`Delta I_(1)I_(2)I_(3)` is an excentral triangle of an equilateral triangle `Delta ABC` such that `I_(1)I_(2)=4` unit, if `DeltaDEF` is pedal triangle of `DeltaABC`, then `(Ar(Delta I_(1)I_(2)I_(3)))/(Ar (DeltaDEF))=`

Options

A. 16
B. 4
C. 2
D. 1

Answer

To solve the problem, we need to find the ratio of the areas of two triangles: the excentral triangle \( \Delta I_1I_2I_3 \) and the pedal triangle \( \Delta DEF \) of the equilateral triangle \( \Delta ABC \).

### Step-by-Step Solution:

1. **Understanding the Triangles**:
- We have an equilateral triangle \( \Delta ABC \).
- The excentral triangle \( \Delta I_1I_2I_3 \) has a side \( I_1I_2 = 4 \) units.
- The pedal triangle \( \Delta DEF \) is formed by dropping perpendiculars from the vertices of \( \Delta ABC \) to the opposite sides.

2. **Properties of the Excentral Triangle**:
- Since \( \Delta I_1I_2I_3 \) is an equilateral triangle (as it is an excentral triangle of \( \Delta ABC \)), all sides are equal. Thus, \( I_1I_2 = I_2I_3 = I_3I_1 = 4 \) units.

3. **Finding the Area of \( \Delta I_1I_2I_3 \)**:
- The area \( A \) of an equilateral triangle with side length \( s \) is given by the formula:
\[ A = \frac{\sqrt{3}}{4} s^2 \]
- For \( \Delta I_1I_2I_3 \):
\[ A_{I_1I_2I_3} = \frac{\sqrt{3}}{4} (4^2) = \frac{\sqrt{3}}{4} \times 16 = 4\sqrt{3} \]

4. **Finding the Area of \( \Delta DEF \)**:
- The pedal triangle \( \Delta DEF \) is also equilateral. The side length of \( \Delta DEF \) can be determined from the properties of the pedal triangle in relation to \( \Delta ABC \).
- Since \( \Delta ABC \) is equilateral and \( I_1I_2I_3 \) is an excentral triangle, the side length of \( \Delta DEF \) is half of that of \( \Delta I_1I_2I_3 \) because the pedal triangle is inscribed within the original triangle.
- Thus, the side length of \( \Delta DEF \) is \( 1 \) unit.
- Now, we can find the area of \( \Delta DEF \):
\[ A_{DEF} = \frac{\sqrt{3}}{4} (1^2) = \frac{\sqrt{3}}{4} \]

5. **Finding the Ratio of Areas**:
- We need to find the ratio of the areas of \( \Delta I_1I_2I_3 \) to \( \Delta DEF \):
\[ \text{Ratio} = \frac{A_{I_1I_2I_3}}{A_{DEF}} = \frac{4\sqrt{3}}{\frac{\sqrt{3}}{4}} = 4\sqrt{3} \times \frac{4}{\sqrt{3}} = 16 \]

### Final Answer:
\[ \frac{Ar(\Delta I_1I_2I_3)}{Ar(\Delta DEF)} = 16 \]