Points S and T are on side CD of rectangle ABCD such that AS and AT trisect DAB. If CT = 2 * (sqrt 3) - 3 and DS=1, then what is the area of ABCD?

xXxTenTacion Oct 26, 2019

#1**0 **

Since CT = 2 sqrt(3) - 3 and SD = 1, the area of rectangle ABCD is (2 sqrt(3) - 3 + 1)*3 = (2*sqrt(3) - 2)*3 = 6*sqrt(3) - 6.

Guest Oct 26, 2019

#2**+2 **

I am confused on how you got the 3 out of nowhere. Could you please explain more?

ThatOnePerson
Oct 27, 2019

#3**+2 **

No wait, Guest's answer may be wrong if my calculations are correct. Since AT and AS trisect angle DAB, that means that angle SAD = 30 degrees and so triangle SAD is a 30-60-90 triangle. Since it's a 30-60-90 triangle, AD = sqrt3. Looking at TAD now, we get that it's a 30-60-90 triangle as well. That means that TD = sqrt3 * sqrt3 = 3. Adding CT and TD gives us that CD = 2sqrt3. Multiplying that by sqrt3, we get that the area of ABCD = 2sqrt3 * sqrt3 = **6**. Hence, the area of ABCD is 6.

ThatOnePerson Oct 27, 2019