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 a trapezium such that the area is maximized, where three sides are given as 9 cm, we can follow these steps:

### Step 1: Understand the trapezium configuration
We have a trapezium with three sides of length 9 cm. Let’s denote the lengths of the sides as follows:
- Side 1 (base) = 9 cm
- Side 2 (non-parallel side) = 9 cm
- Side 3 (non-parallel side) = 9 cm
- Side 4 (the unknown length we need to find)

### Step 2: Set up the trapezium
Let’s denote the fourth side (the unknown side) as \( x \). The trapezium can be visualized with one base being 9 cm and the other base being \( x \).

### Step 3: Area of the trapezium
The area \( A \) of a trapezium can be calculated using the formula:
\[ A = \frac{1}{2} \times (b_1 + b_2) \times h \]
where \( b_1 \) and \( b_2 \) are the lengths of the two bases and \( h \) is the height.

In our case:
- \( b_1 = 9 \) cm
- \( b_2 = x \) cm

### Step 4: Express the height in terms of \( x \)
To maximize the area, we need to express the height \( h \) in terms of \( x \). We can use trigonometric relationships in the trapezium.

Let’s drop perpendiculars from the endpoints of the base \( x \) to the base of length 9 cm. The height \( h \) can be expressed as:
\[ h = 9 \sin(\theta) \]
where \( \theta \) is the angle between the non-parallel side and the base of length 9 cm.

### Step 5: Express the area in terms of \( x \)
Now substituting \( h \) into the area formula:
\[ A = \frac{1}{2} \times (9 + x) \times (9 \sin(\theta)) \]
This simplifies to:
\[ A = \frac{9(9 + x)}{2} \sin(\theta) \]

### Step 6: Maximize the area
To maximize the area, we need to find the value of \( x \) that maximizes \( A \). We can differentiate \( A \) with respect to \( x \) and set the derivative to zero. However, we can also use geometric properties to find that the area is maximized when the trapezium is isosceles.

### Step 7: Find the optimal \( x \)
For maximum area, the trapezium should be isosceles, meaning the two non-parallel sides should be equal. Therefore, the fourth side \( x \) should be equal to the base \( b_1 \):
\[ x = 9 \text{ cm} \]

### Step 8: Calculate the fourth side
However, since we need the fourth side such that the area is maximum, we can use the property of isosceles trapezium:
\[ x = 9 + 9 = 18 \text{ cm} \]

### Final Answer
Thus, the length of the fourth side such that the area of the trapezium is maximum is:
\[ \boxed{18 \text{ cm}} \]