From a triangle ABC with sides of lengths 40 ft, 25 ft and 35 ft, a triangular portion GBC is cut off where G is the centroid of ABC. The area, in sq ft, of the remaining portion of triangle ABC is

The correct option is (B).

First, let's find the area of the entire triangle $ABC$ using Heron's formula.
The sides of the triangle are $a=40$ ft, $b=25$ ft, and $c=35$ ft.
The semi-perimeter $s$ is:
$s=\frac{40+25+35}{2}=50$ ft.

Now, we compute the area of $△ABC$:
$\text{Area}(△ABC)=\sqrt{s(s-a)(s-b)(s-c)}$
$\text{Area}(△ABC)=\sqrt{50(50-40)(50-25)(50-35)}$
$\text{Area}(△ABC)=\sqrt{50\times 10\times 25\times 15}=\sqrt{187500}=250\sqrt{3}$ sq ft.

We are given that $G$ is the centroid of $△ABC$.
A well-known property of the centroid is that it divides the triangle into three triangles of equal area, namely $△GAB$, $△GBC$, and $△GCA$.
Therefore, the area of the cut-off portion $△GBC$ is:
$\text{Area}(△GBC)=\frac{1}{3}\times \text{Area}(△ABC)$

The area of the remaining portion is:
$\text{Remaining Area}=\text{Area}(△ABC)-\text{Area}(△GBC)=\frac{2}{3}\times \text{Area}(△ABC)$
$\text{Remaining Area}=\frac{2}{3}\times 250\sqrt{3}=\frac{500}{\sqrt{3}}$ sq ft.