Let the lengths of the altitudes drawn from the vertices of `Delta ABC` to the opposite sides are 2, 2 and 3. If the area of `Delta ABC " is " Delta`, then find the area of triangle

Explanation

To solve the problem, we need to find the area of triangle ABC given the lengths of the altitudes from its vertices to the opposite sides. The altitudes are given as follows:

- \( h_a = 2 \) (altitude from vertex A to side BC)
- \( h_b = 2 \) (altitude from vertex B to side AC)
- \( h_c = 3 \) (altitude from vertex C to side AB)

Let the area of triangle ABC be denoted as \( \Delta \).

### Step 1: Express the area in terms of the sides and altitudes

The area \( \Delta \) of triangle ABC can be expressed using the formula for the area in terms of the base and height:

1. Using \( BC \) as the base:
\[ \Delta = \frac{1}{2} \times a \times h_a = \frac{1}{2} \times a \times 2 = a \]
(Equation 1)

2. Using \( AC \) as the base:
\[ \Delta = \frac{1}{2} \times b \times h_b = \frac{1}{2} \times b \times 2 = b \]
(Equation 2)

3. Using \( AB \) as the base:
\[ \Delta = \frac{1}{2} \times c \times h_c = \frac{1}{2} \times c \times 3 = \frac{3}{2}c \]
(Equation 3)

### Step 2: Relate the sides to the area

From the equations derived, we can express the sides \( a, b, \) and \( c \) in terms of the area \( \Delta \):

- From Equation 1: \( a = \Delta \)
- From Equation 2: \( b = \Delta \)
- From Equation 3: \( c = \frac{2}{3} \Delta \)

### Step 3: Calculate the semi-perimeter

The semi-perimeter \( s \) of triangle ABC is given by:
\[ s = \frac{a + b + c}{2} = \frac{\Delta + \Delta + \frac{2}{3} \Delta}{2} = \frac{2\Delta + \frac{2}{3} \Delta}{2} = \frac{\frac{6}{3}\Delta + \frac{2}{3}\Delta}{2} = \frac{\frac{8}{3}\Delta}{2} = \frac{4}{3}\Delta \]

### Step 4: Apply Heron's formula

According to Heron's formula, the area of the triangle can also be expressed as:
\[ \Delta = \sqrt{s(s-a)(s-b)(s-c)} \]

Substituting the values we found:
\[ \Delta = \sqrt{\left(\frac{4}{3}\Delta\right)\left(\frac{4}{3}\Delta - \Delta\right)\left(\frac{4}{3}\Delta - \Delta\right)\left(\frac{4}{3}\Delta - \frac{2}{3}\Delta\right)} \]

This simplifies to:
\[ \Delta = \sqrt{\left(\frac{4}{3}\Delta\right)\left(\frac{1}{3}\Delta\right)\left(\frac{1}{3}\Delta\right)\left(\frac{2}{3}\Delta\right)} \]

### Step 5: Simplify the expression

Calculating the terms inside the square root:
\[ \Delta = \sqrt{\frac{4}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} \Delta^4} \]
\[ = \sqrt{\frac{8}{81} \Delta^4} = \frac{2\sqrt{2}}{9} \Delta^2 \]

### Step 6: Solve for \( \Delta \)

Setting the two expressions for \( \Delta \) equal to each other:
\[ \Delta = \frac{2\sqrt{2}}{9} \Delta^2 \]

Rearranging gives:
\[ \Delta^2 - \frac{9}{2\sqrt{2}} \Delta = 0 \]
Factoring out \( \Delta \):
\[ \Delta \left( \Delta - \frac{9}{2\sqrt{2}} \right) = 0 \]

Thus, we have:
\[ \Delta = 0 \quad \text{or} \quad \Delta = \frac{9}{2\sqrt{2}} \]

Since the area cannot be zero, we conclude:
\[ \Delta = \frac{9}{2\sqrt{2}} = \frac{9\sqrt{2}}{4} \]

### Final Answer
The area of triangle ABC is \( \frac{9\sqrt{2}}{4} \).

---