The lenghts of arectangle is two inches more than its width.
If the lenghth is increased by four inches and the width is doubled, a new rectangle is formed whose area is 75 square units more than the old area. What are the dimensions of the original rectangle ?
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Let l and w be the length and width of the rectangle.
The 1st condtion is:
l = 2+w ..........(1)
The area of the rectangle = lw.
The new mearements : l+4 and 2w .
Then the area = (l+4)(2w) = 75+lw ............(2) by 2nd condition.
Now put l = w+2 from first condition in eq(2):
(w+2+4)(2w) = 75 +(w+2)w
(w+6)(2w) = w^2+2w+75
2w^2+12w = w^2+2w+75
2w^2-w^2 +12w-2w -75 = 0
w^2+10w -75 = 0
(w+15)(w-5) = 0
W-5 = 0 or w+15 = 0
So l = 2+w =2+5 =7
l = 5 inch and w = 7 inch. And the area lw = 35 sq units.
We'll establish the dimensions of the original rectangle:
- the width: x
- the length: x + 2
The area of the original rectangle is:
A1 = x*(x+2)
We'll establish the dimensions of the new formed rectangle:
- the width: 2x
- the length: (x + 2) + 4 = x + 6
The area of the new rectangle is:
A2 = 2x(x+6)
We know from enunciation that the new area is 75 more than the area of the original rectangle.
A2 = 75 + A1
2x(x+6) = 75 + x(x+2)
We'll remove the brackets:
2x^2 + 12x = 75 + x^2 + 2x
We'll move all terms to one side:
2x^2 + 12x - 75 - x^2 - 2x = 0
We'll combine like terms:
x^2 + 10x - 75 = 0
We'll apply the quadratic formula:
x1 = [-10+sqrt(100 + 300 )]/2
x1 = (-10+20)/2
x1 = 5
x2 = (-10-20)/2
x2 = -15
Since the measure of a side cannot be negative, we'll reject the second negative value.
So, the width of the original rectangle is:
x = 5 inches
the length: x + 2 = 7 inches
Let the width of the original rectangle be W. As the length is 2 more than the width L= W +2.
The area of the original rectangle is W*(W+2)
Now when the length is increased by 4 it becomes W + 6
When the width is doubled it becomes 2*W
The new area becomes 2W (W+6)
This is 75 more than the original
=> 2W (W+6) = W*(W+2) + 75
=> 2W^2 + 12W = W^2 + 2W + 75
=> W^2 + 10W - 75 = 0
=> W^2 + 15W - 5W - 75 = 0
=> W ( W + 15) - 5 ( W + 15) =0
=> (W - 5)(W + 15) =0
So W can be 5 or -15
We can eliminate -15 as width can't be negative.
W= 5 and L = 5 + 2 = 7
So the original width is 5 and the original length is 7.
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