In rectangle WXYZ, A is on side WX such that AX = 4, B is on side YZ such that BY = 18, and C is on side XY such that angle ACB= 90 degrees and CY = 2CX. Find AB.

Guest Jun 20, 2020

#2**+2 **

**In rectangle \(WXYZ\), \(A\) is on side \(WX\) such that \(AX = 4\), \(B\) is on side \(YZ\) such that \(BY = 18\), and \(C\) is on side \(XY\) such that angle \(ACB= 90^\circ\) and \(CY = 2CX\). Find \(AB\).**

\(\begin{array}{|lrcll|} \hline &AC^2 &=& 4^2+cx^2 \\ &CB^2 &=& (2cx)^2+18^2 \\ &x^2= AC^2+CB^2 &=& 4^2+cx^2+(2cx)^2+18^2 \\ &x^2 &=& 16+cx^2+ 4cx^2+ 324 \\ &x^2 &=& 5cx^2+ 340 \qquad (1) \\ \hline &x^2 &=& (18-4)^2 + (3cx)^2 \\ &x^2 &=& 14^2 + 9cx^2\\ &x^2 &=& 9cx^2 + 196 \qquad (2) \\ \hline (2)=(1): & 9cx^2 + 196 &=& 5cx^2+ 340 \\ & 9cx^2 - 5cx^2 &=& 340 - 196 \\ & 4cx^2 &=& 144 \\ & cx^2 &=& 36 \\ & \mathbf{cx} &=& \mathbf{6} \\ \hline (1): &x^2 &=& 5cx^2+ 340 \quad | \quad cx=6 \\ &x^2 &=& 5*6^2+ 340 \\ &x^2 &=& 520 \\ &x &=& 2\sqrt{130} \\ &\mathbf{x} &=& \mathbf{22.8035085020} \\ \hline \end{array}\)

AB is \(\approx \mathbf{22.8}\)

heureka Jun 20, 2020