16. A rectangle has length \( L \) and width \( W \) \( L \) is increased by \( 20 \% \) \( W \) is decreased by \( 35 \% \) Calculate the percentage reduction in the area of the rectangle.

To determine the percentage reduction in the area of the rectangle, let's follow these steps: 1. **Original Area:** \[ \text{Area}_{\text{original}} = L \times W \] 2. **New Dimensions:** - **Length (\( L \)) is increased by \( 20\% \):** \[ L_{\text{new}} = L + 0.20L = 1.20L \] - **Width (\( W \)) is decreased by \( 35\% \):** \[ W_{\text{new}} = W - 0.35W = 0.65W \] 3. **New Area:** \[ \text{Area}_{\text{new}} = L_{\text{new}} \times W_{\text{new}} = 1.20L \times 0.65W = 0.78L \times W \] 4. **Change in Area:** \[ \Delta \text{Area} = \text{Area}_{\text{original}} - \text{Area}_{\text{new}} = LW - 0.78LW = 0.22LW \] 5. **Percentage Reduction:** \[ \text{Percentage Reduction} = \left( \frac{\Delta \text{Area}}{\text{Area}_{\text{original}}} \right) \times 100 = \left( \frac{0.22LW}{LW} \right) \times 100 = 22\% \] **Conclusion:** The area of the rectangle is reduced by **22%**.