Geometry problem#2

Let $A$ be the cross sectional area of the roll of paper. Then $A=\pi(6^2-2^2)=32\pi$ . Since half of the roll of paper was consumed, then the remaining area is $\dfrac{A}{2}$ . Let $x$ be the thickness of the remaining roll of paper, then

$\dfrac{A}{2}=\pi[(x+2)^2-2^2]$

$16\pi=\pi(x^2+4x+4-4)$

$x^2+4x-16=0$

By the quadratic formula, we have

$x=\dfrac{-4\pm \sqrt{16-4(-16)}}{2}=-2\pm 2\sqrt{5}$

$x\approx 2.472$ or $x \approx -6.472$ (reject the negative value)

So the new radius of the cylinder (including the paper) is $x+2=2.472+2=4.472$ . Finally, the diameter is $2(4.472) \approx$ $\boxed{8.944}$

Assuming that the paper has constant thickness and width, the the cross-sectional area of the paper roll is proportional to the length of the paper roll.

A full roll of paper has an internal and external diameters of $4$ cm and $12$ cm respectively. The cross-sectional area of the full roll is:

$A_0 = \pi \left( (\frac {12}{2})^2 - (\frac {4}{2})^2 \right) = \pi (36 - 4) = 32\pi$

Let the external diameter of the roll be $D$ then is given by:

$A_{\frac{1}{2}} = \pi \left( (\frac {D}{2})^2 - 2^2 \right) = \dfrac {32 \pi}{2}\quad \Rightarrow \dfrac {D^2}{4} = 16 + 4 \quad \Rightarrow D = \sqrt{80} = \boxed{8.94}$