A rectangular sheet of paper 40cm x 22cm, is rolled to form a hollow cylinder ofheight 40cm. The radius of the cylinder(in cm) is

A rectangular sheet of paper 44 cm × 20 cm is rolled along its length and a cylinder isformed. Find the total surface area of cylinder thus formed.

A rectangular sheet of metal with length 220 m and width 30 m is rolled along its length toform a cylinder. Find the volume of the cylinder thus formed.

A paper sheet is of rectangular shape of 24 cm by 13 cm from each of its corners a square of 4 cm is cut off. An open box is made of the remaining sheet, what is the volume of the box.

How much toilet paper is left on the roll? The volume of a cylinder with length ℓℓ and radius r𝑟 is given by the formula:πℓr2𝜋ℓ𝑟2  You've got some time on your hands, so you carefully measure your toilet roll and discover its dimensions to the nearest mm:length ℓ=100ℓ=100 mm,radius of the inner cardboard cylinder r=30𝑟=30 mm, andradius of the outer cylinder R=51𝑅=51 mm.              The volume of toilet paper on the roll is thus given by the formula:V(ℓ,r,R)=πℓ(R2−r2)𝑉(ℓ,𝑟,𝑅)=𝜋ℓ(𝑅2−𝑟2) .  According to your measurements, the volume of toilet paper is (to the nearest cubic mm):V(100,30,51)=𝑉(100,30,51)= mm3.However, since your initial measurements were only accurate to the nearest mm, the true volume of toilet paper may be different. The total differential approximation can be used to estimate how the measured volume V(100,30,51)𝑉(100,30,51) differs from the true value of V(ℓ,r,R)𝑉(ℓ,𝑟,𝑅) as: V(ℓ,r,R)≈V(100,30,51)+∂V∂ℓ(100,30,51)(ℓ−100)+∂V∂r(100,30,51)(r−30)+∂V∂R(100,30,51)(R−51)𝑉(ℓ,𝑟,𝑅)≈𝑉(100,30,51)+∂𝑉∂ℓ(100,30,51)(ℓ−100)+∂𝑉∂𝑟(100,30,51)(𝑟−30)+∂𝑉∂𝑅(100,30,51)(𝑅−51) . We calculate the partial derivatives (to the nearest integer):∂V∂ℓ(100,30,51)=∂𝑉∂ℓ(100,30,51)= ∂V∂r(100,30,51)=∂𝑉∂𝑟(100,30,51)= ∂V∂R(100,30,51)=∂𝑉∂𝑅(100,30,51)= .Since we measured ℓ,rℓ,𝑟 and R𝑅 to the nearest mm,|ℓ−100|<0.5,|r−30|<0.5|ℓ−100|<0.5,|𝑟−30|<0.5 and |R−51|<0.5|𝑅−51|<0.5 .Using the integer approximations above together with the triangle inequality, |V(ℓ,r,R)−V(100,30,51)|≤∣∣∂V∂ℓ(100,30,51)∣∣|(ℓ−100)|+∣∣∂V∂r(100,30,51)∣∣|(r−30)|+∣∣∂V∂R(100,30,51)∣∣|(R−51)||𝑉(ℓ,𝑟,𝑅)−𝑉(100,30,51)|≤|∂𝑉∂ℓ(100,30,51)||(ℓ−100)|+|∂𝑉∂𝑟(100,30,51)||(𝑟−30)|+|∂𝑉∂𝑅(100,30,51)||(𝑅−51)| . And so our measured volume  V(100,30,51)𝑉(100,30,51) differs from the true volume by|V(ℓ,r,R)−V(100,30,51)|≤|𝑉(ℓ,𝑟,𝑅)−𝑉(100,30,51)|≤ mm3.To two decimal places, this represents an approximate error of no more than|V(ℓ,r,R)−V(100,30,51)|V(100,30,51)≤5.26%|𝑉(ℓ,𝑟,𝑅)−𝑉(100,30,51)|𝑉(100,30,51)≤5.26% .

A square piece of paper is folded in half tomake a rectangle. The perimeter of therectangle is 24.9 cm. What is the side lengthof the square piece of paper?