Q403 | Math Olympiad | Geometry | 2021 Fall AMC 10B Problem 15 | Triple Perpendicular Congruence

[@xualain3129]

Let sin 2a=12/13 ->cos 2a=5/13 ->1-2*sin(a)^2=5/13 ->
sin a=2/sqrt(13)
x=6/sin a=3*sqrt(13) -> area(ABCD)=x*x=117

[@reachthestars98]

Yes, that also works.