Question

the lenght of the rectangle is 8 m less than twice its breadth.If the perimeter of the rectangle is 56m. Find its dimensions​

Answer 1

Given :

• The lenght of the rectangle is 8 m less than twice its breadth.
• Perimeter of the rectangle is 56m.

To find :

• Dimensions of the rectangle =?

Formula Used :

• Perimeter of the rectangle = 2(length + breadth)

Step-by-step explanation :

Let, the breadth of the rectangle be, x.

Then, the length of the rectangle be, 2x.

It is Given that,

The lenght of the rectangle is 8 m less than twice its breadth.

So, length of the rectangle = 2x - 8

As We know that,

Perimeter of the rectangle = 2(length + breadth)

Substituting the values in the above formula, we get,

➟ 56 = 2(x + 2x - 8)

➟ 56 = 2 (3x - 8)

➟ 56 = 6x - 16

➟ 6x = 56 + 16

➟ 6x = 72

➟ x = 72/6

➟ x = 12.

Therefore, We got the value of, x = 12.

Hence,

Breadth of the rectangle, x = 12 m.

Length of the rectangle, 2x - 8 = 2 × 12 - 8 = 16 m.

Answer 2

Given that ,

• The length of the rectangle is 8 m less than twice its breadth

• The perimeter of rectangle is 56 m

Let ,

Breadth of rectangle be " x "

Then , length of rectangle = 2x - 8

We know that , the perimeter of rectangle is given by

$\large \sf \fbox{Perimeter = 2(l + B) }$

Thus ,

56 = 2{(2x - 8) + x}

28 = 2x - 8 + x

3x = 36

x = 12

$\therefore \sf \underline{The \: length \: and \: breadth \: of \: rectangle \: are \: 16 \: m \: and \: 12 \: m}$