Question

The length of a rectangle is 1 m more than twice the width, and the area of the rectangle is 45 m . Find the dimensions of the rectangle.

Answer 1

Answer: 10cm,4.5cm

Step-by-step explanation:

Let the length be x

Let the breadth be y

The length is 1m more than twice the breadth

I.e

Length = 1+ 2breadth

I.e

X=1+2y......equation 1

Area of the rectangle is

Length ×breadth

I.e area= length×breadth

45= x×y

Substitute for X=1+2y

45= (1+2y)×y

45=y+2y^2

2y2+y-45=0

Factorise

Factors of -90 that will give 1 is 10&9

So we can have -10 &+9 or +10 &-9

So we will go with +10 &-9

Substitute for it

2y2+10y-9y-45

2y(y+5)-9(y+5)

(2y-9)(y+5)

2y-9=0

2y=9

Y=9/2=4.5 or

Y+5=0

Y= -5

Substitute for y in equation 1

X= 1+2y

X=1+2(4.5)

X=1+9

X= 10 or

X= 1+2(-5)

X= 1+(-10)

X= 1-10

X= -9

The positive values will be taken,

So length which is x is 10cm

And width which is y is 4.5cm

Answer 2

The length and width of the rectangle is 10 m and 4.5 m respectively.

Step-by-step explanation:

Given,

The length of a rectangle is 1 m more than twice the width.

Area of the rectangle = 45 sq m

To find the length and width of the rectangle.

Formula

If the length and width of a rectangle be l and b respectively, the area will be lb.

Let, the width (b) be = x and

Length (l) = 2x+1

According to the problem,

(2x+1)x = 45

or, $2x^{2} +x = 45$

or, $2x^{2}+x-45 = 0$

or, $2x^{2}+(10-9)x-45 = 0$

or, $2x^{2}+10x-9x-45 = 0$

or, 2x(x+5)-9(x+5) = 0

or, (2x-9)(x+5) =

Hence,

2x-9 = 0 and x+5 = 0

or, x = $\frac{9}{2}$ and x= -5

We will take x =  $\frac{9}{2}$  as the value of length cannot be negative.

So,

Width = 4.5 m and Length = 2×4.5+1 = 10 m