document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1197946>Question 1197946</A>: <I><SPAN STYLE=\"word-break: break-all;\"> There is a circle of diameter 24 cm. It is folded along vertical chord AB so that point D on the circumference coincides with the centre C. The area of the shaded region, in cm^2 is:<BR>\n" );
document.write( "a) 192π - 72√3<BR>\n" );
document.write( "b) 24π - 18√3<BR>\n" );
document.write( "c) 48π - 18√3<BR>\n" );
document.write( "d) 192π - 144√3<BR>\n" );
document.write( "e) 48π - 36√3\r<BR>\n" );
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document.write( "I have tried to draw out some angles and such to find ratios to no avail. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #831428 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=831428\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "It depends on where the shaded region is located. Is it to the left of the mirror line? Or to the right? Unfortunately you haven't specified.\r<BR>\n" );
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document.write( "But luckily there are answer choices to pick from which will help narrow down the correct region.\r<BR>\n" );
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document.write( "This is probably what the diagram looks like<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" src = \"https://i.imgur.com/RlkbKVm.png\"><BR>\n" );
document.write( "I'm placing point A as the center, rather than C. But feel free to swap those letters if you prefer.<BR>\n" );
document.write( "The colored regions aren't necessarily what is shaded in the diagram of your textbook. <BR>\n" );
document.write( "They're shaded that way to break the figure into manageable pieces.\r<BR>\n" );
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document.write( "d = 24 = diameter<BR>\n" );
document.write( "r = d/2 = 24/2 = 12 = radius<BR>\n" );
document.write( "The radius goes from A to B, and also from A to D.<BR>\n" );
document.write( "This explains how AB = 12 and AD = 12.\r<BR>\n" );
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document.write( "The dashed vertical mirror line will have point D reflect over it to land on point A, which is the center of the circle.<BR>\n" );
document.write( "Because of this mirror symmetry, we know that AE = ED<BR>\n" );
document.write( "Furthermore, AE = AD/2 = 12/2 = 6\r<BR>\n" );
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document.write( "Focus on triangle AEB. <BR>\n" );
document.write( "This is a 30-60-90 triangle because of the fact hypotenuse AB = 12 is exactly double that of the short leg AE = 6. <BR>\n" );
document.write( "This makes the long leg BE = 6*sqrt(3).\r<BR>\n" );
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document.write( "Recall that <BR>\n" );
document.write( "longLeg = shortLeg*sqrt(3)<BR>\n" );
document.write( "which applies to 30-60-90 triangles exclusively.\r<BR>\n" );
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document.write( "Since we know we're dealing with a 30-60-90 triangle, we know that angle EAB is 60 degrees. <BR>\n" );
document.write( "This doubles to 2*60 = 120 degrees to represent angle BAC.<BR>\n" );
document.write( "This doubling process is valid because triangles EAB and EAC are congruent, which in turn makes angle EAB = angle EAC.\r<BR>\n" );
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document.write( "The portion of the circle not shaded red (including the blue region) is 120/360 = 1/3 of the circle. <BR>\n" );
document.write( "Therefore, the red shaded region is the remaining 2/3 of the circle.\r<BR>\n" );
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document.write( "The area of the full circle of radius r = 12 is:<BR>\n" );
document.write( "A = pi*r^2<BR>\n" );
document.write( "A = pi*12^2<BR>\n" );
document.write( "A = pi*144<BR>\n" );
document.write( "A = 144pi\r<BR>\n" );
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document.write( "The red shaded area, shaped like pacman, is 2/3 of that full area.<BR>\n" );
document.write( "red shaded area = (2/3)*(144pi) = 96pi\r<BR>\n" );
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document.write( "Now let's find the area of triangle EAB shaded in blue.<BR>\n" );
document.write( "area = 0.5*base*height<BR>\n" );
document.write( "area = 0.5*AE*BE<BR>\n" );
document.write( "area = 0.5*6*6*sqrt(3)<BR>\n" );
document.write( "area = 18*sqrt(3)\r<BR>\n" );
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document.write( "Double this result to get the area of triangle CAB.<BR>\n" );
document.write( "2*18*sqrt(3) = 36*sqrt(3)<BR>\n" );
document.write( "again this doubling process is valid because triangles EAB and EAC are congruent.\r<BR>\n" );
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document.write( "Then combine this with the 96pi found earlier to get the area of everything to the left of the dashed mirror line.\r<BR>\n" );
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document.write( "96pi+36*sqrt(3)\r<BR>\n" );
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document.write( "This is not listed as one of the answer choices, so we'll see if we can find the area of the region to the right of the mirror line.\r<BR>\n" );
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document.write( "We subtract from 144pi, which was the area of the full circle<BR>\n" );
document.write( "144pi - (96pi+36*sqrt(3))<BR>\n" );
document.write( "144pi - 96pi - 36*sqrt(3)<BR>\n" );
document.write( "48pi - 36*sqrt(3)<BR>\n" );
document.write( "This is listed as one of the answer choices. So this is likely the correct answer. <BR>\n" );
document.write( "Of course this is assuming the shaded region in your textbook is to the right of the mirror line.\r<BR>\n" );
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document.write( "Answer: <font color=red>choice E) 48pi - 36*sqrt(3)</font><BR>\n" );
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