A trapezium is such that three of its sides have lengths as 9cm, then the length (in cm ) of the fourth side such that the area of trapezium is maximum, is

Answer

To find the length of the fourth side of the trapezium such that the area is maximized, we can follow these steps:

### Step 1: Understand the trapezium configuration
We have a trapezium with three sides of length 6 cm each. Let’s denote the lengths of the sides as follows:
- Side 1 = 6 cm
- Side 2 = 6 cm
- Side 3 = 6 cm
- Side 4 = x cm (the side we need to find)

### Step 2: Set up the area formula
The area \( A \) of a trapezium can be calculated using the formula:
\[ A = \frac{1}{2} \times (a + b) \times h \]
where \( a \) and \( b \) are the lengths of the two parallel sides, and \( h \) is the height.

For our case, we can consider two of the sides (let's take the two sides of length 6 cm) as the parallel sides, and the height \( h \) can be expressed in terms of the angle \( \theta \) formed by the trapezium.

### Step 3: Express height in terms of \( \theta \)
Using trigonometry, the height \( h \) can be expressed as:
\[ h = 6 \sin \theta \]
where \( \theta \) is the angle between one of the sides of length 6 cm and the base.

### Step 4: Express the area in terms of \( \theta \)
Now, substituting the height into the area formula:
\[ A = \frac{1}{2} \times (6 + x) \times (6 \sin \theta) \]
This simplifies to:
\[ A = 3(6 + x) \sin \theta \]

### Step 5: Express \( x \) in terms of \( \theta \)
Using the cosine function, we can express \( x \):
\[ x = 6 \cos \theta + 6 \cos \theta = 12 \cos \theta \]
Thus, we can substitute \( x \) back into the area formula:
\[ A = 3(6 + 12 \cos \theta) \sin \theta \]
This simplifies to:
\[ A = 18 \sin \theta + 36 \sin \theta \cos \theta \]

### Step 6: Maximize the area
To maximize the area, we need to take the derivative of \( A \) with respect to \( \theta \) and set it to zero:
\[ \frac{dA}{d\theta} = 18 \cos \theta + 36 (\cos^2 \theta - \sin^2 \theta) = 0 \]
This can be rewritten using the double angle identity:
\[ 18 \cos \theta + 36 \cos 2\theta = 0 \]
Dividing through by 18 gives:
\[ \cos \theta + 2 \cos 2\theta = 0 \]

### Step 7: Solve the equation
Using the identity \( \cos 2\theta = 2\cos^2 \theta - 1 \), we can rewrite the equation:
\[ \cos \theta + 2(2\cos^2 \theta - 1) = 0 \]
This leads to:
\[ 4\cos^2 \theta + \cos \theta - 2 = 0 \]
Using the quadratic formula:
\[ \cos \theta = \frac{-1 \pm \sqrt{1 + 32}}{8} = \frac{-1 \pm 6}{8} \]
This gives us:
\[ \cos \theta = \frac{5}{8} \quad \text{or} \quad \cos \theta = -\frac{7}{8} \]
Since \( \cos \theta \) must be positive in our case, we take:
\[ \cos \theta = \frac{5}{8} \]

### Step 8: Calculate the length of the fourth side
Now substituting back to find \( x \):
\[ x = 12 \cos \theta = 12 \times \frac{5}{8} = 7.5 \text{ cm} \]

### Conclusion
The length of the fourth side such that the area of the trapezium is maximum is:
\[ \boxed{7.5 \text{ cm}} \]