Find the area of this parallelogram

Here is their argument. given the obtuse angle x, we make a quadrilateral abcd with ∠dab = x, and ∠abc = 90◦, and ad = bc. say the perpendicular bisector to dc meets the perpendicular bisector to ab at p. then pa = pb and pc = pd. so the triangles pad and pbc have equal sides and are congruent. thus ∠pad = ∠pbc. but pab is isosceles, hence ∠pab = ∠pba. subtracting, gives x = ∠pad−∠pab = ∠pbc −∠pba = 90◦. this is a preposterous conclusion – just where is the mistake in the "proof" and why does the argument break down there?

What is the reflection image of (5,-3) across the line y = -x?

Which of the following equations is of the parabola whose vertex is at (2, 3), axis of symmetry parallel to the y-axis and p = 4? a.)y-3 = 1/16 (x-2)^2 b.)y+3 = -1/16 (x+2)^2 c.)x-2 = 1/16 (y-3)^2