Question 1109048
{{{120^o/360^o=1/3}}}
The sector is {{{1/3}}} of the circle,
so its arc is {{{1/3}}} of the circumference of radius {{{9cm}}} .
When the sector is folded, that arc will be
the circumference of the base of the cone,
so that circumference will be {{{1/3}}} of the circumference of radius {{{9cm}}} ,
and its radius will be {{{1/3}}} of {{{9cm}}} ,
or {{{r[cone]=(1/3)(9cm)=3cm}}} .
The slant height of the cone formed is the {{{9cm}}} radius of the sector.
A cross section of the cone passing through apex and center of the base will look like this
{{{drawing(320,400,-4,4,-1,9,
green(triangle(0,0,3,0,0,8.485)),
green(rectangle(0,0,0.3,0.3)),
triangle(-3,0,3,0,0,8.485),
locate(1,0,3cm),locate(-2,0,3cm),
locate(0.1,3,h),locate(1.5,4.8,9cm)
)}}} As per Pythagorean theorem, the exact value of the height is
{{{h=sqrt((9cm)^2-(3cm)^2)}}}{{{"="}}}{{{sqrt(72)}}}{{{cm}}}{{{"="}}}{{{6sqrt(2)}}}{{{cm}}} .
An approximate value is {{{8.485cm}}}
Then, the volume is
{{{V[cone]}}}{{{"="}}}{{{(pi/3)h*r^2}}}{{{"="}}}{{{(pi/3)6sqrt(2)3}}}{{{cm^3}}}{{{"="}}}{{{18sqrt(2)*pi}}}{{{cm^3}}} .
That is the exact expression for the volume.
{{{79.97cm^3}}} is an approximate value.