The three sides of a trapezium are equal each being 6 cms long. Let `Delta cm^(2)` be the maximum area of the trapezium. The value of `sqrt(3) Delta` is :

Answer

To solve the problem of finding the maximum area of a trapezium with three equal sides of 6 cm each, we can follow these steps:

### Step 1: Understand the trapezium configuration
Given that three sides of the trapezium are equal, we can denote the trapezium as ABCD where AB is the base, CD is the top side, and AD = BC = 6 cm. Let the length of CD be x cm.

### Step 2: Determine the height of the trapezium
To find the height (h) of the trapezium, we can use the Pythagorean theorem. The height can be expressed in terms of x:
- The length of the segment from A to the foot of the height (let's call it E) is (6 - x) / 2, as AE + EB = AB = 6 cm.
- Therefore, using the Pythagorean theorem:
\[ h = \sqrt{6^2 - \left(\frac{6 - x}{2}\right)^2} \]
\[ h = \sqrt{36 - \left(\frac{6 - x}{2}\right)^2} \]

### Step 3: Express the area of the trapezium
The area (A) of the trapezium can be calculated using the formula:
\[ A = \frac{1}{2} \times (AB + CD) \times h \]
Substituting the values:
\[ A = \frac{1}{2} \times (6 + x) \times h \]
Substituting h from the previous step:
\[ A = \frac{1}{2} \times (6 + x) \times \sqrt{36 - \left(\frac{6 - x}{2}\right)^2} \]

### Step 4: Differentiate the area with respect to x
To find the maximum area, we differentiate A with respect to x and set the derivative to zero:
\[ \frac{dA}{dx} = 0 \]
This will involve using the product and chain rule of differentiation.

### Step 5: Solve for x
After differentiating, we will solve the resulting equation for x. This will give us the value of x that maximizes the area.

### Step 6: Calculate the maximum area
Substituting the value of x back into the area formula will give us the maximum area, denoted as Δ.

### Step 7: Find the value of \(\sqrt{3} \Delta\)
Finally, we multiply the maximum area Δ by \(\sqrt{3}\) to find the required value.

### Final Calculation
After performing the calculations:
1. We find that the maximum area Δ is \(27\sqrt{3}\) cm².
2. Therefore, \(\sqrt{3} \Delta = 27\sqrt{3} \times \sqrt{3} = 81\).

Thus, the final answer is:
\[ \sqrt{3} \Delta = 81 \]