Question

the length of a rectangle is 4 feet More than twice the width. the perimeter is 22 feet. what is the width of the rectangle?​

Answer 1

Answer:

Step-by-step explanation:

Formula

P = 2L + 2w

Givens

w = x

L = 4 + 2*w

P = 22

Solution

22 = 2L + 2w                                Substitute for the Length and width

22 = 2*(2*x + 4) + 2x                    Remove Brackets

22 = 4x + 8 + 2x                           Combine

22 = 6x + 8                                   Subtract 8 from both sides

22 - 8 = 6x + 8 - 8                        Combine

6x = 14                                          Divide both sides by 6

6x/6 = 14/6

x = 2.333

Answer

w = 2 1/3 or

w = 2.333

Answer 2

Question -:

The length of a rectangle is 4 feet More than twice the width. the perimeter is 22 feet. what is the width of the rectangle?

Explanation -:

In this question we are given that the length of a rectangle 4 feet more than the twice of width. And the perimeter is 22 feet. We are asked to calculate the width of the rectangle.

Assuming:

It is given that the length is 4 feet more than the twice of width.

Width = x

Twice of width = 2x.

And 4 feet more than twice of width = 4 + 2x

Length = 4 + 2x

Solution:

We know,

$\small \boxed{\sf{ Perimeter_{(RECTANGLE)} = 2(l + w)}}$

22 = 2(4 + 2x + x )

→ 22/2 = (4 + 3x)

→ 11 = (4 + 3x)

→ 11 - 4 = 3x

→ 7 = 3x

→ 7/3 = x

→ x = 2.33

Hence the width of the rectangle is 2.33 feet.