Question 1206736: what is the area of the largest semi-circle which can be inscribed within rhombus abcd with edge length 2 cm if the measure of angle a is 60 degrees
Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!
I will take a crack at this....
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....
In the figure....
ABCD is the rhombus with side length 2
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't know how to draw the semicircle....)
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)
BK is the perpendicular bisector of EF.
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.
With that picture, the side length of each of the equilateral triangles, and thus the radius of the inscribed semicircle, is 
The area of the semicircle is then
ANSWER: 
NOTE: I leave it to the student to determine that the radius of the semicircle is , so that the student gets to work at least part of the problem by himself.
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.
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