Four identical cylindrical chalk-sticks, each of radius \( r = 0.5 \, {cm} \) and length \( l = 10 \, {cm} \), are bound tightly together using a duct tape as shown in the following figure.
The width of the duct tape is equal to the length of the chalk-stick. The area (in cm\(^2\)) of the duct tape required to wrap the bundle of chalk-sticks once, is:
Show Hint
When dealing with composite geometric shapes, divide them into simpler parts and apply standard formulas for perimeter, area, or volume systematically.
Step 1: Evaluate the surface to be covered.
The four cylinders are arranged to form a rectangular shape with semicircular ends. The total perimeter to be covered consists of:
\[
{Perimeter} = 2 \times ({length of rectangle}) + 2 \times ({semicircular arcs})
\]
The rectangle's length is calculated as \( 2r + 2r = 4r \), where \( r = 0.5 \) cm, giving:
\[
4(0.5) = 2 \, {cm}
\]
The combined length of the semicircular arcs equals the circumference of a full circle:
\[
2\pi r = 2\pi (0.5) = \pi \, {cm}
\]
Thus, the total perimeter is:
\[
2(2) + \pi = 4 + \pi \, {cm}
\]
Step 2: Compute the required surface area.
Given that the length of the chalk-stick is \( l = 10 \) cm, the area required is determined by:
\[
{Area} = {Perimeter} \times {Length} = (4 + \pi) \times 10 = 10(4 + \pi) \, {cm}^2
\]
Final Answer:
\[
\boxed{{(4) \( 10 (4 + \pi) \)}}
\]
