Sunday, 31 March 2013

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


The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units andbreadth is increased by 3 units. If we increase the length by 3units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
Let length of rectangle = x units
And width of rectangle = y units
Area of rectangle = length * width = x*y
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.
So
Decrease the length by 5 unit so new length = x - 5
Increase the width by 3 unit so new width = y + 3
New area is reduced by 9 units
So new area  = xy – 9
Plug the value in formula length * width = area we get
(x  - 5)(y + 3) = xy  - 9
Xy  + 3x – 5y – 15  = xy – 9
Subtract xy both side we get
3x - 5y  = 6                 …(1)
If we increase the length by 3units and the breadth by 2 units, the area increases by 67 square units.
Increase the length by 3 unit so new length = x +3
Increase the width by 2 unit so new width = y + 2
New area is increased  by 67 units
So new area  = xy + 67
Plug the value in formula length * width = area we get
(x  +3)(y + 2) = xy  +  67
Xy  + 2x  +  3y  +  6  = xy + 67
Subtract xy both side we get
2x  +  3y = 61                         …(2)*3
3x - 5y  = 6                 …(1)*2
Cross multiply the coefficient of x we get
6x + 9 y = 183
6x -10y  =12
Subtract now we get
19 y = 171
Y = 171/19 = 9
Plug this value of y in equation first we get
2x + 3* 9 = 61
2x  = 61 – 27
2x = 34
X = 34/2 = 17
So length is 17 units and width is 9 units

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