SOLUTION: The length of a rectangle is twice its width. If the width is reduced by 1 cm and the length is also reduced by 2 cm, the area will be 15cm². Find the dimensions of the original

Question 1204174: The length of a rectangle is twice its width. If the width is reduced by 1 cm and the length is also reduced by 2 cm, the area will be 15cm².
Find the dimensions of the original rectangle.
Found 2 solutions by Manan28IN, mananth:
Answer by Manan28IN(1) (Show Source): You can put this solution on YOUR website!
let width of original rectangle be = x
length of original rectangle = 2x
reduced width = x-1
reduced length = 2x-2
reduced area = length*width = (2x-2)*(x-1)
15 = 2x^2-4x+2
x=3.738612 cm
Area of original rectangle = x*2x
= 2x^2
= 27.954 cm^2

Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
The length of a rectangle is twice its width. If the width is reduced by 1 cm and the length is also reduced by 2 cm, the area will be 15cm².
Find the dimensions of the original rectangle.
Let width be x cm
Length will be 2x cm .
(I have reduced length also by 1 and not 2)
If the width is reduced by 1 cm and the length is also reduced by 1 cm, the area will be 15cm².
(x-1)(2x-1)=15
2x^2-x-2x+1=15
2x^2-3x -14=0
2x^2-7x+4x-14=0
x(2x-7)+2(2x-7)=0
(2x-7)(x+2)=0
x= 7/2 0r -2
-2 not possible
x=3.5 = width
length = 2x = 7
Area = 7*3.5 = 24.5 cm^2
So, the dimensions of the original rectangle are:
Width (w) = 3.5cm
Length (l) = 7 cm
Check
(7-1)(3.5-1)= 15

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