vertices \(P,Q,R\)) \(\sim\triangle STU\) by the SSS (Side - Side - Side) Similarity theorem, and the ratio value is \(2\).