Points A and B are on side YZ of rectangle WXYZ such that WA and WB trisect ZWX . If BY=3 and AZ=6, then what is the area of rectangle WXYZ?
Since ZWX is trisected, then ∠AWZ=30 and ∠BWZ=60. So triangle AZW is a 30-60-90 right triangle. So WZ=AZ∗3=83. And triangle BZW is a 30 -60-90 right triangle. BZ = WZ* 3=83∗3=24. So [WXYZ]=[WZ∗(BZ+BY)]=[8(3)∗(24+3)]=216(3) units$^2$.
