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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 the area of triangle $WAZ$ is 6, what is the area of triangle $WAB$?

 

 Dec 20, 2023
 #1
avatar+126978 
+1

Triangle WAZ  is a 30-60-90  right triangle

Angle AWZ = 30

So angle  BWZ = 60

 

WZ / AZ   = sqrt 3

WZ = sqrt 3 AZ

 

Triangle  BWZ is a 30 - 60 -90 right triangle

BZ =  sqrt (3) WZ

 

Area of WAZ  = (1/2) AZ * WZ

6 =  (1/2) AZ * WZ

12 = AZ * WZ

 

So area of   triangle BWZ =  (1/2) BZ *WZ   = (1/2) [sqrt (3) * WZ]  *  [sqrt (3) AZ ]  =

(1/2) (sqrt 3)^2 * AZ * WZ =

(3/2) * 12  =   18

 

[ WAB ] =  [ BWZ ] - [WAZ] =  18 - 6  =   12

 

 

cool cool cool

 Dec 20, 2023

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