The area of a rectangle gets reduced by 9 square units, if its length

1.	The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and the breadth is increased by 3 units. The area isincreased by 67 square units if length is increased by 3 units and breadth isincreased by 2 units. Find the perimeter of the rectangle. [CBSE 2012]
Answer» $\huge\red{\tt{\underline{Solution:-}}}$ Let LENGTH and breadth of rectangle be x unit and y unit. Area = xy According to the question, ⇒ (x - 5) (y + 3) = xy - 9 ⇒ 3x - 5y - 6 = 0 ... (i) ⇒ (x + 3) (y + 2) = xy + 67 ⇒ 2x - 3y – 61 = 0 ... (ii) By CROSS multiplication, we get ⇒ x/305 - (-18) = y/-12 -(-183) = 1/9 - (-10) ⇒ x/323 = y/171 = 1/19 ${\fboxed{ x = <klux>17</klux>, y = 9}}$ Length of the rectangle = 17 UNITS. Breadth of the rectangle = 9 units.

The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and the breadth is increased by 3 units. The area isincreased by 67 square units if length is increased by 3 units and breadth isincreased by 2 units. Find the perimeter of the rectangle. [CBSE 2012]

Answer»

$\huge\red{\tt{\underline{Solution:-}}}$

Let LENGTH and breadth of rectangle be x unit and y unit.

Area = xy

According to the question,

⇒ (x - 5) (y + 3) = xy - 9

⇒ 3x - 5y - 6 = 0 ... (i)

⇒ (x + 3) (y + 2) = xy + 67

⇒ 2x - 3y – 61 = 0 ... (ii)

By CROSS multiplication, we get

⇒ x/305 - (-18) = y/-12 -(-183) = 1/9 - (-10)

⇒ x/323 = y/171 = 1/19

${\fboxed{ x = <klux>17</klux>, y = 9}}$

Length of the rectangle = 17 UNITS.

Breadth of the rectangle = 9 units.