Question 742370
NOTE: Since you are not specifying some overlap, I am assuming this is not a real life problem. In classroom problems, we assume that the edges can be somehow glued together without overlap. I will calculate that way.
 
I imagine the factory that makes those rolls cuts those strips of cardboard in large numbers, one next to the other, on a slant, from a long roll of cardboard.
If you tried to cut just one strip it the same way (on a slant) from a rectangle of cardboard, there would be a lot of waste:
{{{drawing(400,200,-1,11,-0.5,5.5,
green(rectangle(0,0,9.5,4.9)),
line(0,0,8.5,4.9),line(1,0,9.5,4.9),
locate(8.7,5,c),locate(4.1,4.9,8.5c),
locate(7,5,30^o),locate(4,2.9,L)
)}}} {{{c=3.8*pi}}} is the circumference of the roll, but it does not matter for the percentage waste. {{{L=8.5c/cos(30^o)}}}
 
Your best bet for cutting just one strip would be to cut it along the length of a very long and thin cardboard rectangle, like this (drawing not to scale):
{{{drawing(400,100,-0.5,11.5,-0.5,2.5,
green(rectangle(0,0,10.8,2)),line(0,2,4,0),
line(6.8,2,10.8,0),locate(1.9,1,c),
locate(8.8,2.1,x),locate(3.4,2,L),locate(0.1,1.3,green(h)),
locate(1.2,2.1,30^o)
)}}} L should have been much, much longer, but I chose to make it not to scale.
The waste in this case would be the two triangles at the ends.
Their total area is {{{x*h}}}.
The area of the green cardboard rectangle used is {{{L+x)*h}}}
The fraction wasted is {{{xh/((L+x)*h)}}}
and we can express all those lengths (L, x, c, and h) as a function of c.
We knew that {{{L=8.5c/cos(30^o)}}}.
We know that {{{cos(30^o)=sqrt(3)/2}}} , but I do not want to mess with that square root unless I have to.
{{{x=c*cos(30^o)}}} and {{{h=c*sin(30^o)=0.5c}}} are the legs of those right triangles with hypotenuse c.
{{{L+x}}}={{{8.5c/cos(30^o)+c*cos(30^o)}}}={{{c(8.5/cos(30^o)+cos(30^o))}}}={{{c*((8.5+cos^2(30^o))/cos(30^o))}}}
Since {{{cos^2(30^o)=(sqrt(3)/2)^2=3/4=0.75}}},
{{{L+x}}}={{{c*((8.5+0.75)/cos(30^o))=9.25c/cos(30^o)}}}
The wasted fraction is
{{{xh/((L+x)*h)=x/(L+x)}}}={{{c*cos(30^o)/((9.25c/cos(30^o)))}}}={{{c*cos^2(30^o)/9.25c}}}={{{c*0.75/9.25c=0.75/9.25=3/37}}}={{{0.081}}}(rounded)
That means that 8.1% of the long and thin cardboard rectangle would be wasted.
I don't think we can reduce the waste any further, but you would need a cardboard rectangle measuring {{{L+x=9.25c/cos(30^o)}}} by {{{h=0.5c}}}.
With {{{c=3.8*pi=11.94cm}}} (rounded), that would be a strip 6cm wide by 127.5cm long.