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

Options

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

Answer

To solve the problem of finding the angle \( a \) that maximizes the area of a trapezium with three equal sides of 6 cm, we can follow these steps:

### Step 1: Understand the Geometry
We have a trapezium where three sides are equal to 6 cm. Let's denote the trapezium as \( ABCD \) where \( AB \) and \( CD \) are the parallel sides, and \( AD \) and \( BC \) are the non-parallel sides. We can assume \( AB = 6 \) cm, \( AD = 6 \) cm, and \( BC = 6 \) cm. The angle between the sides \( AD \) and \( AB \) is \( a \).

### Step 2: Define the Height and Base
Let \( k \) be the length of the base \( CD \). The height \( h \) of the trapezium can be expressed using the cosine of angle \( a \):
\[ h = 6 \sin(a) \]
The length of the base \( CD \) can be expressed using the cosine of angle \( a \):
\[ k = 6 \cos(a) \]

### Step 3: 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:
\[ A = \frac{1}{2} \times (6 + k) \times h = \frac{1}{2} \times (6 + 6 \cos(a)) \times (6 \sin(a)) \]
This simplifies to:
\[ A = \frac{1}{2} \times 6 \times (1 + \cos(a)) \times 6 \sin(a) = 18 \sin(a)(1 + \cos(a)) \]

### Step 4: Differentiate the Area
To find the maximum area, we differentiate \( A \) with respect to \( a \):
\[ \frac{dA}{da} = 18 \left( \cos(a)(1 + \cos(a)) + \sin(a)(-\sin(a)) \right) \]
Setting the derivative equal to zero to find critical points:
\[ \cos(a)(1 + \cos(a)) - \sin^2(a) = 0 \]

### Step 5: Solve for \( a \)
Using the identity \( \sin^2(a) = 1 - \cos^2(a) \), we can substitute:
\[ \cos(a)(1 + \cos(a)) - (1 - \cos^2(a)) = 0 \]
This simplifies to:
\[ \cos(a) + \cos^2(a) - 1 + \cos^2(a) = 0 \]
\[ 2\cos^2(a) + \cos(a) - 1 = 0 \]
Using the quadratic formula:
\[ \cos(a) = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4} \]
This gives us:
\[ \cos(a) = \frac{1}{2} \quad \text{or} \quad \cos(a) = -1 \]
Since \( a \) is in \( (0, \frac{\pi}{2}) \), we take \( \cos(a) = \frac{1}{2} \), which gives:
\[ a = \frac{\pi}{3} \quad \text{or} \quad 60^\circ \]

### Conclusion
Thus, the angle \( a \) that maximizes the area of the trapezium is:
\[ \boxed{\frac{\pi}{3}} \text{ or } 60^\circ \]