Question

A point is chosen at random from a square region whose sides are of \(2\) metres. If the centre of the square is defined to be the point of intersection of its two diagonals, then the probability that the randomly chosen point is closer to the centre of the square than to any of its four corners equals (rounded off to two decimal places).

Hint:

When a point must be closer to one fixed point than another, compare squared distances to avoid square roots.

Answer 1

Correct Answer: 0.5

Step 1: Place the square.
Centre it at the origin so it covers $-1\le x\le1$, $-1\le y\le1$, with corners at $(\pm1,\pm1)$.

Step 2: Compare squared distances.
Being closer to the centre than to $(1,1)$ means $x^2+y^2<(x-1)^2+(y-1)^2$, which simplifies to $x+y<1$.

Step 3: Combine all four corners.
The four corner conditions together give $|x|+|y|<1$, a diamond inside the square.

Step 4: Take the area ratio.
The diamond has area $\frac12\cdot2\cdot2=2$ and the square has area $4$, so the probability is $\frac24=0.5$.

Step 5: Conclude.
\[ \boxed{0.50} \]