Question

The area of a rectangle gets reduced by 9 square units, if its length is reduced by
5 units and breadth is increased by 3 units. If we increase the length by 3 units and
the breadth by 2 units, the area increases by 67 square units. Find the dimensions
of the rectangle. cross multiplication
​

Answer 1

Given :-

Area reduce by 9 sq unit when

• Length reduced by 5 units .

• Breadth increased by 3 units .

Area increase by 67 sq units when

• Length increased by 3 units.

• Breadth increased by 2units.

To Find :-

Dimensions of Rectangle.

Solution :-

Let's assume that the length of rectangle is x units and breadth y units .

Area of rectangle = length×breadth .

$\sf{\implies Area = x \times y }$

When length is reduced by 5 units and breadth is increased by 3 units .The area will reduced by 9 sq units

$\sf{\implies ( x.y - 9) = (x-5)(y+3) }$

$\sf{\implies (x.y -9 ) = xy + 3x - 5y - 15 }$

$\sf{\implies - 9 + 15 = x.y - x.y + 3x - 5y}$

$\sf{\implies 0 = 3x - 5y - 6 ..........eq \: 1st }$

When length increased by 3 unit and breadth is increased by 2 units. Area will increase by 67 units .

$\sf{\implies (x.y + 67) = (x+3)(y+2) }$

$\sf{\implies x.y + 67 = xy + 2x + 3y + 6 }$

$\sf{\implies 67 - 6 = x.y - x.y +2x + 3y}$

$\sf{\implies 0 = 2x + 3y - 61......... eq\; 2nd }$

Multiplying eq 1st with 2 and eq 2nd with 3 .

$\sf{\implies 0 = (3x - 5y - 6)2 }$

$\sf{\implies 0 = 6x - 10y - 12 ..........eq \: 3rd }$

$\sf{\implies 0 = ( 2x + 3y - 61) 3 }$

$\sf{\implies 0 = 6x + 9y - 183............eq\;4th }$

Substracting eq 4th from 3rd

$\sf{\implies 0 = 6x - 10y - 12 - 6x - 9y + 183 }$

$\sf{\implies y = \frac{171}{19} = 9 }$

Putting value of y in eq 1st

$\sf{\implies 0 = 3x - 45 - 6 }$

$\sf{\implies 51 = 3x }$

$\sf{\implies X = \frac{51}{3} = 17 }$

Length is 17 units and breadth is 9 units