Find the area of a trapezium ABCD in which AB || DC, AB = 26cm, BC = 25cm, CD = 40cm and DA = 25cm.

Options

A. 648`cm^(2)`
B. 792`cm^(2)`
C. 660`cm^(2)`
D. 798`cm^(2)`

Answer

To find the area of trapezium ABCD where AB || DC, we can use the following steps:

### Step 1: Identify the given dimensions
We have:
- AB = 26 cm (one of the parallel sides)
- DC = 40 cm (the other parallel side)
- BC = 25 cm (one of the non-parallel sides)
- DA = 25 cm (the other non-parallel side)

### Step 2: Calculate the height of the trapezium
To find the height (h) of the trapezium, we can drop perpendiculars from points B and A to line DC, meeting at points E and F respectively. This will create two right triangles, ADF and BCE.

Using the Pythagorean theorem in triangle ADF:
- AF = h (height)
- AD = 25 cm
- DF = DC - AB = 40 cm - 26 cm = 14 cm (the base of triangle ADF)

Thus, we can write:
\[ AD^2 = AF^2 + DF^2 \]
\[ 25^2 = h^2 + 14^2 \]
\[ 625 = h^2 + 196 \]
\[ h^2 = 625 - 196 \]
\[ h^2 = 429 \]
\[ h = \sqrt{429} \approx 20.7 \, \text{cm} \]

### Step 3: Calculate the area of the trapezium
The area (A) of a trapezium can be calculated using the formula:
\[ A = \frac{1}{2} \times (AB + DC) \times h \]
Substituting the values we have:
\[ A = \frac{1}{2} \times (26 + 40) \times \sqrt{429} \]
\[ A = \frac{1}{2} \times 66 \times \sqrt{429} \]
\[ A = 33 \times \sqrt{429} \]
Calculating \( \sqrt{429} \approx 20.7 \):
\[ A \approx 33 \times 20.7 \approx 683.1 \, \text{cm}^2 \]

### Final Answer
The area of trapezium ABCD is approximately **683.1 cm²**.

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