Quadrilateral PQRS is a square with a side length of 13. Points X and Y lie outside the square so that PX=RY=12 and QX=SY=5. Find XY^2.
We can drop perpendiculars from X to PQ and from Y to SR. Then by applying the Pythagorean Theorem, we get an answer of XY^2 = 6^2 + 22^2 = 520.
+26367 Quadrilateral PQRS is a square with a side length of \(13\).
Points \(X\) and \(Y\) lie outside the square so that \(PX=RY=12\) and \(QX=SY=5\).
Find \(\overline{XY}^2\).
\(\begin{array}{|rcll|} \hline \mathbf{\overline{XY}^2} &=& \mathbf{(13-2z)^2+(13+2h)^2} \quad &| \quad z=\sqrt{5^2-h^2} \\\\ &=& \left(13-2\sqrt{5^2-h^2}\right)^2+(13+2h)^2 \quad &| \quad h= \dfrac{5*12}{13} \\ &=& \left(13-2\sqrt{5^2-\dfrac{5^2*12^2}{13^2} }\right)^2 +\left(13+ \dfrac{10*12}{13}\right)^2 \\ &=& \left(13-2*5\sqrt{\dfrac{13^2-12^2}{13^2} }\right)^2 +\left(13+ \dfrac{120}{13}\right)^2 \\ &=& \left(13-2*5\sqrt{\dfrac{5^2}{13^2} }\right)^2 +\left(\dfrac{13^2+120}{13}\right)^2 \\ &=& \left(13-\dfrac{50}{13}\right)^2+\left(\dfrac{13^2+120}{13}\right)^2 \\ &=& \left( \dfrac{13^2-50}{13}\right)^2+\left(\dfrac{13^2+120}{13}\right)^2 \\ &=& \dfrac{ \left(13^2-50\right)^2+ \left(13^2+120\right)^2}{13^2} \\ &=& \dfrac{ 13^4-2*13^2*50+50^2+13^4+2*13^2*120+120^2}{13^2} \\ &=& \dfrac{ 13^2 \left( 2*13^2-100+240\right) } {13^2} +\dfrac{50^2+120^2}{13^2} \\ &=& 2*13^2-100+240 + 100 \\ &=& 2*13^2 +240 \\ &=& \mathbf{578} \\ \hline \end{array}\)
