Question 1209055
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In the diagram, ABCD is a parallelogram with AB=37 cm. AD is a semi- circle. 
The area of the parallelogram is 518 cm^2, and < ADC = 60°. 
Find the area of the shaded region in cm^2.
https://ibb.co/xs0HGmk
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<pre>
Use the formula for the area of a parallelogram

    area = {{{a*b*sin(alpha)}}},


where "a" and "b" are any two adjacent sides of the parallelogram and  {{{alpha}}}  is
the concluded angle between these sides.


In our case AB = 37, AD is unknown, {{{alpha}}} = 180°-60° = 120°. So, we have this equation

    518 = 37*AD*sin(120°) = 37*{{{sqrt(3)/2)}}}*AD,

hence

    AD = {{{518/(37*(sqrt(3)/2))}}} = {{{28/sqrt(3)}}} cm.


AD is the diameter, the shaded area is the semi-circle.


Hence, the area of the shaded area is  

    {{{(pi/2)*((AD)^2/4)}}} = {{{(pi/2)*((28^2)/(3*4))}}} = {{{pi*((7^2*4^2)/(2*3*4))}}} = {{{pi*(49*2)/3}}} = {{{(98/3)*pi}}} = {{{(98/3)*3.14159}}} = 102.625 cm^2  (approximately).


<U>ANSWER</U>.  The area of the shaded area is  {{{(98/3)*pi}}}  cm^2,  or  about  102.625 cm^2.
</pre>

Solved.