Question 947419
It is not absolutely necessary to sketch the rhombus,
but it would help visualize the relations between angles.
{{{drawing(240,300,-1.2,1.2,-1.5,1.5,
line(-1,0,0,1.327),line(1,0,0,1.327),
line(-1,0,0,-1.327),line(1,0,0,-1.327),
green(line(-1,0,1,0)),green(line(0,-1.327,0,1.327)),
locate(-1.1,0,A),locate(1,0,C),
locate(0.05,1.4,B),locate(0.05,-1.3,D),
red(arc(1,0,1,1,127,233)),
locate(0.55,0.03,red(106^o))
)}}}
Angles BCD and BAD are opposite angles, so they have the same measure:
{{{BAD=DCD=106^o}}} .
Angle BAC (with side BA and diagonal AC) is half of angle BAD,
so {{{BAC=BAD/2=BCD/2=106^o/2=53^o}}} .
Without the drawing, I can tell that
ADB is one of the base angles of isosceles triangle ADB
(with diagonal BD as its base,
and angle BAD as its vertex angle),
so {{{ADB=(180^o-BAD)/2=(180^o-106^o)/2=74^o/2=37^o}}} .
With the drawing, you can tell that
since the diagonals of a rhombus are perpendicular to each other,
angles ADB and BAC are acute angles in a right triangle,
and therefore complementary, so {{{ADB=90^o-BAC=90^o-53^o=37^o}}} .
 
With the angle measures in hand, the equations to find {{{x}}} and {{{y}}} are:
{{{4y+5=37}}} and {{{6x-7=53}}}
{{{4y+5=37}}}-->{{{4y=37-5}}}-->{{{4y=32}}}-->{{{y=32/4}}}-->{{{highlight(y=8)}}}
{{{6x-7=53}}}-->{{{6x=53+7}}}-->{{{6x=60}}}-->{{{x=60/6}}}-->{{{highlight(x=10)}}}
 
You are not telling me what/where point E is,
but I am guessing that it is the intersection of the diagonals of the rhombus.
Since B and D are opposite vertices, BD is a diagonal,
and if E is the intersection of both diagonals,
then E is the midpoint of both diagonals and {{{ED=BE}}} .
That gives us the equation
{{{5z-7=3z-1}}}-->{{{5z-3z=7-1}}}-->{{{2z=6}}}-->{{{z=6/3}}}-->{{{highlight(z=3)}}} .
That makes
{{{ED=5*3-7=15-7=8}}} and {{{BE=3*3-1=9-1=8}}} , which verifies.
 
Most importantly, with {{{BE=8}}} and {{{EC=6}}}
are the legs of right triangle BEC, so
the Pythagorean theorem says that the length of the hypotenuse is
{{{BC=sqrt(8^2+6^2)=sqrt(64+36)=sqrt(100)=10}}} .
That makes the length of each side of ABCD {{{10}}} (in whatever units),
and the perimeter of ABCD is {{{4*10=highlight(40)}}} .
 
NOTE:
The teacher proposing this problem used a 2X scaled up version of the very popular right triangle with side lengths 3, 4, and 5 to design the rhombus.
That would make {{{tan(BAC)=8/6=4/3=1.333(rounded)}}} .
However, the solution to the problem involves {{{BAC=53^o}}} ,
and {{{tan(53^o)=1.327(rounded)}}} , which is not quite equal,
but very close to {{{4/3=1.333(rounded)}}} .
Using the fact that {{{sin(53^o)=0.7986(rounded)}}} ,
you would not have needed the length of EC,
and could have calculated {{{BC=approximately}}}{{{8/0.7986=10.02}}} .
That tells me that you were obviously not expected to use trigonometric functions, but the problem designer tried to make it believable by choosing {{{106^o}}} for angle BCD.