Question

. The perimeter of a rectangle is 38 m. If its length is 5 m less than twice its width, find the dimensions of the rectangle.
Ans-11 length & 8 width . Pls tell proper process . Irrelevant answer will be reported.​

Answer 1

✬ Length = 11 m ✬

✬ Width = 8 m ✬

Step-by-step explanation:

Given:

• Perimeter of rectangle is 38 m.
• Length is 5 m less than twice its width.

To Find:

• Length of dimensions of rectangle.

Solution: Let the width of rectangle be x m. Therefore,

➙ Length = 5 less than two time of x.

➙ Length = (2x – 5) m

As we know that for rectangle

★ Perimeter = 2 × ( Length + Width ) ★

• Perimeter is 38 m.

$\implies{\rm }$ 38 = 2 ( 2x – 5 + x )

$\implies{\rm }$ 38/2 = ( 2x – 5 + x )

$\implies{\rm }$ 19 = 3x – 5

$\implies{\rm }$ 19 + 5 = 3x

$\implies{\rm }$ 24/3 = x

$\implies{\rm }$ 8 = x

So measure of width of rectangle is 8 m and measure of length = 2(8) – 5 = 11 m.

Answer 2

Let the width of the rectangle be X.

Now, According to the question length of the rectangle should be 2X - 5

We know that peremeter of a rectangle = 2( L + B )

Now,

It can be written in word as follows,

• 2 ( L + B ) = 38

• Or, 2 ( X + ( 2X - 5) = 38
• Or, 2 ( 3X - 5 ) = 38

• Or, ( 3X - 5 ) = 38/2

• Or, ( 3X - 5 ) = 19

• Or, 3X = 19 + 5
• Or, 3X = 24
• Or, X = 24/ 3
• Or, X = 8 Metres

Now,

• Width of the rectangle = X = 8 Metres

• Length of the rectangle = 2X - 5 = ( 8 x 2 ) - 5 = 16 - 5 = 11 Metres