Question

The length of a rectangle is

2
meters more than
2
times the width. If the area is
40
square meters, find the width and the length.

Answer 1

Step 1

Let length and breadth be x and y respectively

Given

x=2+2y

Step 2

Area of rectangle is lengthxbreadth

Therefore

area=x times y

Area=(2+2y)*y

=2y+2y^2

Area=40

Step 3

40=2y+2y^2

2y^2+2y-40=0

The roots are - 8 and 10

The equation is:

2y^2+10y-8y-40=0

2y(y+5)-8(y+5)=0

Which gives

y=-5 and y=4

Y=5 Rejected,

Therefore

y=4,breadth=4

So length=10(2*4+2)

Answer 2

Explanation:

From the problem statement, we can set up the following two equations:

`L = 2W + 2`

`40 = L * W`

Plugging the first equation into the second, we can solve for
`W`:

`40 = L * W`

`40 = (2W + 2) * W`

`40 = 2W^(2) + 2W`

`2W^(2) + 2W - 40 = 0`

`W^(2) + W - 20 = 0`

`(W + 5)(W - 4) = 0`

`W = -5, 4`

Since the length must be a positive number, then we know that
`W = 4`. We can now plug this number into the second equation to get
`L`:

`40 = L * W`

`40 = L * (4)`

`L = 10`