Question

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

• A. \(225\sqrt{3}\)
• B. \(\frac{500}{\sqrt{3}}\)
• C. \(\frac{275}{\sqrt{3}}\)
• D. \(\frac{250}{\sqrt{3}}\)

Answer 1

The Correct Option is B

Step 1: Formula for area using Heron’s formula

For a triangle with sides \( a, b, c \) and semi‑perimeter \( s \): \[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \]

Step 2: Given dimensions

Triangle \( ABC \) has: \[ a = 40, \quad b = 35, \quad c = 25 \] Semi‑perimeter: \[ s = \frac{40 + 35 + 25}{2} = 50 \]

Step 3: Area of triangle ABC

Substitute into Heron’s formula: \[ \text{Area}_{ABC} = \sqrt{50 \times (50 - 40) \times (50 - 35) \times (50 - 25)} \] \[ = \sqrt{50 \times 10 \times 15 \times 25} \] \[ = \sqrt{187,500} = 250\sqrt{3} \]

Step 4: Using centroid property

The centroid divides each median in a \( 2 : 1 \) ratio. This means the area of \( \triangle GBC \) is: \[ \frac{1}{3} \times \text{Area}_{ABC} \] However, the required portion here is \( \frac{2}{3} \) of \( \triangle ABC \): \[ \text{Required Area} = \frac{2}{3} \times 250\sqrt{3} \] \[ = \frac{500}{\sqrt{3}} \]

Final Answer:

\[ \boxed{\frac{500}{\sqrt{3}}} \]