Question

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. \begin{center} \includegraphics[width=6cm]{7.png} \end{center} 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:

• A. \( 20 (4 + \pi) \)
• B. \( 20 (8 + \pi) \)
• C. \( 10 (8 + \pi) \)
• D. \( 10 (4 + \pi) \)

Hint:

When dealing with composite geometric shapes, divide them into simpler parts and apply standard formulas for perimeter, area, or volume systematically.

Answer 1

The Correct Option is D

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) \( 10 (4 + \pi) \)}} \]