Triangle QSR is 45,45,90 triangle and hence QS and SR must be equal to \(\sqrt{2}\)


Triangle PQS is a has angles 20,70 and 90. Since the side having \(\sqrt{2}\) has a corresponding angle of 70 I guess the side corresponding to angle 20 must be less than \(\sqrt{2}\)

Now \(\sqrt{8}\) equals 2\(\sqrt{2}\) and the Quantity B must equal \(\sqrt{2}\) + quantity less than \(\sqrt{2}\). Hence, I guess Quantity A is the correct answer.

I think A is the correct answer.

can someone please shed some on this, it's very interesting to know the logic behind solving this

What I did was imagine the second triangle to be equal to a 30-60-90. If this were the case, you would have the 60 degree angle face root2, which implies that the 30 degree angle facing PS would be root2/root3 -> something less than 1. Since this would actually make PS bigger (since PS is facing 30 rather 20), this implies that, even in the biggest case, we have root2+some decimal which can never reach root8, since that is just under 3 and root 2 is 1.41 roughly. Therefore, I would select A to be bigger.