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