The three sides of a trapezium are equal, each being 6 cm long, find the area of the trapezium when it is maximum.

Answer

To find the maximum area of a trapezium with three equal sides of length 6 cm, we can follow these steps:

### Step 1: Understand the trapezium configuration
Given that three sides of the trapezium are equal (each measuring 6 cm), we can denote the lengths of the sides as follows:
- Let the lengths of the two non-parallel sides be \( a = 6 \) cm each.
- Let the length of the parallel side (the base) be \( b \).

### Step 2: Express the area of the trapezium
The area \( A \) of a trapezium can be expressed as:
\[ A = \frac{1}{2} \times (b_1 + b_2) \times h \]
where \( b_1 \) and \( b_2 \) are the lengths of the two parallel sides and \( h \) is the height. In our case, we can consider one of the bases as \( b \) and the other base as \( 6 \) cm (the length of the equal sides).

### Step 3: Relate the height to the sides
To find the height \( h \), we can use the Pythagorean theorem. The height can be expressed in terms of \( x \), where \( x \) is the distance from the top vertex to the foot of the height:
\[ h = \sqrt{6^2 - x^2} = \sqrt{36 - x^2} \]

### Step 4: Set up the area function
The area can now be expressed in terms of \( x \):
\[ A = \frac{1}{2} \times (b + 6) \times h \]
Substituting \( h \):
\[ A = \frac{1}{2} \times (b + 6) \times \sqrt{36 - x^2} \]

### Step 5: Find the maximum area
To maximize the area, we can differentiate \( A \) with respect to \( x \) and set the derivative to zero:
1. Differentiate \( A \) with respect to \( x \):
\[ \frac{dA}{dx} = \frac{1}{2} \left( (b + 6) \cdot \frac{d}{dx}(\sqrt{36 - x^2}) \right) \]
Using the chain rule:
\[ \frac{d}{dx}(\sqrt{36 - x^2}) = \frac{-x}{\sqrt{36 - x^2}} \]
Thus,
\[ \frac{dA}{dx} = \frac{1}{2} (b + 6) \cdot \frac{-x}{\sqrt{36 - x^2}} \]

2. Set \( \frac{dA}{dx} = 0 \):
This implies:
\[ x = 0 \quad \text{or} \quad b + 6 = 0 \quad \text{(not possible)} \]

### Step 6: Evaluate the area at critical points
When \( x = 0 \):
\[ h = \sqrt{36 - 0^2} = 6 \]
The area becomes:
\[ A = \frac{1}{2} \times (6 + 6) \times 6 = \frac{1}{2} \times 12 \times 6 = 36 \text{ cm}^2 \]

### Step 7: Conclusion
The maximum area of the trapezium occurs when \( x = 0 \) and is given by:
\[ \text{Maximum Area} = 36 \text{ cm}^2 \]