Three sides of a trapezium are each equal to 6 cm. Let `alpha in (0,pi/2)` be tha angle between a pair of adjacent sides.
If the area of the trapezium is the maximum possible, then what is `alpha` equal to ?

Options

A. `pi/6`
B. `pi/4`
C. `pi/3`
D. `(2pi)/(5)`

Explanation

To solve the problem, we need to find the angle \( \alpha \) that maximizes the area of a trapezium with three sides equal to 6 cm. Let's go through the solution step by step.

### Step 1: Understanding the trapezium
We have a trapezium \( ABCD \) where \( AB \) is parallel to \( CD \). The lengths of sides \( AB \), \( AD \), and \( BC \) are given as 6 cm. Let \( \alpha \) be the angle between sides \( AD \) and \( AB \).

### Step 2: Setting up the problem
Let \( DE \) be the height from point \( D \) to line \( AB \). We denote the length \( DE \) as \( h \) and \( DE = x \). The length of the base \( CD \) can be expressed in terms of \( x \) as \( CD = 6 + 2x \).

### Step 3: Finding the height
Using the right triangle \( ADE \):
\[ AD^2 = AE^2 + DE^2 \]
Substituting the known values:
\[ 6^2 = h^2 + x^2 \]
Thus,
\[ h = \sqrt{36 - x^2} \]

### Step 4: 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 \( AB = 6 \), \( CD = 6 + 2x \), and \( h = \sqrt{36 - x^2} \):
\[ A = \frac{1}{2} \times (6 + (6 + 2x)) \times \sqrt{36 - x^2} \]
\[ A = \frac{1}{2} \times (12 + 2x) \times \sqrt{36 - x^2} \]
\[ A = (6 + x) \sqrt{36 - x^2} \]

### Step 5: Maximizing the area
To find the maximum area, we need to take the derivative of \( A \) with respect to \( x \) and set it to zero:
\[ \frac{dA}{dx} = \sqrt{36 - x^2} + (6 + x) \cdot \frac{-x}{\sqrt{36 - x^2}} = 0 \]
Multiplying through by \( \sqrt{36 - x^2} \) to eliminate the denominator:
\[ (36 - x^2) + (6 + x)(-x) = 0 \]
\[ 36 - x^2 - 6x - x^2 = 0 \]
\[ -2x^2 - 6x + 36 = 0 \]
Dividing by -2:
\[ x^2 + 3x - 18 = 0 \]
Factoring:
\[ (x + 6)(x - 3) = 0 \]
Thus, \( x = 3 \) (since \( x = -6 \) is not valid).

### Step 6: Finding \( \alpha \)
Now, we need to find \( \alpha \). In triangle \( ADE \):
\[ \cos \alpha = \frac{DE}{AD} = \frac{x}{6} = \frac{3}{6} = \frac{1}{2} \]
Thus,
\[ \alpha = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \]

### Final Answer
The angle \( \alpha \) that maximizes the area of the trapezium is:
\[ \alpha = \frac{\pi}{3} \]