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Points $P$, $Q$, and $R$ are on sides $\overline{AB}$, $\overline{CD}$, and $\overline{BC}$ of rectangle $ABCD$ as shown below such that $BP = 4$, $CQ = 18$, $\angle PRQ = 90^\circ$, and $CR = 2BR$. Find $PQ$. 

Guest Mar 28, 2018
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Points $P$, $Q$, and $R$ are on sides $\overline{AB}$, $\overline{CD}$, and $\overline{BC}$ of rectangle $ABCD$ as shown below such that $BP = 4$, $CQ = 18$, $\angle PRQ = 90^\circ$, and $CR = 2BR$.

Find $PQ$. 

 

 

 

 

\(\text{Let $x = \overline{PQ}$} \\ \text{Let $y = \overline{BC}=\overline{AD}$} \\ \text{Let $u = \overline{RQ}$} \\ \text{Let $v = \overline{RP}$} \)

 

\(\begin{array}{|lrcll|} \hline & x^2 &=& y^2+(18-4)^2 \\ & x^2 &=& y^2+(14)^2 \\ & x^2 &=& y^2+196\\ (1) & y^2 &=& x^2-196 \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline v^2 &=& 4^2+ \left( \dfrac{1}{3}y \right)^2 \\ &=& 16+ \dfrac{1}{9}y^2 \\\\ u^2 &=& 18^2 + \left( \dfrac{2}{3}y \right)^2 \\ &=& 324 + \dfrac{4}{9}y^2 \\\\ x^2 &=& u^2+v^2 \\ &=& 324 + \dfrac{4}{9}y^2 + 16+ \dfrac{1}{9}y^2 \\ &=& 340 + \dfrac{5}{9}y^2 \\ x^2 &=& 340 + \dfrac{5}{9}y^2 \quad &| \quad y^2 = x^2-196 \\ x^2 &=& 340 + \dfrac{5}{9}(x^2-196) \\ x^2 &=& 340 + \dfrac{5}{9}x^2- \dfrac{5\cdot 196}{9} \\ x^2+ \dfrac{5}{9}x^2 &=& 340 - \dfrac{5\cdot 196}{9} \\ \dfrac{4}{9}x^2 &=& 340 - \dfrac{5\cdot 196}{9} \quad &| \quad \cdot \dfrac{9}{4} \\ x^2 &=& \dfrac{9\cdot 340}{4} - \dfrac{5\cdot 196}{4} \\ x^2 &=& \dfrac{9\cdot 340-5\cdot 196}{4} \\ x^2 &=& 520 \\ x^2 &=& 4\cdot 130 \\ x &=& 2\cdot\sqrt{ 130} \\ \mathbf{x} & \mathbf{\approx}& \mathbf{22.80} \\ \hline \end{array}\)

 

laugh

heureka  Mar 28, 2018

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