Question

From an interior point of an equilateral triangle, perpendiculars are drawn on all three sides. The sum of the lengths of the three perpendiculars is s. Then the area of the triangle is

• A. \(\frac{\sqrt{3}s^2}{2}\)
• B. \(\frac{s^2}{\sqrt{3}}\)
• C. \(\frac{2s^2}{\sqrt3}\)
• D. \(\frac{s^2}{2\sqrt3}\)

Answer 1

The Correct Option is B

Let \( D, E, F \) denote the feet of perpendiculars from point \( P \) within equilateral triangle \( \triangle ABC \) to sides \( BC, CA, AB \), respectively.

Let the sum of these perpendiculars be \( s \): \[ PD + PE + PF = s \]

Step 1: Express Area as a sum of smaller triangle areas

The area of \( \triangle ABC \) can be expressed as the sum of the areas of \( \triangle PBC, \triangle PCA, \) and \( \triangle PAB \): \[ \text{Area} = \frac{1}{2} \cdot BC \cdot PD + \frac{1}{2} \cdot AC \cdot PE + \frac{1}{2} \cdot AB \cdot PF \] Given that \( AB = BC = AC \), this simplifies to: \[ \text{Area} = \frac{1}{2} \cdot AB \cdot (PD + PE + PF) = \frac{1}{2} \cdot AB \cdot s \tag{1} \]

Step 2: Apply the standard area formula for an equilateral triangle

The area of an equilateral triangle with side length \( AB \) is: \[ \text{Area} = \frac{\sqrt{3}}{4} \cdot AB^2 \] Equating this with equation (1): \[ \frac{\sqrt{3}}{4} AB^2 = \frac{1}{2} AB \cdot s \] Dividing both sides by \( AB \) (since \( AB eq 0 \)): \[ \frac{\sqrt{3}}{4} AB = \frac{1}{2} s \Rightarrow AB = \frac{2s}{\sqrt{3}} \]

Step 3: Substitute the derived side length into equation (1)

Substitute the expression for \( AB \) back into equation (1): \[ \text{Area} = \frac{1}{2} \cdot \left( \frac{2s}{\sqrt{3}} \right) \cdot s = \frac{s^2}{\sqrt{3}} \]

Final Answer:

✅ The area of the equilateral triangle, in terms of the sum \( s \) of the perpendiculars from an interior point to its sides, is: \[ \boxed{\frac{s^2}{\sqrt{3}}} \] Thus, the correct option is: (B)