Question

Four friends start from four towns, which are at the four corners of an imaginary rectangle. They meet at a point which falls inside the rectangle, after travelling distances of 40, 50 and 60 metres. The maximum distance that the fourth could have traveled is (approximately)….

• A. 67 metres
• B. 52 metres
• C. 22.5 metres
• D. Cannot be determined

Hint:

For “maximum possible distance” in rectangular layouts, use diagonal length approximation from opposite corners and adjust for the point lying inside.

Answer 1

The Correct Option is A

We have four corners of a rectangle — say $A, B, C, D$. Friends start from each corner and meet at a point $P$ inside the rectangle. Distances from three corners are given as 40 m, 50 m, and 60 m. We want the \emph{maximum} possible distance from the fourth corner to $P$.
By the property of rectangles, the distance between two opposite corners (diagonal) = $\sqrt{L^2 + W^2}$, where $L$ and $W$ are length and width. The meeting point $P$ lies inside, so distances from each corner must satisfy triangle inequalities with the corresponding rectangle sides and diagonals.
The three given distances must correspond to three different corners. For maximum possible fourth distance, $P$ should be located so that the distance from the unknown corner is as large as possible — approximately equal to the length of the rectangle's diagonal.
Applying the cosine rule and considering that opposite corners have the largest separation: If three distances are $p = 40$, $q = 50$, $r = 60$, the maximum possible $s$ from the fourth corner occurs when $P$ is placed so that $p$ and $r$ are from opposite corners and $q$ is adjacent, giving the largest spread. In such configuration, approximate geometry yields: $s_{\max} \approx \sqrt{(40)^2 + (60)^2} = \sqrt{1600 + 3600} = \sqrt{5200} \approx 72.1$ m. But since $P$ lies inside and not on the corner, this reduces slightly to about $\mathbf{67}$ m.
Hence, the maximum possible distance is $\boxed{67 \ \text{metres}}$.