Question

The length of a rectangle is 3 units less than twice its width. If its perimeter is 54units, find its dimensions.

Answer 1

Solution:

Here we are given a condition based on the relation between length and breadth of the rectangle. And also the perimeter of the rectangle is given.

• Length is 3 units less than twice its width.
• Perimeter of the rectangle = 54 units.

And we have to determine the dimension that is the length and breadth of the rectangle.

So, let's start solving....

Let, the breadth be x. Then, length will be 2x - 3 because of the first condition mentioned in the question.

Now Perimeter of the rectangle is given which can be calculated by using the length and breadth of the rectangle because:

• P = 2(l + b)

Plugging the required values,

➝ 54 = 2(2x - 3 + x)

➝ 54 = 2(3x - 3)

➝ 3x - 3 = 27

➝ 3x = 30

➝ x = 10

Then 2x - 3 = 2(10) - 3 = 17

Thus, the required dimensions are:

• Length = 17 m
• Breadth = 10 m

And we are done !

Answer 2

Given :-

• The length of a rectangle is 3 units less than twice its width.

• Perimeter is 54 units.

To Find :-

• The dimension.

Solution :-

$\dashrightarrow{\boxed{\red{\sf{Perimeter = 2(l + b)}}}}$

Values :-

$\longrightarrow\sf{54 = (2x - 3 + x)} \\ \\ \longrightarrow\sf{54 = 2(3x - 3)} \\ \\ \longrightarrow\sf{3x - 3 = 27} \\ \\ \longrightarrow\sf{3x = 30} \\ \\ \longrightarrow\sf{x = \: }{\textsf{\textbf{10}}} \\ \\ \sf{Then, \: 2x - 3 = 2(10) - 3 = \: }{\textsf{\textbf{17.}}}$

Hence,

• Breadth = 10.

• Length = 17.