In a trapezium ABCD, AB || DC, AB = a cm, and DC = b cm. If M and N are the midpoints of the nonparallel sides, AD and BC respectively then find the ratio of ar(DCNM) and ar(MNBA).

Answer

To find the ratio of the areas of trapeziums DCNM and MNBA in trapezium ABCD where AB || DC, AB = a cm, and DC = b cm, and M and N are midpoints of non-parallel sides AD and BC, we can follow these steps:

### Step 1: Understand the trapezium and identify midpoints
In trapezium ABCD, we have:
- AB || DC
- AB = a cm
- DC = b cm
- M is the midpoint of AD
- N is the midpoint of BC

### Step 2: Calculate the length of MN
Since M and N are midpoints, the length of MN can be calculated using the formula for the length of the segment connecting the midpoints of two sides of a trapezium:
\[ MN = \frac{AB + DC}{2} = \frac{a + b}{2} \]

### Step 3: Calculate the area of trapezium DCNM
The area of trapezium DCNM can be calculated using the formula for the area of a trapezium:
\[ \text{Area}_{DCNM} = \frac{1}{2} \times (DC + MN) \times h \]
Where h is the height of the trapezium. Substituting the values we have:
\[ \text{Area}_{DCNM} = \frac{1}{2} \times \left(b + \frac{a + b}{2}\right) \times h \]
\[ = \frac{1}{2} \times \left(b + \frac{a + b}{2}\right) \times h \]
\[ = \frac{1}{2} \times \left(\frac{2b + a + b}{2}\right) \times h \]
\[ = \frac{1}{2} \times \left(\frac{a + 3b}{2}\right) \times h \]
\[ = \frac{(a + 3b)h}{4} \]

### Step 4: Calculate the area of trapezium MNBA
Similarly, the area of trapezium MNBA can be calculated:
\[ \text{Area}_{MNBA} = \frac{1}{2} \times (MN + AB) \times h \]
Substituting the values we have:
\[ \text{Area}_{MNBA} = \frac{1}{2} \times \left(\frac{a + b}{2} + a\right) \times h \]
\[ = \frac{1}{2} \times \left(\frac{a + b + 2a}{2}\right) \times h \]
\[ = \frac{1}{2} \times \left(\frac{3a + b}{2}\right) \times h \]
\[ = \frac{(3a + b)h}{4} \]

### Step 5: Find the ratio of the areas
Now, we can find the ratio of the areas of trapezium DCNM to trapezium MNBA:
\[ \text{Ratio} = \frac{\text{Area}_{DCNM}}{\text{Area}_{MNBA}} = \frac{\frac{(a + 3b)h}{4}}{\frac{(3a + b)h}{4}} \]
The \(h\) and \(\frac{1}{4}\) cancel out:
\[ \text{Ratio} = \frac{a + 3b}{3a + b} \]

### Final Answer
The ratio of the areas of trapezium DCNM to trapezium MNBA is:
\[ \frac{a + 3b}{3a + b} \]