<PRE><B>Question 1206736</B>
&lt;br&gt;
I will take a crack at this....&lt;br&gt;
Without proof, it seems to me that the largest area will be if the diameter of the semicircle is parallel to the long diagonal of the rhombus.  We get a picture like this....&lt;br&gt;
{{{drawing(400,400,-1,3,-1,3
,line(0,0,1,sqrt(3)),line(1,sqrt(3),3,sqrt(3)),line(3,sqrt(3),2,0),line(2,0,0,0)
,red(line(2/3,0,8/3,2sqrt(3)/3))
,red(line(5/3,sqrt(3)/3,2/3,2sqrt(3)/3)),red(line(5/3,sqrt(3)/3,5/3,sqrt(3)))
,locate(0,0,A),locate(0.9,1.9,B),locate(3,1.9,C),locate(2,0,D)
,locate(0.5,1.2,E),locate(5/3,1.9,F),locate(8/3,2sqrt(3)/3,G),locate(2/3,0,H)
,locate(5/3,sqrt(3)/3,O)
,green(line(2/3,0,2/3,2sqrt(3)/3))
,green(line(2/3,2sqrt(3)/3,5/3,sqrt(3)))
,green(line(5/3,sqrt(3),8/3,2sqrt(3)/3))
,blue(line(1,sqrt(3),7/6,5sqrt(3)/6))
,locate(7/6,5sqrt(3)/6,K)
)}}}&lt;br&gt;
In the figure....&lt;br&gt;
ABCD is the rhombus with side length 2&lt;br&gt;
E, F, G, and H are the points on the rhombus where the inscribed semicircle touches.  GH is the diameter of the semicircle; O is the center of the semicircle; OE, OF, OG, and OH are radii  (I don&#39;t know how to draw the semicircle....)&lt;br&gt;
OE is perpendicular to AB and OF is perpendicular to BC (radii to the points of tangency are perpendicular to the sides of the rhombus)&lt;br&gt;
BK is the perpendicular bisector of EF.&lt;br&gt;
With angle A being 60 degrees, triangles AEH, EKB, FBK, and FCG are 30-60-90 right triangles, and triangles HEO, EFO, and OFG are equilateral triangles. &lt;br&gt;
With that picture, the side length of each of the equilateral triangles, and thus the radius of the inscribed semicircle, is {{{2sqrt(3)/3}}}&lt;br&gt;
The area of the semicircle is then&lt;br&gt;
{{{A=(1/2)(pi)(r^2)=(1/2)(pi)(12/9)=(2/3)pi}}}&lt;br&gt;
ANSWER: {{{(2/3)pi}}}&lt;br&gt;
NOTE: I leave it to the student to determine that the radius of the semicircle is {{{2sqrt(3)/3}}}, so that the student gets to work at least part of the problem by himself.&lt;br&gt;
One way to find the radius is to let the radius (side length of each of the equilateral triangles) be 2x.  Then use 30-60-90 right triangles BFK and FCG to find the lengths of BF and CF in terms of x; then solve for x knowing that BF+FC=2.&lt;br&gt;
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