Question

the length of a rectangle is three more than twice the width. Determine the dimensions that will give you a total area of 27 m2

Answer 1

Hey buddy! Here's your answer!

Width = x

Length = 2x + 3

Area = 27 sq.m

x(2x + 3) = 27

2x² + 3x -27 = 0

2x² - 6x + 9x - 27 = 0

2x (x - 3) + 3 (x - 3) = 0

(2x + 3)(x - 3) = 0

x = 3 [as it can't be negative]

Width = 3 m

Length = 9 m

Hope this helps! :)

Answer 2

Answer:

The dimension of the rectangle is 9 m by 3 m.

Step-by-step explanation:

Given:-

The length of a rectangle is $3$ more than twice the width.

The area of a rectangle = 27 $m^2$

To find:-

The dimension of the rectangle.

Step 1 of 1

Let the length of the rectangle be $l$.

And the breadth of the rectangle is $b$.

According to the question,

The length of a rectangle is $3$ more than twice the width, i.e.,

$l=2b+3$ _____ (1)

Also,

The area of a rectangle = 27

⇒ $l\times b= 27$

⇒ $(2b+3)\times b =27$

⇒ $2b^2+3b-27=0$

Using the middle-term splitting method, we have

$2b^2+9b-6b-27=0$

$b(2b + 9) - 3(2b + 9) = 0$

(2b + 9)(b - 3) = 0

b = 3 and b = -9/2

Since the breadth of the rectangle cannot $be$ negative.

So, b = -9/2 is not possible.

Substituting the value b = 3 in the equation (1), we get

$l$ = 2(3) + 3

 = 6 + 3

 = 9 m

Therefore, the dimension of the rectangle is 9 m by 3 m.

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