If the length of a rectangle is increased by $$\frac{1}{3}$$​rd and the width is decreased by $$\frac{1}{3}$$​rd, then the area of the rectangle is decreased by the fraction

Correct option is C

Given:

1. Original length = L
2. Original width = W
3. Original area = A = L $$\times$$​ W

Changes:
- New length = $$L + \frac{1}{3}L = \frac{4}{3}L$$​
- New width = $$W - \frac{1}{3}W = \frac{2}{3}W$$​
- New area = A' = $$\text{New length} \times \text{New width}$$​

Solution:

1. Calculate the original area:

A = L $$\times$$​ W

2. Calculate the new area:

A' = $$\frac{4}{3}L \times \frac{2}{3}W = \frac{8}{9}(L \times W)$$​

3. Find the decrease in area:

​$$\text{Decrease in area} = A - A' = L \times W - \frac{8}{9}(L \times W)$$​
​$$\text{Decrease in area} = \frac{1}{9}(L \times W)$$​

4. Calculate the fractional decrease in area:

​$$\text{Fractional Decrease} = \frac{\text{Decrease in area}}{\text{Original Area}} = \frac{\frac{1}{9}(L \times W)}{L \times W} = \frac{1}{9}$$​

Final Answer:

​$$\mathbf{C. \; \frac{1}{9}}$$​