document.write( "Question 1206673: A sector of angle 65° is removed to form a thin circular sheet of radius 15cm and folded to form a right circular cone. Calculate the volume of the cone \n" ); document.write( "
Algebra.Com's Answer #844239 by mananth(16946)\"\" \"About 
You can put this solution on YOUR website!
A sector of angle 65° is removed to form a thin circular sheet of radius 15cm and folded to form a right circular cone. Calculate the volume of the cone\r
\n" ); document.write( "\n" ); document.write( "Volumeof cone = 1/3 * pi*r^2*h\r
\n" ); document.write( "\n" ); document.write( "we need to find h the height and radius of base of cone (r)\r
\n" ); document.write( "\n" ); document.write( "Area of sector = ((theta)/360 )* pi*r^2\r
\n" ); document.write( "\n" ); document.write( "A=(65/360)*pi*15^2\r
\n" ); document.write( "\n" ); document.write( "A= 127.627 cm^2\r
\n" ); document.write( "\n" ); document.write( "When folded the area of sector will be the lateral surface area of cone = pi*r*l.
\n" ); document.write( " radius of sector will become the slant height of cone\r
\n" ); document.write( "\n" ); document.write( "pi*r*15= 127.627\r
\n" ); document.write( "\n" ); document.write( "r=127.627/ (15*pi)
\n" ); document.write( "r=2.708 cm\r
\n" ); document.write( "\n" ); document.write( "h= sqrt(l^2-r^2)\r
\n" ); document.write( "\n" ); document.write( "h = sqrt(15^2-2.708)^2)= 14.75\r
\n" ); document.write( "\n" ); document.write( "Volume of cone = (1/3) *pi*(2.708)^2*14.75
\n" ); document.write( "= 113.27 cm^3\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "
\n" );