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

Answer:

Step-by-step 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 $L$:

$40 = L * W$

$40 = L * (4)$

$L = 10$

Answer 2

Answer:

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)