A field is in the shape of a trapezium whoseparallel sides are 200 m and 400 m long, whereas each of other two sides is 260 m long. What is the area (in `m^2`)of the field?

Options

A. 78000
B. 52000
C. 72000
D. 60000

Answer

To find the area of the trapezium-shaped field, we can follow these steps:

### Step 1: Identify the trapezium dimensions
We know the lengths of the parallel sides:
- AB (top parallel side) = 200 m
- CD (bottom parallel side) = 400 m
- The lengths of the non-parallel sides (AD and BC) = 260 m each.

### Step 2: Draw the trapezium
Draw trapezium ABCD with AB and CD as the parallel sides. Label the sides accordingly:
- AB = 200 m
- CD = 400 m
- AD = BC = 260 m

### Step 3: Drop perpendiculars
Drop perpendiculars from points A and B to line CD. Let these perpendiculars meet CD at points E and F respectively. Thus, AE and BF are the heights of the trapezium.

### Step 4: Calculate the length of DE and FC
Since AB is parallel to CD, we can find the lengths DE and FC:
- DE + EF + FC = CD
- EF = AB = 200 m
- Therefore, DE + FC = 400 m - 200 m = 200 m.

Since DE = FC (because of symmetry), we can set DE = FC = x:
- 2x = 200 m
- x = 100 m.

Thus, DE = 100 m and FC = 100 m.

### Step 5: Use the Pythagorean theorem
Now, we can use the Pythagorean theorem in triangle AED (or triangle BFC):
- AD² = AE² + DE²
- 260² = AE² + 100²
- 67600 = AE² + 10000
- AE² = 67600 - 10000
- AE² = 57600
- AE = √57600
- AE = 240 m.

### Step 6: Calculate the area of the trapezium
Now that we have the height (AE = 240 m), we can use the formula for the area of a trapezium:
\[ \text{Area} = \frac{1}{2} \times (AB + CD) \times \text{height} \]
Substituting the values:
\[ \text{Area} = \frac{1}{2} \times (200 + 400) \times 240 \]
\[ = \frac{1}{2} \times 600 \times 240 \]
\[ = 300 \times 240 \]
\[ = 72000 \, m^2. \]

### Final Answer
The area of the field is **72000 m²**.

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