# Rectangle lengthThe 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...

Rectangle length

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

w =5.

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.**