document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1182637>Question 1182637</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A rectangular lot 80 m x 40 m long is divided into two <BR>\n" );
document.write( "areas by an arc whose center is at the mid-point of the <BR>\n" );
document.write( "shorter side. If the radius of the arc is 30 m, what is the <BR>\n" );
document.write( "ratio of the area of the smaller part to the area of the <BR>\n" );
document.write( "bigger part? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #812743 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13200)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><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/cgi-bin/show-face.mpl?user=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=812743\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "Here is a sketch....<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=drawing%28800%2C400%2C0%2C100%2C0%2C60%0D%0A%2Cline%2810%2C10%2C90%2C10%29%2Cline%2890%2C10%2C90%2C50%29%2Cline%2890%2C50%2C10%2C50%29%2Cline%2810%2C50%2C10%2C10%29%0D%0A%2Cline%2810%2C30%2C25%2C10%29%2Cline%2810%2C30%2C25%2C50%29%2Cline%2810%2C30%2C30%2C30%29%0D%0A%2Ccircle%280%2C30%2C30%29%0D%0A%2Clocate%2810%2C53%2CA%29%2Clocate%2810%2C9%2CB%29%2Clocate%2890%2C9%2CC%29%2Clocate%2890%2C53%2CD%29%0D%0A%2Clocate%288%2C30%2CE%29%2Clocate%2825%2C9%2CF%29%2Clocate%2825%2C53%2CG%29%2Clocate%2831%2C30%2CH%29%0D%0A%2Clocate%287%2C40%2C20%29%2Clocate%2818%2C40%2C30%29%0D%0A%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "Rectangle ABCD has side lengths 40 and 80; the circular arc is centered at E, the midpoint of AB. AE=20 and EG=30; the Pythagorean Theorem gives us AG=10*sqrt(5).<br><BR>\n" );
document.write( "The area of the smaller portion of the lot is the area of the two congruent triangular regions EAG and EBF, plus the area of the circle sector EFG.<br><BR>\n" );
document.write( "The areas of the triangles are easy to calculate.<br><BR>\n" );
document.write( "For the area of the circle sector, you need to find what fraction of a circle the sector is, so you need to find the measure of angle FEG.  Angle FEG is twice angle AGE; angle AGE is arcsin(20/30).  Then the area of the sector is that fraction, times the area of the whole circle, which is (pi)r^2 = 900pi.<br><BR>\n" );
document.write( "I leave it to you to do the calculations....<br><BR>\n" );
document.write( "To check your calculations, here are my figures, slightly rounded:<BR>\n" );
document.write( "Angle FEG: 83.62 degrees<BR>\n" );
document.write( "Area of circle sector: 656.76 m^2<BR>\n" );
document.write( "Area of smaller portion of lot (circle sector plus two triangles): 1103.97 m^2<BR>\n" );
document.write( "Area of larger portion: 3200-1103.97 = 2096.03 m^2<BR>\n" );
document.write( "Ratio of smaller to larger: 1104:2096<br><BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );