Question 450961
<pre><b>
Since the diagonals of a parallelogram bisect each other,
and the diagonals are 10 and 22, then the halves of the
diagonals are 5 and 11 
{{{drawing(400,550/3,-2,22,-2,9,
locate(8.2,4.3,"65°"), locate(9,6,5),
line(0,0,13.87400479,0),
line(13.87400479,0,20.79342001,7.185658223),
line(6.91941522,7.185658223,20.79342001,7.185658223),
line(6.91941522,7.185658223,0,0),
line(0,0,20.79342001,7.185658223),
line(6.91941522,7.185658223,13.87400479,0),
locate(6,2.1,11), locate(11.2,2,5),
locate(13.5,4.5,11)
 )}}}

Look at the red triangle:

{{{drawing(400,550/3,-2,22,-2,9,
locate(8.2,4.3,"65°"), locate(9,6,5),
line(13.87400479,0,20.79342001,7.185658223),
line(6.91941522,7.185658223,20.79342001,7.185658223),
line(6.91941522,7.185658223,0,0),
line(10.39671001,3.592829112,20.79342001,7.185658223),
line(6.91941522,7.185658223,10.39671001,3.592829112),
locate(6,2.1,11), locate(11.2,2,5),
locate(13.5,4.5,11), red(line(0,0,13.87400479,0)
line(13.87400479,0,10.39671001,3.592829112),line(10.39671001,3.592829112,0,0)),
red(line(13.87400479,0,10.39671001,3.592829112)))}}}

The interior angle at the top of the red triangle is
supplementary to the 65° angle.  So it is 180°-65°=115°

{{{drawing(400,550/3,-2,22,-2,9,
locate(8.2,4.3,"65°"), locate(9,6,5),
line(13.87400479,0,20.79342001,7.185658223),
line(6.91941522,7.185658223,20.79342001,7.185658223),
line(6.91941522,7.185658223,0,0), locate(7.1,0,b),
line(10.39671001,3.592829112,20.79342001,7.185658223),
line(6.91941522,7.185658223,10.39671001,3.592829112),
locate(6,2.1,11), locate(11.2,2,5), locate(9,3,"115°"),
locate(13.5,4.5,11), red(line(0,0,13.87400479,0)
line(13.87400479,0,10.39671001,3.592829112),line(10.39671001,3.592829112,0,0)),
red(line(13.87400479,0,10.39671001,3.592829112)))}}}

We have a case of SAS, so we use the law of cosines to
calculate b, the base of the red triangle:

bē = 11ē + 5ē - 2(11)(5)cos(115°)
bē = 121 + 25 - 110(-.4226182617)
bē = 285.4640264
 b = 16.8956807 inches.

We have the base of the parallelogram.
Now we have to find the height of the parallelogram.

We will use the law of sines to find the angle &#5054;
{{{drawing(400,550/3,-2,22,-2,9,
locate(8.2,4.3,"65°"), locate(9,6,5),
line(13.87400479,0,20.79342001,7.185658223),
line(6.91941522,7.185658223,20.79342001,7.185658223),
line(6.91941522,7.185658223,0,0), locate(7.1,0,b),
line(10.39671001,3.592829112,20.79342001,7.185658223),
line(6.91941522,7.185658223,10.39671001,3.592829112),
locate(6,2.1,11), locate(11.2,2,5), locate(9,3,"115°"),
locate(13.5,4.5,11), red(line(0,0,13.87400479,0), locate(2.6,1,theta),
line(13.87400479,0,10.39671001,3.592829112),line(10.39671001,3.592829112,0,0)),
red(line(13.87400479,0,10.39671001,3.592829112)))}}}

   5        b
 ————— = —————————
 sin&#5054;    sin(115°)

b*sin&#5054;  = 5sin(115°)
            
         5sin(115°)
 sin&#5054;  = ———————————
             b

         5sin(115°)
 sin&#5054;  = ———————————
         16.8956807


 sin&#5054;  = .2682069468

    &#5054;  = 15.55759756°

Now we will extend the base and draw in the height 
of the parallelogram, labeling it h:

{{{drawing(400,550/3,-2,22,-2,9,
locate(8.2,4.3,"65°"), locate(9,6,5),
line(13.87400479,0,20.79342001,7.185658223),
line(6.91941522,7.185658223,20.79342001,7.185658223),
line(6.91941522,7.185658223,0,0), locate(7.1,0,b),
line(10.39671001,3.592829112,20.79342001,7.185658223),
line(6.91941522,7.185658223,10.39671001,3.592829112),
locate(6,2.1,11), locate(11.2,2,5), locate(9,3,"115°"),
locate(13.5,4.5,11), red(line(0,0,13.87400479,0), locate(2.6,1,theta),
line(13.87400479,0,10.39671001,3.592829112),line(10.39671001,3.592829112,0,0)),
red(line(13.87400479,0,10.39671001,3.592829112),
line(20.79342001,7.185658223, 20.79342001,0),line(13.87400479,0,20.79342001,0),

locate(21,3.6,h),rectangle(19.8,0,20.7,1,0))


)}}}
 
The big right triangle that has angle &#5054;  on the left, and
the opposide side h on the right, has a hypotenuse which
is the longer diagonal, 22, so

         h
sin&#5054;  = —————
        22

   h = 22(sin&#5054;)
   h = 22(.2682069468)
   h = 5.90055283

Finally we can calculate the area of the parallelogram,

   A = bh
   A = (16.8956807in)(5.90055283in)
   A = 99.69385657 square inches

Edwin</pre>