Question 1149185
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Let x be the length of AD, the base of the parallelogram.<br>
The height of the parallelogram is<br>
{{{42*sin(60) = 42*(sqrt(3)/2) = 21*sqrt(3)}}}<br>
The area, 504, is base times height:<br>
{{{504 = (x)(21*sqrt(3))}}}
{{{x = 504/(21*sqrt(3)) = 24/sqrt(3) = 8*sqrt(3)}}}<br>
AD is the diameter of the semicircle, length x=8*sqrt(3); the radius of the semicircle is 4*sqrt(3).  The area of the semicircle is<br>
{{{(1/2)(pi)(4*sqrt(3))^2 = 24pi}}}<br>