square on the top of a square

Question

i am stuck with a question of mensuration for three days and i posted it on facebook and asked friends but no correct solution yet. please help .

Answer 1

You must have $|XY|=2\cdot(|WX|/2)$, which is

$(\sin(v)-\sin(45))\cdot14\cdot\sqrt{2} = 2\cos(v)\cdot14\cdot\sqrt{2}$

Which you can solve for $v$ and substitute back into $|XY|$. Then the area is $4|XY|^2$.

Answer 2

Hint: Use symmetry and the Pythagoren Theorem for $\triangle AOT$, where $O$ is the center of the circle and $T$ is the intersection between $AB$ and the line through $O$ perpendicular to $AB$.

This directly gives you everything you need

Answer 3

$(14+a)^2 + (a/2)^2=(14*2^{0.5})^2 $ From geometry...