document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=132596>Question 132596</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 1. A hexagonal right prism has a volume of 500 cubic inches. If the base is a regular hexagon with a side 4 inches. What is the altitude of the prism? Round off your answer to two decimal places.<BR>\n" );
document.write( "2. What is the ratio of the volume of a sphere and a cone with the base diameter of the sphere?<BR>\n" );
document.write( "3.How many cubic inches of material are needed for a solid rubber ball with a diameter of 3 inches? Round off your answer to two decimal places.<BR>\n" );
document.write( "4.What is the approximate area of a segment of circle with a radius 12 meters if the length of the chord is 20 meters? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #98813 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=98813\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The volume of a right prism is given by the area of the base times the height (altitude).  Since you know the volume, divide it by the area of the base.  The area of a regular hexagon in terms of the length of a side is: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=%28%283%2Asqrt%283%29%29%2F2%29t%5E2 ALIGN=MIDDLE  ALT=\"A=%28%283%2Asqrt%283%29%29%2F2%29t%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > where t is the length of the side.\r<BR>\n" );
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document.write( "The volume of a right circular cone is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=V=%281%2F3%29pi%2Ar%5E2h ALIGN=MIDDLE  ALT=\"V=%281%2F3%29pi%2Ar%5E2h\"   DISABLED_style= \"max-width:90%; height:auto;\" > where r is the radius of the base and h is the height.  The volume of a sphere of radius r is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%284%2F3%29pi%2Ar%5E3 ALIGN=MIDDLE  ALT=\"%284%2F3%29pi%2Ar%5E3\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  The volume of a cone can be re-written as: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=V=%281%2F3%29pi%2Ar%5E3%28h%2Fr%29 ALIGN=MIDDLE  ALT=\"V=%281%2F3%29pi%2Ar%5E3%28h%2Fr%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  So the ratio of the volume of a sphere to a cone with a base diameter equal to the diameter of the sphere would be <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28%284%2F3%29pi%2Ar%5E3%29%2F%28%281%2F3%29pi%2Ar%5E3%28h%2Fr%29%29=4r%2Fh ALIGN=MIDDLE  ALT=\"%28%284%2F3%29pi%2Ar%5E3%29%2F%28%281%2F3%29pi%2Ar%5E3%28h%2Fr%29%29=4r%2Fh\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "The formula for the volume of a sphere is in the paragraph above.  Use it.\r<BR>\n" );
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document.write( "Construct the perpendicular bisector of the chord.  It will intersect the circle center.  This forms a right triangle with half the cord as one side and the radius intersecting one endpoint of the cord as the hypotenuse.  The angle between the constructed line and the radius through the cord endpoint is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=arcsin%2810%2F12%29 ALIGN=MIDDLE  ALT=\"arcsin%2810%2F12%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  (10 divided by 12 comes from half the chord divided by the radius) Twice this angle is the central angle defined by the endpoints of the chord.  The area of a circle segment, in terms of the radius of the circle and the central angle is:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%281%2F2%29r%5E2%28alpha-sin%28alpha%29%29 ALIGN=MIDDLE  ALT=\"%281%2F2%29r%5E2%28alpha-sin%28alpha%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > if you calculate <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=arcsin%2810%2F12%29 ALIGN=MIDDLE  ALT=\"arcsin%2810%2F12%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > in radians, or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%281%2F2%29r%5E2%28%28pi%2F180%29alpha-sin%28alpha%29%29 ALIGN=MIDDLE  ALT=\"%281%2F2%29r%5E2%28%28pi%2F180%29alpha-sin%28alpha%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > if you calculate <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=arcsin%2810%2F12%29 ALIGN=MIDDLE  ALT=\"arcsin%2810%2F12%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > in degrees.  Where <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=alpha ALIGN=MIDDLE  ALT=\"alpha\"   DISABLED_style= \"max-width:90%; height:auto;\" > is the central angle and r is the radius. </SPAN>\n" );
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