Points S and T are on side CD of rectangle ABCD such that AS and AT trisect angle DAB . If CT=3 and DS=6, then what is the area of ABCD?

Nice, Melody.....

ST can also be found, thusly :

AD = 6/tan(30) = 6sqrt(3)

And, using similar triangles, we have

DS / AD  ≈ AD / [DS + ST]

DS^2 + DS*ST   = AD^2

ST  = [ AD^2 - DS^2] /  DS

ST = [ 108 - 36] / 6  =  [ 72 / 6] = 12

And the area is :

[DS + ST + TC] * AD =

[6 + 12 + 3] * 6/tan (30) =  126/tan(30) =  126*sqrt(3)  units^2

Hi Mellie,

I love answering your questions. I would of answered this one earlier but it got lost. I thought I saw it but then I could not find it.     

Points S and T are on side CD of rectangle ABCD such that AS and AT trisect angle DAB . If CT=3 and DS=6, then what is the area of ABCD?

Here is the diagram. I answered it just  from a rough sketch but i wanted to check it was correct so I draw it to scale. :)

 DS is given as 6 units and TC is given as 3 units, ST is not given.

The LaTex function is not working so I will have to do it by hand.

tan 30 = DS/H = 6/H

1/sqrt3 = 6/H

H=6sqrt3

tan60 = DT/H

sqrt3 = (6+ST)/6sqrt3

6*3=6+ST

18=6+ST

12=ST

so

DC=6+12+3 = 21

Area = AD * DC

Area = 6sqrt3 * 21

Area = 126 sqrt3      units squared.

That was a great question - thanks Mellie.

Thanks CPhill, that is a good answer also.