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?

Mellie
Nov 14, 2015

#2**+15 **

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

CPhill
Nov 16, 2015

#1**+15 **

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.

Melody
Nov 15, 2015

#2**+15 **

Best Answer

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

CPhill
Nov 16, 2015