If ABCD is an isoceles trapezium. Diagonal BD = 15 cm ,height = 9 cm find it area AB || CD

Answer

To find the area of the isosceles trapezium ABCD with given dimensions, we can follow these steps:

### Step 1: Understand the trapezium structure
In an isosceles trapezium, we have two parallel sides (let's denote them as AB and CD) and two equal non-parallel sides (AD and BC). The height (perpendicular distance between the two parallel sides) is given as 9 cm.

### Step 2: Identify the diagonal and height
We know that the diagonal BD measures 15 cm and the height from the base CD to the top AB is 9 cm.

### Step 3: Use the properties of right triangles
When we drop perpendiculars from points A and B to line CD, we create two right triangles (ABD and BCD). The height of these triangles is 9 cm, and the diagonal BD is 15 cm.

### Step 4: Find the length of the base of the right triangle
Using the Pythagorean theorem in triangle ABD:
- Let the base of triangle ABD be x.
- According to the Pythagorean theorem:
\[ BD^2 = AB^2 + height^2 \]
\[ 15^2 = x^2 + 9^2 \]
\[ 225 = x^2 + 81 \]
\[ x^2 = 225 - 81 \]
\[ x^2 = 144 \]
\[ x = 12 \text{ cm} \]

### Step 5: Calculate the lengths of the parallel sides
Since the trapezium is isosceles, the total length of the parallel sides AB and CD can be calculated as:
- Let the length of AB be \( a \) and the length of CD be \( b \). The distance from the foot of the perpendiculars to the ends of CD will be equal, hence:
\[ a + 2x = b \]
where \( x = 12 \text{ cm} \).

### Step 6: Calculate the area of the trapezium
The area \( A \) of a trapezium can be calculated using the formula:
\[ A = \frac{1}{2} \times (AB + CD) \times height \]
Substituting the known values:
\[ A = \frac{1}{2} \times (a + b) \times 9 \]
Since we do not have the lengths of AB and CD directly, we can assume that the trapezium is symmetric and we can express it in terms of \( a \) and \( b \).

### Step 7: Substitute values
Assuming \( AB = 12 \text{ cm} \) and \( CD = 12 + 2 \times 12 = 36 \text{ cm} \):
\[ A = \frac{1}{2} \times (12 + 36) \times 9 \]
\[ A = \frac{1}{2} \times 48 \times 9 \]
\[ A = 24 \times 9 = 216 \text{ cm}^2 \]

### Final Area Calculation
Thus, the area of the isosceles trapezium ABCD is \( 108 \text{ cm}^2 \).