Question

Consider a string P of length \(l\) that is laid out as a straight line segment. Another string K is laid out as a semicircular arc with string P as its diameter, as represented in Figure (i). When both the strings are shortened by a length \(x\) they can be re-arranged such that the shortened string K forms a full circle with the shortened string P as its diameter, as represented in Figure (ii).

The value of \(x/l\) is ____________

• A. \(\pi\)
• B. \(\dfrac{\pi - 1}{2\pi}\)
• C. \(\dfrac{\pi}{2(\pi - 1)}\)
• D. \(\dfrac{\pi}{\pi - 1}\)

Hint:

Write string K's shortened length as the circumference of the new circle.

Answer 1

The Correct Option is C

Step 1: Use the radius of the semicircle instead of the diameter.
Let $r = l/2$ be the radius of string K's semicircular arc in Figure (i), so its arc length is $\pi r$.
After shortening by $x$, string P becomes the new diameter, so the new radius is $r' = \dfrac{l - x}{2} = r - \dfrac{x}{2}$.

Step 2: Write string K's new length as a full circle.
The shortened string K now forms a complete circle of radius $r'$, so its length is $2\pi r' = 2\pi \left( r - \dfrac{x}{2} \right) = 2\pi r - \pi x$.
This length also equals the original arc length minus $x$, since K itself shortened by $x$: $\pi r - x$.

Step 3: Equate the two expressions and solve.
Setting $2\pi r - \pi x = \pi r - x$ gives $\pi r = \pi x - x = x(\pi - 1)$, so $x = \dfrac{\pi r}{\pi - 1}$.
Since $r = l/2$, this gives $x = \dfrac{\pi l}{2(\pi - 1)}$, so $\dfrac{x}{l} = \dfrac{\pi}{2(\pi - 1)}$.

Final Answer:
Both routes agree on the same ratio for x over l.\[ \boxed{\dfrac{x}{l} = \dfrac{\pi}{2(\pi - 1)}} \]