The area of a rectangle gets reduced by 8m2 , when its length is reduc

1.	The area of a rectangle gets reduced by 8m2 , when its length is reduced by 5m and its breadth is increased by 3m. If we increase the length by 3m and breadth by 2m, the area is increased by 74m2 . Find the length and the breadth of the rectangle.
Answer» 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.

The area of a rectangle gets reduced by 8m2 , when its length is reduced by 5m and its breadth is increased by 3m. If we increase the length by 3m and breadth by 2m, the area is increased by 74m2 . Find the length and the breadth of the rectangle.

Answer»

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.