Question 1143960
<pre>{{{drawing(400,1000/3,-4,20,-2,18,

line(0,0,10,0), line(0,0,18,15.87450787), line(10,0,8, 15.87450787),
line(8,15.87450787,18,15.87450787),locate(5,0,10m),locate(4.96,4.5,12m),
locate(1.4,1.2,theta), red(arc(0,0,6,-6,0,42)),
locate(8.4,4.5,8m),line(0,0,8, 15.87450787),  line(10,0,18, 15.87450787) )}}}

The area of a parallelogram is the base times the height.  We are given the
base as 10m, so we only need the height.  We know that the diagonals of a
parallelogram bisect each other, So the two sides of the triangle are half
the diagonals 16m and 24m, so the two slanted sides of the triangle are
8m and 12m.

We use the law of cosines to calculate angle <font face="symbol">q</font>.

{{{cos(theta) = (10^2+12^2-8^2)/(2(10)(12))=(100+144-64)/240=180/240=3/4}}}

{{{theta=41.40962211^o}}}

Now we extend the base and draw in the height h:

{{{drawing(400,1000/3,-4,20,-2,18,

line(0,0,10,0), line(0,0,18,15.87450787), line(10,0,8, 15.87450787),
line(8,15.87450787,18,15.87450787),locate(5,0,10m),locate(4.96,4.5,12m),
locate(1.4,1.2,theta), red(arc(0,0,6,-6,0,42)),

green(triangle(0,0,18,15.87450787,18,0),locate(18.2,8,h)),

locate(8.4,4.5,8m),line(0,0,8, 15.87450787),  line(10,0,18, 15.87450787) )}}}

Now we have the green right triangle in which:

The OPPOSITE side of theta is height h, and the HYPOTENUSE is the longer given
diagonal, which is 24m.  So

{{{sin(theta) = OPPOSITE/HYPOTENUSE = h/24}}}

Put a 1 under the sine:

{{{sin(theta)/1 = h/24}}}

Cross-multiply:

{{{h=24sin(theta)= 24sin(41.40962211^o)=24(0.6614378278)=15.87450787m}}}

{{{Area=base*height=(10m)(15.87450787m) = 158.7450787m^2}}}

Edwin</</pre>