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Points $A$ and $B$ are on side $\overline{YZ}$ of rectangle $WXYZ$ such that $\overline{WA}$ and $\overline{WB}$ trisect $\angle ZWX$.  If $BY = 2$ and $BZ = 4$, then what is the area of rectangle $WXYZ$?

 Jun 7, 2024
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W                            X

 

 

 

Z      A        B          Y

         4             2

 

Triangle  ZWB  is a 30-60-90  right triangle

Angle ZWB  = 60

Angle WBZ = 30

So WZ = BZ/sqrt 3  = 4 /sqrt 3

 

Area of WXYZ =  ZY * WZ  =  6 * 4/sqrt 3  =  24/sqrt 3 =  8 sqrt 3

 

 

cool cool cool

 Jun 7, 2024

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