Question 1067719
A parallelogram's diagonals bisect each other,
so they intersect at a point that is
{{{5.6cm/2=2.8cm}}} from one end of the {{{4.4cm}}} side, and
{{{7cm/2=3.5cm}}} from the other end.
We can locate that point by drawing
an arc with radius{{{2.8cm}}} centered at one end of a {{{4.4cm}}} side/segment, and
an arc with radius{{{3.5cm}}} centered at the other end of the same {{{4.4cm}}} side/segment
{{{drawing(300,300,-1.5,5,-1,5.5,
green(triangle(0,0,1.699,2.2256,4.4,0)),
line(0,0,4.4,0),
red(arc(0,0,5.6,5.6,300,320)),
red(arc(4.4,0,7,7,210,230)),
locate(1.8,0,4.4cm),
locate(0.85,1.113,green(2.8cm)),
locate(2.4,1.113,green(3.5cm))
)}}}
Connecting that point to the ends of the segment
shows us half of each diagonal.
If we extend the half-diagonal segments by the same length,
we have the diagonals,
and the free ends of those diagonals
are the other two vertices of the parallelogram
{{{drawing(300,300,-1.5,5,-1,5.5,
green(triangle(0,0,3.498,4.4512,4.4,0)),
green(triangle(0,0,-1.002,4.4512,4.4,0)),
line(0,0,4.4,0),line(-1.002,4.4512,3.498,4.4512),
line(0,0,-1.002,4.4512),line(4.4,0,3.498,4.4512)
)}}}