Question 1203371
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It depends if the diagram is this
{{{drawing(400,400,-10,10,-10,10,
line(-6,4,6,4),
line(6,4,6,-4),
line(6,-4,-6,-4),
line(-6,-4,-6,4),
arc(6,0,8,8,-90,90),
locate(0,4,"20 ft"),
locate(-5.5,0,"12 ft"),
locate(-9.8,-8,matrix(1,2,"Diagram","not")),
locate(-9.8,-8.7,matrix(1,2,"to","scale"))
)}}}


OR if it's this
{{{drawing(400,400,-10,10,-10,10,
line(-6,4,6,4),
line(6,4,6,-4),
line(6,-4,-6,-4),
line(-6,-4,-6,4),
arc(0,4,12,12,-180,0),
locate(0,4,"20 ft"),
locate(-5.5,0,"12 ft"),
locate(-9.8,-8,matrix(1,2,"Diagram","not")),
locate(-9.8,-8.7,matrix(1,2,"to","scale"))
)}}}


If it's the 1st diagram, then the semicircular area is 
(1/2)*pi*r^2 = (1/2)*pi*6^2 = 18pi square feet.
That approximates to 18pi = 18*3.14 = 56.52 square feet


Add that onto the rectangular area of 12*20 = 240 square feet.
240+56.52 = 296.52


The total area in the 1st figure is approximately 296.52 square feet.




If on the other hand the diagram was the 2nd one I've shown above, then:
semicircle area = (1/2)*pi*r^2 = (1/2)*pi*10^2 = 50pi = 50*3.14 = 157
rectangular area = 12*20 = 240 


total approximate area = 240+157 = 397 square feet


Both involve 240, but the difference is the semicircular area. 
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