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 _____

Answer

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 onto the sides BC, CA, and AB, respectively. 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 its vertices. The formula for the perimeter of triangle DEF is:
\[ P_{DEF} = a \cos A + b \cos B + c \cos C \]
where \(a\), \(b\), and \(c\) are the lengths of the sides of triangle ABC opposite to angles A, B, and C respectively.

2. **Using the Sine Rule**:
According to the sine rule:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \]
From this, we can express the sides in terms of the circumradius and the sines of the angles:
\[ a = 2R \sin A, \quad b = 2R \sin B, \quad c = 2R \sin C \]

3. **Substituting into the Perimeter Formula**:
Now, substituting the expressions for \(a\), \(b\), and \(c\) into the perimeter formula:
\[ P_{DEF} = (2R \sin A) \cos A + (2R \sin B) \cos B + (2R \sin C) \cos C \]
This simplifies to:
\[ P_{DEF} = 2R (\sin A \cos A + \sin B \cos B + \sin C \cos C) \]

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

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 the Value of \( \sin A \sin B \sin C \)**:
We can use the area of triangle ABC to find \( \sin A \sin B \sin C \):
\[ \Delta = \frac{abc}{4R} \implies abc = 4R \Delta \]
Given \( R = 3 \) cm and \( \Delta = 6 \) cm²:
\[ abc = 4 \cdot 3 \cdot 6 = 72 \]

7. **Using the Area to Find \( \sin A \sin B \sin C \)**:
The area can also be expressed as:
\[ \Delta = \frac{abc}{4R} \implies \sin A \sin B \sin C = \frac{abc}{4R \cdot 2R} = \frac{abc}{8R^2} \]
Substituting the known values:
\[ \sin A \sin B \sin C = \frac{72}{8 \cdot 9} = 1 \]

8. **Final Calculation of the Perimeter**:
Now substituting back into the perimeter formula:
\[ P_{DEF} = 3 \cdot 4 \cdot 1 = 12 \]

Thus, the perimeter of triangle DEF is **12 cm**.