Question 1099501
{{{drawing(300,140,-2,28,-2,12,
triangle(0,0,22,0,4,9.6),
triangle(22,0,26,9.6,4,9.6),
triangle(0,0,13,0,1.923,4.615),
triangle(4,9.6,21,9.6,17.235,2.541),
locate(-0.5,0,A),locate(21.5,0,D),
locate(3.7,11,B),locate(25.8,11,C),
locate(12.5,0,E),locate(1.4,5.9,F),
locate(20.8,11,G),locate(16.8,2.6,H)
)}}}
Right triangle AEF, with a right angle at F,
has hypotenuse AE=13, leg EF=12, and leg AF that we can calculate as
{{{AF=sqrt(13^2-12^2)=sqrt(25)=5}}} .
Right triangle BGH, with a right angle at H,
has hypotenuse BG=17, leg GH=8, and leg BH that we can calculate as
{{{BH=sqrt(17^2-8^2)=sqrt(225)=15}}} .
 
From those two triangles,we could get angles BAD and ADB=DBC,
and with AD=22, we would have angle-side-angle of triangle ABD,
which is half of the parallelogram.
That looks complicated, though.
 
If we find the height {{{h}}}, perpendicular to AD, we could calculate the area as {{{h*AD}}} .
 
{{{drawing(300,140,-2,28,-2,12,
green(triangle(0,0,4,0,4,9.6)),
green(rectangle(4,0,3.5,0.5)),
triangle(0,0,22,0,4,9.6),
triangle(22,0,26,9.6,4,9.6),
triangle(0,0,13,0,1.923,4.615),
triangle(4,9.6,21,9.6,17.235,2.541),
locate(-0.5,0,A),locate(21.5,0,D),
locate(3.7,11,B),locate(25.8,11,C),
locate(12.5,0,E),locate(1.4,5.9,F),
locate(20.8,11,G),locate(16.8,2.6,H),
locate(4.2,5.9,green(h)),locate(3.7,0,I)
)}}}
The height drawn forms right triangle ABI,
and right triangle BDI.
ABI shares the angle at A angle with right triangle AEF,
so they are similar right triangles.
So, {{{AI/BI=AF/EF=5/12}}}
BDI has angle BDI, congruent with angle GBH,
and that makes right triangles BDI similar to right triangle GBH.
So, {{{ID/BI=BH/GH=15/8}}} .
adding {{{AI/BI=5/12}}} and {{{ID/BI=15/8}}} , we get
{{{AI/BI+ID/BI=5/12+15/8}}}
{{{(AI+ID)/BI=55/24}}}
{{{AD/BI=55/24}}}
{{{22/BI=55/24}}}
{{{BI*55=22*24}}}
{{{BI=22*24/55=9.6}}}
As BI-9.6 is the height of the parallelogram relative to base AD=22,
the area of the parallelogram is
{{{BI*AD=9.6*22=highlight(211.2)}}}