Let the length and the breadth of the rectangle be x m and y m, respectively.
∴ Area of the rectangle = (xy) sq.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) = 8
⇒ xy – [xy – 5y + 3x – 15] = 8
⇒ xy – xy + 5y – 3x + 15 = 8
⇒ 3x – 5y = 7 ………(i)
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 = 74
⇒ [xy + 3y + 2x + 6] – xy = 74
⇒ 2x + 3y = 68 ………(ii)
On multiplying (i) by 3 and (ii) by 5, we get:
9x – 15y = 21 ……….(iii)
10x + 15y = 340 ………(iv)
On adding (iii) and (iv), we get:
19x = 361
⇒ x = 19
On substituting x = 19 in (iii), we get:
9 × 19 – 15y = 21
⇒171 – 15y = 21
⇒15y = (171 – 21) = 150
⇒y = 10
Hence, the length is 19m and the breadth is 10m.
