pq and rs are lengths of the parallelogram, each of which has a length of 2x.
qr and sp are widths of the parallelogram, each of which has a length of x.
for purposes of calculating the area of the parallelogram, rs is the base.
the opposite angles of a parallelogram are equal and the adjacent angles of a parallelogram are supplementary.
this means:
angle pqr = 110
angle qrs = 70
andle rsp = 110
angle spq= 70
drop a perpendicular from point q to intersect with line sr at point w.
qw becomes the height of the parallelogram.
this forms right triangle qrw.
qr is the hypotenuse of this triangle whose length is equal to x.
angle qrw is equal to 70 degrees.
qw is the side opposite angle qrw whose length is equal to h.
to find h in terms of x, use trig function of sin(70) = opp/hyp = h/x
solve for h to get h = x*sin(70).
the area of the parallelogram is equal to base * height.
base has a length of 2x (given).
the area is therefore equal to 2x * h
replace h with x*sin(70) to get area = 2*x*x*sin(70).
simplify to get area = 2*x^2*sin(70).
divide both sides of the equation by 2*sin(70) to get area/(2*sin(70)) = x^2.
area = 93.5304 (given).
formula becomes 93.5304/(2*sin(70)) = x^2.
solve for x^2 to get x^2 = 93.5304/(2*sin(70)).
take square root of both sides of the equation to get x = sqrt(93.5304/(2*sin(70))).
this makes x = 7.05453658
the diagram of your parallelogram is shown below:
point w is not marked (an oversight).
it is the intersection of the line from point q to the line sr.
qw is perpendicular to sr, therefore angle rwq is 90 degrees.
your solution is that x = 7.05453658 which is the same as x = sqrt(93.5304/(2*sin(70)))