2. The area of a rectangle remains the same if the length is increased by 7 metres and the breadth is decreased by 3 metres. The area remains unaffected if the length is decreased by 7 metres and the breadth is increased by 5 metres. Find the dimensions of the rectangle.

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2. The area of a rectangle remains the same if the length is increased by 7 metres and the breadth is decreased by 3 metres. The area remains unaffected if the length is decreased by 7 metres and the breadth is increased by 5 metres. Find the dimensions of the rectangle.

Solution:

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

Hence, the area of rectangle = xy sq.units

Length is increased by 7 m

⇒ now, the new length is

$x+7$

Breadth is decreased by 3 m

⇒ now, the new breadth is

$y-3$

The area of the rectangle = xy.

Calculation,

$xy\text{ }=\text{ }\left( x+7 \right)\left( y-3 \right)$

$xy\text{ }=\text{ }xy\text{ }+\text{ }7y\text{ }-\text{ }3x\text{ }-\text{ }21$

$3x\text{ }\text{ }7y\text{ }+\text{ }21\text{ }=\text{ }0\text{ }\ldots \ldots \ldots .\text{ }\left( i \right)$

Length is decreased by 7 m

⇒ The new length is

$x-7$

Breadth is increased by 5 m

⇒ The new breadth is

$y+5$

The area of the rectangle = xy.

Calculation,

$xy\text{ }=\text{ }\left( x-7 \right)\left( y+5 \right)$

$xy\text{ }=\text{ }xy\text{ }-\text{ }7y\text{ }+\text{ }5x\text{ }-\text{ }35$

$5x\text{ }\text{ }7y\text{ }\text{ }35\text{ }=\text{ }0\text{ }\ldots \ldots \ldots .\text{ }\left( ii \right)$

By cross-multiplication method, Solve (i) & (ii),

$\frac{x}{\left( -7\times -5 \right)-\left( -7\times 21 \right)}=\frac{y}{\left( -3\times -35 \right)-\left( -5\times 21 \right)}=\frac{1}{\left( 3\times -7 \right)-\left( 5\times -7 \right)}$