Find the length of a rectangle whose area and breadth are respectively: \( (15-18) \) 15. \( (1+a+b+a b) \mathrm{sq} \) unit, \( (a+1) \mathrm{unit} \) 16. \( 184 \mathrm{sq} \mathrm{m} 50 \mathrm{sq} \mathrm{dm}, 12 \mathrm{~m} 3 \mathrm{dm} \). 17. \( 85 \mathrm{sq} \mathrm{m} 86 \mathrm{sq} \mathrm{dm}, 8 \mathrm{~m} .1 \mathrm{dm} \). 18. \( 49 \mathrm{sq} \mathrm{m} .93 \mathrm{sq} \mathrm{dm} 80 \mathrm{sq} \mathrm{cm}, 6 \mathrm{~m} 15 \mathrm{~cm} \).

To find the length of a rectangle given its area and breadth, we can use the formula: \[ \text{Length} = \frac{\text{Area}}{\text{Breadth}} \] Let's solve each of the problems one by one. ### Problem 15 **Area:** \( (1 + a + b + ab) \) sq unit **Breadth:** \( (a + 1) \) unit Using the formula: \[ \text{Length} = \frac{1 + a + b + ab}{a + 1} \] ### Problem 16 **Area:** \( 184 \) sq m \( 50 \) sq dm **Breadth:** \( 12 \) m \( 3 \) dm First, we need to convert the area into a single unit. 1 sq m = 100 sq dm, so: \[ 184 \text{ sq m} = 18400 \text{ sq dm} \] Thus, the total area is: \[ 18400 \text{ sq dm} + 50 \text{ sq dm} = 18450 \text{ sq dm} \] Now, convert the breadth into decimeters: \[ 12 \text{ m} = 120 \text{ dm} \quad \text{and} \quad 3 \text{ dm} = 3 \text{ dm} \] So, the total breadth is: \[ 120 \text{ dm} + 3 \text{ dm} = 123 \text{ dm} \] Now, we can find the length: \[ \text{Length} = \frac{18450 \text{ sq dm}}{123 \text{ dm}} \approx 150 \text{ dm} \] ### Problem 17 **Area:** \( 85 \) sq m \( 86 \) sq dm **Breadth:** \( 8 \) m \( 1 \) dm Convert the area: \[ 85 \text{ sq m} = 8500 \text{ sq dm} \] Thus, the total area is: \[ 8500 \text{ sq dm} + 86 \text{ sq dm} = 8586 \text{ sq dm} \] Convert the breadth: \[ 8 \text{ m} = 80 \text{ dm} \quad \text{and} \quad 1 \text{ dm} = 1 \text{ dm} \] So, the total breadth is: \[ 80 \text{ dm} + 1 \text{ dm} = 81 \text{ dm} \] Now, we can find the length: \[ \text{Length} = \frac{8586 \text{ sq dm}}{81 \text{ dm}} \approx 106 \text{ dm} \] ### Problem 18 **Area:** \( 49 \) sq m \( 93 \) sq dm \( 80 \) sq cm **Breadth:** \( 6 \) m \( 15 \) cm Convert the area: \[ 49 \text{ sq m} = 4900 \text{ sq dm} \quad \text{and} \quad 93 \text{ sq dm} = 93 \text{ sq dm} \quad \text{and} \quad 80 \text{ sq cm} = 0.8 \text{ sq dm} \] Thus, the total area is: \[ 4900 \text{ sq dm} + 93 \text{ sq dm} + 0.8 \text{ sq dm} = 4993.8 \text{ sq dm} \] Convert the breadth: \[ 6 \text{ m} = 60 \text{ dm} \quad \text{and} \quad 15 \text{ cm} = 1.5 \text{ dm} \] So, the total breadth is: \[ 60 \text{ dm} + 1.5 \text{ dm} = 61.5 \text{ dm} \] Now, we can find the length: \[ \text{Length} = \frac{4993.8 \text{ sq dm}}{61.5 \text{ dm}} \approx 81.2 \text{ dm} \] ### Summary of Lengths - **Problem 15:** \( \frac{1 + a + b + ab}{a + 1} \) - **Problem 16:** \( 150 \text{ dm} \) - **Problem 17:** \( 106 \text{ dm} \) - **Problem 18:** \( 81.2 \text{ dm} \)