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: Set up the coordinate system.
Let the square be centered at the origin. Since the side length is \(2\), the square is
\[ -1\leq x\leq 1,\qquad -1\leq y\leq 1 \] The centre is
\[ (0,0) \] and the four corners are
\[ (1,1),\;(1,-1),\;(-1,1),\;(-1,-1) \]

Step 2: Use distance comparison.
A point \((x,y)\) is closer to the centre than to the corner \((1,1)\) if
\[ x^2+y^2 < (x-1)^2+(y-1)^2 \] Simplifying,
\[ x+y<1 \] Similarly, comparing with the other three corners gives
\[ x-y<1,\qquad -x+y<1,\qquad -x-y<1 \] Together, these inequalities form the region
\[ |x|+|y|<1 \] This is a diamond-shaped region inside the square.

Step 3: Find the required area.
The diamond has diagonal lengths \(2\) and \(2\).
So, its area is
\[ \frac12 \times 2 \times 2=2 \] The area of the square is
\[ 2\times 2=4 \] Therefore, the required probability is
\[ \frac{2}{4}=\frac12=0.50 \]

Step 4: Final conclusion.
Hence, the required probability is
\[ \boxed{0.50} \]