Question

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.

Answer 1

Answer:2*sqrt(130)

Step-by-step explanation:$From right triangle AXC , we have \angle XAC = 90^\circ - \angle XCA . We also must have \angle YCB + \angle ACB + \angle XCA = 180^\circ , so \angle YCB = 180^\circ - 90^\circ - \angle XCA = 90^\circ - \angle XCA , which means \angle YCB = \angle XAC . Combining this with \angle X = \angle Y , we have \triangle XCA \sim \triangle YBC by AA Similarity, and\[\frac{CX}{AX} = \frac{BY}{CY},\]so\[\frac{CX}{4} = \frac{18}{2CX}.\]\\\\This gives us 2CX^2 = 72 , so CX^2 =36 and CX=6 .$$Therefore, CY = 12 .Applying the Pythagorean Theorem to right triangles XCA and CYB gives us\begin{align*}CA^2 &= XA^2 + XC^2 = 16 + 36 = 52,\\BC^2 &= CY^2 + BY^2 = 144 + 324 = 468.\end{align*}Applying the Pythagorean Theorem to right triangle ABC gives\[AB = \sqrt{AC^2 + BC^2} = \sqrt{52 + 468} = \sqrt{520} = \boxed{2\sqrt{130}}.\](We can also compute AB by dropping a perpendicular from A to \overline{YZ} , which creates a right triangle.)$