Question 1011125
The room, with the pole, {{{red(AC)}}} ,
and the pole's shadow, {{{blue(AB)}}} , would look like this:
{{{drawing(400,300,-2,18,-2,13,
rectangle(0,0,10,5),
rectangle(7,6,16,10.5),
line(10,5,16,10.5),line(10,0,16,6),
line(0,5,7,10.5),green(line(0,0,7,6)),
green(line(7,10.5,7,6)),green(line(16,6,7,6)),
blue(line(16,6,0,0)),locate(8,3.2,blue(s)),
red(line(0,0,16,10.5)),locate(8.7,6,red(p)),
locate(-0.2,0,A),locate(16,6,B),locate(16,11.5,C),
locate(4.5,0,10),locate(13,3,10),locate(16.1,9,5)
)}}} {{{blue(s)}}}=length of AB, and {{{red(p)}}}=length of AB, both in meters.
On the floor of the room, {{{blue(s)}}} is the diagonal of a 10m×10m square
(or the hypotenuse of a right triangle with legs measuring {{{10m}}} ,
so the Pythagorean theorem tells us that
{{{(blue(s))^2=10^2+10^2=200}}} .
On an imaginary vertical plane, triangle {{{ABC}}} is a right triangle,
and its legs' length, in meters, are
{{{blue(s)}}} and {{{5}}} .
So the length, in meters, of hypotenuse {{{AC}}} is
{{{red(p)=sqrt((blue(s))^2+5^2)=sqrt(200+25)=sqrt(225)=highlight(15)}}} .
That is the length of the longest pole that can be put into a room of dimensions 10m×10m×5m,
provided that the pole is pencil-thin or thinner.
 
If the pole is wider than pencil-thin its ends will not go all the way into corners {{{A}}} and {{{C}}} , =
so it would need to be a bit shorter,
but I do not think that you are expected to do real world calculations,
as an engineer would do,
so the expected answer to the second question is the volume of a pole/cylinder
that is {{{15m}}} long and {{{0.7m=70cm}}} in radius.
For a cylinder of radius {{{R}}} and height {{{L}}} , the volume is
{{{Volume=pi*R^2*L}}} ,
so for your problem, the volume of the pole in cubic meters is
{{{Volume=pi*0.7^2*15=7.35pi}}} .
That is about {{{highlight(23.09m^3)}}} .