Circumradius of `DeltaABC` is 3 cm and its area is `6 cm^(2)`. If DEF is the triangle formed by feet of the perpendicular drawn from A,B and C on the sides BC, CA and AB, respectively, then the perimeter of `DeltaDEF` (in cm) is _____

Explanation

To solve the problem, we need to find the perimeter of triangle DEF formed by the feet of the perpendiculars from vertices A, B, and C of triangle ABC to the opposite sides. We are given the circumradius (R) of triangle ABC as 3 cm and its area (Δ) as 6 cm².

### Step-by-step Solution:

1. **Understanding the Relationship**: The perimeter of triangle DEF can be expressed in terms of the sides of triangle ABC and the angles at the vertices. The formula for the perimeter of triangle DEF (the pedal triangle) is given by:
\[ P_{DEF} = a \cos A + b \cos B + c \cos C \]
where \( a, b, c \) are the lengths of the sides of triangle ABC opposite to angles A, B, and C respectively.

2. **Using the Sine Rule**: From the sine rule, we know:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \]
This means:
\[ a = 2R \sin A, \quad b = 2R \sin B, \quad c = 2R \sin C \]

3. **Substituting in the Perimeter Formula**:
Substitute the expressions for \( a, b, c \) into the perimeter formula:
\[ P_{DEF} = (2R \sin A) \cos A + (2R \sin B) \cos B + (2R \sin C) \cos C \]
Factor out \( 2R \):
\[ P_{DEF} = 2R (\sin A \cos A + \sin B \cos B + \sin C \cos C) \]

4. **Using the Double Angle Identity**: We can use the identity \( \sin A \cos A = \frac{1}{2} \sin(2A) \):
\[ P_{DEF} = 2R \left( \frac{1}{2} \sin(2A) + \frac{1}{2} \sin(2B) + \frac{1}{2} \sin(2C) \right) \]
Simplifying gives:
\[ P_{DEF} = R (\sin(2A) + \sin(2B) + \sin(2C)) \]

5. **Using the Property of Sine**: We know that:
\[ \sin(2A) + \sin(2B) + \sin(2C) = 4 \sin A \sin B \sin C \]
Therefore:
\[ P_{DEF} = R \cdot 4 \sin A \sin B \sin C \]

6. **Finding \( \sin A \sin B \sin C \)**: From the area formula, we know:
\[ \Delta = \frac{abc}{4R} \]
Rearranging gives:
\[ abc = 4R \Delta \]
Substituting \( R = 3 \) cm and \( \Delta = 6 \) cm²:
\[ abc = 4 \cdot 3 \cdot 6 = 72 \]

7. **Using the Area to Find \( \sin A \sin B \sin C \)**: We can also express the area in terms of the sides and angles:
\[ \Delta = \frac{abc}{4R} \implies \sin A \sin B \sin C = \frac{abc}{4R} \]
Therefore:
\[ \sin A \sin B \sin C = \frac{72}{4 \cdot 3} = 6 \]

8. **Final Calculation**: Now substituting back into the perimeter formula:
\[ P_{DEF} = R \cdot 4 \cdot 6 = 3 \cdot 24 = 72 \]

Thus, the perimeter of triangle DEF is **4 cm**.