Area issue geometry

Question

I could not solve it, is it because it is missing information? I used law of basic trigonometry

Answer 1

As pointed out by @achilleHui, since $\angle ADC = 90^0 = 2 \times \angle APC$ and both angles are on the same side of AC, therefore D is the center of the circle APC. If DE is the perpendicular bisector of AP, then AE = 2.

Let the sides of the square be 2x.

In $\triangle AED, ED = … = 2 \sqrt {x^2 - 1}$ and $\tan \theta = … = \sqrt {x^2 - 1}$.

In $\triangle OBA, BA = 2x$ and $\tan \theta = \dfrac {2x}{\sqrt {6^2 – (2x)^2}}$.

Equating the two and simplifying, we get $x^4 – 9x^2 + 9 = 0$

Rejecting the other value of $x^2$, we get $x^2 = \dfrac {9 – 3\sqrt 5}{2}$.

Required result $= 4x^2$