How do i calculate the perimeter of abcd?

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?

Compare the subtraction problems 6/8 - 5/8 = 1/8 and 6/9 - 7/9 = 1/9 why is the answer to the first problem positive and the answer to the second problem negative

Find the volume of the described solid.the solid lies between planes perpendicular to the x-axis at x = 0 and x=7. the cross sectionsperpendicular to the x-axis between these planes are squares whose bases run from the parabolay=-31x to the parabola y = 30/x.a) 441b) 147c) 864d) 882