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Points $S$ and $T$ are on side $\overline{CD}$ of rectangle $ABCD$ such that $\overline{AS}$ and $\overline{AT}$ trisect $\angle DAB$. If $CT = 2\sqrt{3}-3$ and $DS = 1$, then what is the area of $ABCD$?

https://latex.artofproblemsolving.com/c/c/e/cce3c90dcfa09daf5610e4efbe8d333bcd0dafb5.png

 Nov 4, 2019
 #1
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I'm getting 6*sqrt(3) - 6.

 Nov 4, 2019
 #2
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not correct, sorry.

Guest Nov 4, 2019
 #3
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Angle SAD  must  = 30°

So...triangle  SAD  is a 30-60-90  right triangle....and if DS  =  1 ....then  AD  =  sqrt (3) * DS  = sqrt (3)

And  AS=  2

 

And angle AST  =  120°   And angle TAS  = 30°   so angle ATS  = 30°

 

So....AS  = TS   = 2

 

So  CD   =  CT  + TS + SD   =   [ 2sqrt (3) - 3 ]  + [ 2 ] + [1]    =  2sqrt (3)

 

So  [ ABCD ]  =   AD * CD  =  2sqrt (3)   *  sqrt (3)   =   6 units^2

 

 

cool cool cool

 Nov 4, 2019
 #4
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+1

ohh... i forgot about 30-60-9, lol!

Guest Nov 4, 2019

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