Question

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 is
increased by 67 square units if length is increased by 3 units and breadth is
increased by 2 units. Find the perimeter of the rectangle. [CBSE 2012]​

Answer 1

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

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

${\fboxed{ x = 17, y = 9}}$

Length of the rectangle = 17 units.

Breadth of the rectangle = 9 units.

Answer 2

$\huge\mathcal{\boxed{\fcolorbox{lime}{yellow}{ANSWER}}}$

Let the length and breadth of the rectangle be x and y units respectively. Then,

⭐ Area = xy sq. units .

☃️ If length is reduced by 5 units and the breadth is increases by 3 units, then area is reduced by 9 square units .

→ xy - 9 = (x - 5) (y + 3)

→ xy - 9 = xy + 3x - 5y - 15

→ 3x - 5y - 6 = 0 ----(1)

☃️ When length is increased by 3 units and breadth by 2 units, the area is increased by 67 sq. units .

∴ xy + 67 = (x+3) (y+2)

⇒ xy + 67 = xy + 2x + 3y + 6

⇒ 2x + 3y − 61=0 ----(2)

Thus, we get the following system of linear equations ;

→ 3x − 5y − 6 = 0

→ 2x + 3y − 61 = 0

By using cross-multiplication, we have

$\frac{x}{305 + 18} = \frac{ - y}{ - 183 + 12} = \frac{1}{9 + 10} \\ \\ = > x = \frac{323}{19} = 17 \:\:and \:\: y = \frac{171}{19} = 19$

✔️ Hence, the length and breadth of the rectangle are 17 units and 19 units respectively.