Christmas party with rabbits

The pattern is the triangular numbers, think about lattice points on a graph,.

What we're effectively doing by removing people from the boundaries is shortening the sidelength by 2. Because the people will be in the shape of an isosceles triangle in the bigger isoceles triangle.

We have the formula for triangular numbers to be $\dfrac{n(n+1)}{2}$

Thus after shortening the sidelengths we have $\dfrac{(n-1)(n-2)}{2}$

Solving

$210=\dfrac{(n-1)(n-2)}{2}$

$0=n^2-3n-418$

$0=(n-22)(n+19)$

Since n must be positive, -19 is extraneous.

Thus we have the sidelength of the triangle to be 22 so the area is $22^2/2=\boxed{242}$

Imagine a square with side equals the length of the side of the island. For the sake of convenience the length will be measured by number of integer points The number of integer point inside that square will be (n-2)^2, notice that the hypotenuse takes up n-2 points and divide the square into two parts with equal number of points Thus we have the equation (n-2)^2-(n-2) = 210, solving for n taken into account that n is positive yields n = 23 Thus the side length of the triangle is 22, so S = 22*22/2 = 242

Here is my somewhat complicated approach

Let L be the length of the perpendicular sides of that triangle.

Now as  we can see the sum of x and y coordinates of the given situation should add up being less than or equal to L-2

$x+y≤ L-2$

By removing the equality we get

$x+y< L-3$

Now adding a fake variable $λ$

to remove the sign of inequlity

It gives

$x+y+λ = L-3$

This implies total number of combination of x and y becomes $\dbinom{L-3+2}{2}\ = 210$

From here we get $L =22$

This implies area = $22×22/2=242$

210 is 20th triangle number, so triangle muddy be 22 long.

As an aside, the wording of the question is weird. Rather than adding the note at the end, include in the question that guests can't stand on the beach.