Let a point P lies inside an equilateral `triangle ABC` such that its perpendicular distances from sides are `P_(1),P_(2),P_(3)`.If side length of `triangle ABC` is 2 unit then

Options

A. `(P_(1)+P_(2)+P_(3))` is equal to `sqrt(3)`
B. `(P_(1)+P_(2)+P_(3))` is equal to `4sqrt(3)`
C. Area of `triangle=sqrt(3)` Sq unit
D. Area of `triangle ABC =3` Sq unit

Answer

To solve the problem, we need to find the sum of the perpendicular distances from point P inside an equilateral triangle ABC to its sides, given that the side length of the triangle is 2 units.

### Step-by-Step Solution:

1. **Understand the Geometry**:
- We have an equilateral triangle ABC with side length \( a = 2 \) units.
- Let \( P \) be a point inside the triangle, and let \( P_1, P_2, P_3 \) be the perpendicular distances from point \( P \) to sides \( BC, CA, \) and \( AB \) respectively.

2. **Calculate the Area of Triangle ABC**:
- The area \( A \) of an equilateral triangle can be calculated using the formula:
\[ A = \frac{\sqrt{3}}{4} a^2 \]
- Substituting \( a = 2 \):
\[ A = \frac{\sqrt{3}}{4} \times 2^2 = \frac{\sqrt{3}}{4} \times 4 = \sqrt{3} \]

3. **Express the Area in Terms of Distances**:
- The area of triangle ABC can also be expressed as the sum of the areas of the three smaller triangles formed by point \( P \):
\[ A = \text{Area}(APB) + \text{Area}(BPC) + \text{Area}(CPA) \]
- The area of each triangle can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
- For triangle \( APB \):
\[ \text{Area}(APB) = \frac{1}{2} \times AB \times P_1 = \frac{1}{2} \times 2 \times P_1 = P_1 \]
- For triangle \( BPC \):
\[ \text{Area}(BPC) = \frac{1}{2} \times BC \times P_2 = \frac{1}{2} \times 2 \times P_2 = P_2 \]
- For triangle \( CPA \):
\[ \text{Area}(CPA) = \frac{1}{2} \times CA \times P_3 = \frac{1}{2} \times 2 \times P_3 = P_3 \]

4. **Combine the Areas**:
- Thus, we can write:
\[ A = P_1 + P_2 + P_3 \]
- Since we already found \( A = \sqrt{3} \), we have:
\[ P_1 + P_2 + P_3 = \sqrt{3} \]

5. **Conclusion**:
- The sum of the perpendicular distances from point \( P \) to the sides of triangle ABC is:
\[ P_1 + P_2 + P_3 = \sqrt{3} \]

### Final Answers:
- The area of triangle ABC is \( \sqrt{3} \).
- The sum of the perpendicular distances \( P_1 + P_2 + P_3 \) is \( \sqrt{3} \).