Rectangle Math Question

Question

A rectangle is 8 inches longer than twice the width. If the width doubled and the length is decreased by 4 inches, a new rectangle is formed. Find the dimensions of the original rectangle if the perimeter of the new rectangle is 4 inches less than the perimeter of the original rectangle.

Answer 1

original rectangle width = 2

original rectangle length = 2w + 8

new rectangle width = double width: 2w

new rectangle length = double length increased by 4: 2w+4

perimeter = 2*length + 2 * width, so:

perimeter of old rectangle = 2w + 2(2w+8)

perimeter of new rectangle = 4w + 2(2w+4)

perimeter of new rectangle is 4 less than perimeter of old rectangle, so perimeter of new rectangle equals perimeter of old rectangle - 4:

4w + 2(2w+4) = 2w+ 2(2w+8) - 4

8w + 8 = 6w + 12

2w = 4

w=2

l = 2w + 8 = 12

Check answer:

perimeter of old rectangle = 2*2 + 2*12 = 28

perimeter of new rectangle = 2*4 + 2*8 = 24, which is 4 less than old rectangle

Answer 2

Original:

width=x

length=2x+8

perimeter= 2x+2(2x+8)=6x+16

New rectangle:

width=2x

length=2x+8-4=2x+4

perimeter=2(2x)+2(2x+4)=4x+4x+8=8x+8

8x+8=6x+16-4

2x=4

x=2

original dimensions:

width=2

length=2(2)+8=12

Answer 3

Let the original rectangle dimensions are: L₁ and W₁ [Length and width].

Given:

A rectangle is 8 inches longer than twice the width.

=> twice the width.: 2W₁

=> 8 inches longer than twice the width : L₁ = 2W₁ + 8 -------1

If the width doubled and the length is decreased by 4 inches, a new rectangle is formed.

=> If the width doubled : W₂ = 2W₁. --------2

=> length is decreased by 4 inches : L₂ = L₁ - 4. -------3

perimeter of the new rectangle is 4 inches less than the perimeter of the original rectangle.

=> P₂ = P₁ - 4 -------4

To Find : the dimensions of the original rectangle : that is, L₁ and W₁

Solution: First find the perimeters of new and original rectangles.

P = 2(L+W) [Perimeter of a rectangle formula].

P₁ = 2(L₁ + W₁)

P₁ = 2(2W₁ +8 + W₁) [ From equation ----------1 ]

P₁ = 2(3W₁ +8 )

P₁ = 6W₁ +16 --------5

P₂ = 2(L₂ + W₂)

P₂ = 2(2W₁ + L₁ - 4 ) [ From equation ----------2 and 3]

P₂ = 4W₁ + 2L₁ - 8 -------6

Plug in equations----------- 5 and 6 in equation-------4

=> P₂ = P₁ - 4

4W₁ + 2L₁ - 8 = 6W₁ +16 - 4

4W₁ + 2( 2W₁ + 8 ) - 8 = 6W₁ +16 - 4 [ From equation-----------1 , plug-in L₁ values ].

4W₁ + 4W₁ + 16 - 8 = 6W₁ +12.

8W₁ + 8 = 6W₁ +12.

2W₁ = 4.

W₁ = 2 --------7

Plug-in equation-------- 7 in equation--------1

L₁ = 2W₁ + 8

L₁ = 2(2) + 8

L₁ = 4 + 8

L₁ = 12.

Hence, the dimensions of the original rectangle are: L₁ = 12, W₁ = 2.

Hope this helps! for any further doubts, i am available to clarify!