Question 1108960
The problem suggests that the shape of the flower vase is a right cone frustum.
A cross section though the axis of that cone (with measurements in cm) would look like this:
{{{drawing(500,300,-12.5,12.5,-2,13,
green(triangle(-7,0,-7,12,-12,12)),
green(rectangle(-7,12,-7.5,11.5)),
green(triangle(7,0,7,12,12,12)),
green(rectangle(7,12,7.5,11.5)),
green(arrow(0,5.5,0,16)),
green(arrow(0,5.5,0,-5)),
line(-7,0,7,0),line(-12,12,12,12),
line(-7,0,-12,12),line(7,0,12,12),
locate(3,1,7),locate(3,12,7),locate(9,12,x),
locate(7.1,7,green(h=12)),locate(9.5,6,13),
locate(-6.9,7,green(12)),locate(0.1,7,green(12)),
arrow(7,-1.5,5,0),locate(7.1,-1.1,base),
arrow(2,-1,0,0),locate(2.1,-0.7,center),
locate(2.1,-1.4,of),locate(3.1,-1.4,base)
)}}}
Applying the Pythagorean theorem to the right triangle on the right side,
{{{x^2+12^2=13^2}}}
{{{x^2+144=169}}}
{{{x^2=169-144}}}
{{{x^2=25}}}
{{{x=5}}}
So the radius of the lower base (in cm) is {{{r=7}}} ,
and the radius of the upper base of the cone frustum, in cm, is
{{{R=7+5=12}}} .
If you are supposed to apply the formula
{{{V=(pi/3)h(R^2+r^2+Rr)}}} to find the volume of the frustum,
the volume of the vase, in cubic cm, is
{{{V=(pi/3)12(12^2+7^2+12*7)}}}
{{{V=4pi(144+49+84)}}}
{{{V=4pi(277)}}}
{{{V=1108pi}}}
{{{V=3480.884669(rounded)}}}
Being reasonably accurate, we would say that the flower base can hold
{{{3481cm^3}}} or {{{3481mL}}} of water.
Filling it to the brim, we could still add another
{{{3481mL-500mL=highlight(2981mL)}}} .
 
Without using a memorized formula, or an equivalent one
we would have to calculate the volume of the frustum as
the volume of a "parent" cone minus
the volume of another cone, cut off the parent cone's apex.
We would see that the shape of the vase is a frustum
cut from a "parent" cone with a base radius of 12cm,
and a height (in cm) of
{{{12+12*7/5=12(1+7/5)=12*(12/5)=144/5=28.8}}} .
The volume of such a parent cone (in cubic cm) would be
{{{(pi/3)*(144/5)*12^2=12^4*pi/15}}} ,
and the cone cut off would have a height of
{{{28.8cm-12cm=16.8cm}}} .
That height is {{{16.8/28.8=7/12}}} of the height of the larger cone,
so the cone cut off would have a volume
{{{(7/12)^3=7^3/12^3=343/1728}}} times as large that of the parent cone.
That would make the volume (in cubic cm) of the vase
{{{(1-343/1728)*(12^4*pi/15)=((1728-343)/12^3)(12^4*pi/15)=1385*12*pi/15=277*4*pi=1108pi}}}