ABCD is a trapezium with parallel sides AB = a cm and DC = b cm, E and Fare the mid-points of the non-parallel sides. Find the ratio of ar(ABFE) and ar(EFCD).

Options

A. (3b + a) : (3a + b)
B. (3a + b) : (3b + a)
C. (2a + 3b): (3a + b)
D. (3a + 2b) : (2a + 3b)

Answer

To find the ratio of the areas of trapezium ABFE and trapezium EFCD, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the trapezium and its dimensions**:
- Let the lengths of the parallel sides be \( AB = a \) cm and \( DC = b \) cm.
- Let \( E \) and \( F \) be the midpoints of the non-parallel sides \( AD \) and \( BC \), respectively.

2. **Calculate the length of line segment EF**:
- Since \( E \) and \( F \) are midpoints, the length of \( EF \) can be calculated as:
\[ EF = \frac{AB + DC}{2} = \frac{a + b}{2} \]

3. **Determine the height of the trapezium**:
- Let the height of the trapezium \( ABCD \) be \( H \). The heights from \( A \) to \( E \) and from \( D \) to \( F \) will also be \( H \).

4. **Calculate the area of trapezium ABFE**:
- The area of trapezium \( ABFE \) can be calculated using the formula:
\[ \text{Area}_{ABFE} = \frac{1}{2} \times (AB + EF) \times H \]
- Substituting the values:
\[ \text{Area}_{ABFE} = \frac{1}{2} \times \left( a + \frac{a + b}{2} \right) \times H \]
- Simplifying:
\[ = \frac{1}{2} \times \left( \frac{2a + a + b}{2} \right) \times H = \frac{1}{2} \times \left( \frac{3a + b}{2} \right) \times H = \frac{3a + b}{4} \times H \]

5. **Calculate the area of trapezium EFCD**:
- The area of trapezium \( EFCD \) can be calculated as:
\[ \text{Area}_{EFCD} = \frac{1}{2} \times (EF + DC) \times H \]
- Substituting the values:
\[ \text{Area}_{EFCD} = \frac{1}{2} \times \left( \frac{a + b}{2} + b \right) \times H \]
- Simplifying:
\[ = \frac{1}{2} \times \left( \frac{a + b + 2b}{2} \right) \times H = \frac{1}{2} \times \left( \frac{a + 3b}{2} \right) \times H = \frac{a + 3b}{4} \times H \]

6. **Find the ratio of the areas**:
- Now, we can find the ratio of the areas:
\[ \text{Ratio} = \frac{\text{Area}_{ABFE}}{\text{Area}_{EFCD}} = \frac{\frac{3a + b}{4} \times H}{\frac{a + 3b}{4} \times H} \]
- Canceling out \( \frac{1}{4} \) and \( H \):
\[ \text{Ratio} = \frac{3a + b}{a + 3b} \]

### Final Result:
The ratio of the areas \( \text{ar}(ABFE) \) to \( \text{ar}(EFCD) \) is:
\[ \text{ar}(ABFE) : \text{ar}(EFCD) = 3a + b : a + 3b \]