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

Let: \[ PD + PE + PF = s \]

Step 1: Area as sum of smaller triangle areas

Using perpendiculars: \[ \text{Area} = \frac{1}{2} \cdot AB \cdot PE + \frac{1}{2} \cdot BC \cdot PD + \frac{1}{2} \cdot AC \cdot PF \] Since \( AB = BC = AC \), we write: \[ \text{Area} = \frac{1}{2} \cdot AB \cdot (PE + PD + PF) = \frac{1}{2} \cdot AB \cdot s \tag{1} \]

Step 2: Use standard area formula of equilateral triangle

\[ \text{Area} = \frac{\sqrt{3}}{4} \cdot AB^2 \] Equating this with equation (1): \[ \frac{\sqrt{3}}{4} AB^2 = \frac{1}{2} AB \cdot s \] Divide both sides by \( AB \): \[ \frac{\sqrt{3}}{4} AB = \frac{1}{2} s \Rightarrow AB = \frac{2s}{\sqrt{3}} \]

Step 3: Substitute into equation (1)

\[ \text{Area} = \frac{1}{2} \cdot \frac{2s}{\sqrt{3}} \cdot s = \frac{s^2}{\sqrt{3}} \]

Final Answer:

✅ The area of the triangle in terms of the sum \( s \) of the perpendiculars is: \[ \boxed{\frac{s^2}{\sqrt{3}}} \] So, the correct option is: (B)

Answer 2

Concept:

From any interior point in an equilateral triangle, if we draw perpendiculars to the three sides, the **sum of the lengths of the perpendiculars is equal to the altitude** of the triangle.

Let the side of the equilateral triangle be \( a \), and the altitude be: \[ s = \frac{\sqrt{3}}{2}a \Rightarrow a = \frac{2s}{\sqrt{3}} \]

Step 1: Use the Area Formula

The standard formula for the area of an equilateral triangle is: \[ \text{Area} = \frac{\sqrt{3}}{4} a^2 \] Substituting \( a = \frac{2s}{\sqrt{3}} \): \[ \text{Area} = \frac{\sqrt{3}}{4} \left( \frac{2s}{\sqrt{3}} \right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{4s^2}{3} = \frac{s^2}{\sqrt{3}} \]

Final Answer:

✅ The area of the triangle in terms of the sum \( s \) of the perpendiculars is: \[ \boxed{\frac{s^2}{\sqrt{3}}} \] (Option B)