document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=36023>Question 36023</A>This question is from textbook <A HREF=http://www.amazon.com/exec/obidos/ASIN/0395977274/algebrahomeworkh>geometry</A><BR>\n" );
document.write( ": <I><SPAN STYLE=\"word-break: break-all;\"> A sphere is inscribed in a cone with radius 6 and height 8. Find the volume of the sphere. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #22059 by </B><A HREF=/tutors/aboutme.mpl?userid=venugopalramana><B>venugopalramana(3286)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=venugopalramana><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.geocities.com/venugopalramana/FREE_MATHS_ASSISTANCE.html valign=middle><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=22059\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">         A sphere is inscribed in a cone with radius 6 and height 8. Find the volume of the sphere.<BR>\n" );
document.write( "DRAW A SECTION...TRIANGLE ABC REPRESENTS THE SECTION OF THE CONE,WITH A AS VERTEX AND BC AS BASE.HENCE BC =6+6=12<BR>\n" );
document.write( "DRAW AD PERPENDICULAR FROM A TO BC.AD IS HEIGHT OF CONE =8<BR>\n" );
document.write( "DRAW A CIRCLE WITH CENTRE AT O ON AD \r<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=drawing%28600%2C600%2C-5%2C15%2C-5%2C15%2Cgreen%28+line%28+0%2C0%2C+12%2C0+%29+%29%2Cgreen%28line%286%2C0%2C6%2C8%29%29%2Clocate%28+0%2C+0%2C+B+%29%2Clocate%286%2C0%2CD%29%2Cgreen%28+line%28+12%2C0%2C6%2C8+%29+%29%2Clocate%286%2C3%2CO%29%2Clocate%286%2C8%2CA%29%2Clocate%288.5%2C4.5%2CE%29%2Clocate%283.5%2C4.5%2CF%29%2Cgreen%28line%286%2C3%2C8.5%2C4.5%29%29%2Clocate%2812%2C0%2CC%29%2Ccircle%286%2C3%2C3%29%2Cgreen%28line%280%2C0%2C6%2C8%29%29%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "\n" );
document.write( "AS SECTION OF SPHERE TOUCHING BC,CA,AB AT D,E,F.<BR>\n" );
document.write( "HENCE OD=OE=OF=RADIUS OF SPHERE=R SAY<BR>\n" );
document.write( "NOW TRIANGLES ADC,AND AEO ARE RIGHT ANGLED<BR>\n" );
document.write( "ANGLE ADC=90=ANGLE AEO...AS AEC IS TANGENT..SPHERE TOUCHING CONE.<BR>\n" );
document.write( "ANGLE DAC=ANGLE OAE=SAME ANGLE<BR>\n" );
document.write( "HENCE THE 2 TRIANGLES ADC AND AEO ARE SIMILAR.<BR>\n" );
document.write( "   HENCE DC/EO=AD/AE<BR>\n" );
document.write( "DC=6.....AD=8......EO=R=RADIUS OF SPHERE.....AE=SQRT(AO^2-OE^2)<BR>\n" );
document.write( "=SQRT{(AD-OD)^2-R^2}<BR>\n" );
document.write( "=SQRT{(8-R)^2-R^2}=SQRT{64+R^2-16R-R^2}<BR>\n" );
document.write( "=SQRT(64-16R)<BR>\n" );
document.write( "HENCE WE HAVE <BR>\n" );
document.write( "6/R=8/SQRT(64-16R)<BR>\n" );
document.write( "6SQRT(64-16R)=8R<BR>\n" );
document.write( "6*4SQRT(4-R)=8R<BR>\n" );
document.write( "3SQRT(4-R)=R....SQUARING....<BR>\n" );
document.write( "9(4-R)=R^2<BR>\n" );
document.write( "R^2+9R-36=0<BR>\n" );
document.write( "(R-3)(R+12)=0<BR>\n" );
document.write( "R=3 </SPAN>\n" );
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