Question

In a rectangle, if the length is increased by 3 meters and breadth is decreased by 4 metres, the area of the rectangle is reduced by 67 square metres. If length is reduced by 1 metre and breadth in increased by 4 metres, the area is increased by 89 sq. metres. Find the dimensions of the rectangle.

Answer 1

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Answer 2

Given :

• In a rectangle, if the length is increased by 3 m and breadth is decreased by 4 m, the area of the rectangle is reduced by 67 m².
• If length is reduced by 1 m and breadth in increased by 4 m, the area is increased by 89 m².

To find :

• Dimensions of the rectangle.

Solution :

Consider,

• Length of the rectangle = x m
• Breadth of the rectangle = y m

Then,

• Area of the rectangle = xy m²

According to the 1st condition :-

• if the length is increased by 3 m and breadth is decreased by 4 m, the area of the rectangle is reduced by 67 m².

Therefore ,

• Length = (x+3) m
• Breadth = (y-4) m

$\implies\sf{(x+3)(y-4)=xy-67}$

$\implies\sf{xy-4x+3y-12=xy-67}$

$\implies\sf{-4x+3y=-55}$

$\implies\sf{4x-3y=55................eq[1]}$

According to the 2nd condition :-

• If length is reduced by 1 m and breadth in increased by 4 m, the area is increased by 89 m².

Therefore,

• Length = (x-1) m
• Breadth = (y+4) m

$\implies\sf{(x-1)(y+4)=xy+89}$

$\implies\sf{xy+4x-y-4=xy+89}$

$\implies\sf{4x-y=93.............eq[2]}$

★ By elimination :-

4x - 3y = 55

4x - y = 93

(-). (+). (-)

______________

→ - 2y = - 38

→ y = 19

Now put y = 19 in eq[2] for getting the value of x.

$\implies\sf{4x-y=93}$

$\implies\sf{4x-19=93}$

$\implies\sf{4x=93+19}$

$\implies\sf{4x=112}$

$\implies\sf{x=28}$

• Length = 28 m
• Breadth = 19 m

Therefore, the dimensions of the rectangle are 28 m and 19 m.