Question

The area of a rectangle is 54 m2 , and the length of the rectangle is 3 m more than twice the width. find the dimensions of the rectangle. length : m width : m

Answer 1

l = 3 + 2w

Find the width
Area = 54
l × w = 54
(3 + 2w) × w = 54
3w + 2w^2 = 54
2w^2 + 3w - 54 = 0
(2w - 9)(w + 6) = 0
w = 9/2 or w = -6 (width shouldn't be negative)
w = 9/2
w = 4.5 m

Find the length
l = 3 + 2w
l = 3 + 2(4.5)
l = 3 + 9
l = 12 m

The width is 4.5 m, the length is 12 m

Answer 2

The dimension of the length and width be 12m and 4.5m respectively.

Given that,

• The area of the rectangle is $54m^2$.
• The length should be 3 more than twice of width i.e. l = 3 + 2w.
• Here length be l and width be w.

Based on the above information, the calculation is as follows:

As we know that

$Area = l\times w$

$(3 + 2w) \times w = 54\\\\3w + 2w^2 = 54\\\\2w^2 + 3w - 54 = 0\\\\(2w - 9)(w + 6) = 0$

$w = 9\div 2$

= 4.5

So, the length be

= 3 + 2(4.5)

= 3 + 9

= 12m

Therefore we can conclude that the dimension of the length and width be 12m and 4.5m respectively.

Learn more: brainly.com/question/16167300