1. **State the problem:** A farmer has two adjacent rectangular fields sharing a common lengthwise fence. The total length of the combined fields is $50 \frac{1}{8}$ meters and the total width is $30 \frac{1}{9}$ meters. We need to find the total length of fencing used, including the external fence and the internal dividing fence. 2. **Convert mixed numbers to improper fractions:** - Length: $50 \frac{1}{8} = \frac{(50 \times 8) + 1}{8} = \frac{400 + 1}{8} = \frac{401}{8}$ meters - Width: $30 \frac{1}{9} = \frac{(30 \times 9) + 1}{9} = \frac{270 + 1}{9} = \frac{271}{9}$ meters 3. **Understand the fencing layout:** - The two fields are side by side along the width. - The total length is $\frac{401}{8}$ meters. - The total width is $\frac{271}{9}$ meters. - The fence encloses the entire perimeter and includes one internal fence dividing the two fields lengthwise (parallel to the width). 4. **Calculate the total fencing:** - External fencing includes two lengths and two widths: $2 \times \text{length} + 2 \times \text{width}$ - Internal fence divides the width into two fields, so it is equal to the length: $\text{length}$ Total fencing $= 2 \times \frac{401}{8} + 2 \times \frac{271}{9} + \frac{401}{8}$ 5. **Calculate each term:** - $2 \times \frac{401}{8} = \frac{802}{8} = \frac{401}{4}$ - $2 \times \frac{271}{9} = \frac{542}{9}$ - Internal fence $= \frac{401}{8}$ 6. **Sum all parts:** $$\frac{401}{4} + \frac{542}{9} + \frac{401}{8}$$ 7. **Find common denominator:** - Denominators are 4, 9, and 8. - Least common denominator (LCD) is 72. 8. **Convert fractions to denominator 72:** - $\frac{401}{4} = \frac{401 \times 18}{72} = \frac{7218}{72}$ - $\frac{542}{9} = \frac{542 \times 8}{72} = \frac{4336}{72}$ - $\frac{401}{8} = \frac{401 \times 9}{72} = \frac{3609}{72}$ 9. **Add numerators:** $$7218 + 4336 + 3609 = 15163$$ 10. **Total fencing:** $$\frac{15163}{72}$$ meters 11. **Convert to mixed number:** - Divide 15163 by 72: - $72 \times 210 = 15120$ - Remainder $= 15163 - 15120 = 43$ So, $$210 \frac{43}{72}$$ meters **Final answer:** The farmer uses $210 \frac{43}{72}$ meters of fencing.