Question

The area of rectangle is reduced by 9 sq.cm if its length is reduced by 5 cm and the breadth is increased by 3 cm. If we increase the length by 3 cm and breadth by 2 cm , the area is increased by 67 sq. cm. Find the length and breadth of the rectangle. ​

Answer 1

The length is 17 m and the breadth is 9 m.

Step-by-step explanation:

Given:

Let us assume that the length and the breadth of the rectangle be x m and y m, respectively.  

∴ Area of the rectangle = (xy) square m

Case 1:  

When the length is reduced by 5m and the breadth is increased by 3 m:  

New length = (x - 5) m  

New breadth = (y + 3) m  

∴ New area = (x - 5) (y + 3) sq.m  

∴ xy - (x - 5) (y + 3) = 9  

⇒ xy - [xy - 5y + 3x - 15] = 9  

⇒ xy - xy + 5y - 3x + 15 = 9

⇒ - (3x-5y) = 9 - 15

⇒ - (3x - 5y) = - 6

⇒ 3x – 5y = 6                   [1]

Case 2:  

When the length is increased by 3 m and the breadth is increased by 2 m:  New length = (x + 3) m

New breadth = (y + 2) m  

∴ New area = (x + 3) (y + 2) sq.m  

⇒ (x + 3) (y + 2) - xy = 67  

⇒ [xy + 3y + 2x + 6] - xy = 67

⇒ xy + 3y +2x + 6 -xy = 67

⇒ 2x + 3y = 67 - 6

⇒ 2x + 3y = 61  

⇒ 2x + 3y = 61          [2]

On multiplying Eq (1) by 3 and Eq (2) by 5, we get

3(3x - 5y) = 3(6)  

⇒ 9x - 15y = 18              [3]

⇒ 5(2x + 3y) = 5(61)

⇒ 10x + 15y = 305          [4]

On adding Eq (3) and Eq (4), we get

9x - 15y + 10x + 15y = 18 + 305

⇒ 19x = 323

⇒ $x = \frac{323}{19}$

⇒ x = 17  

On substituting x = 17 in Eq (3), we get:  

⇒ $9\times 17 - 15y = 18$

⇒ 153 - 15y = 18  

⇒ -15y = 18 - 153

⇒- 15y = -135

⇒ 15y = 135

⇒ $y = \frac{135}{15}$

⇒ y = 9

Hence, the length is 17 m and the breadth is 9 m.