**Question 16:**

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

- $225\sqrt{3}$
- $\frac{500}{\sqrt{3}}$
- $\frac{275}{\sqrt{3}}$
- $\frac{250}{\sqrt{3}}$

**Explanation:**

The lengths are given as 40, 25 and 35.

The perimeter = 100

Semi-perimeter, s = 50

Area = $ \sqrt{50 × 10 × 25 × 15}$ = $250\sqrt{3}$

The triangle formed by the centroid and two vertices is removed.

Since the cenroid divides the median in the ratio 2 : 1

The remaining area will be two-thirds the area of the original triangle.

Remaining area = $\frac{2}{3}× 250\sqrt{3}$ = $\frac{500}{\sqrt{3}}$

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