Question

The perimeter of a rectangle is 48 m. If the width were doubled and the length were increased by 24 m, the perimeter would be 112 m. What is the length of the original rectangle?

Answer 1

The answer is 16

Explanation:

Firstly, we have to determine two equations that represent the perimeter before and after of changings.

Let

W=width of the rectangle

L=length of the rectangle

P=perimeter of the rectangle

Before of changings:

`P=2*W+2*L\\P=48\\48=2*W+2*L`

After of changings:

`P=2*(2*W)+2*(L+24)\\\\P=4*W+2*L+48\\P=112\\112=4*W+2*L+48\\64=4*W+2*L`

Finally, we resolve both equations:

`(-1)*(48)=(-1)*(2*L+2*W)\\-48=-2*W-2*L\\\\64=4*W+2*L\\\\`

Adding both equations:

`-48+64=-2*W-2*L+4*W+2*L\\16=2*W\\W=16/2=8\\\\`

Replacing W value in any equation:

`64=4*8+2*L\\64=32+2*L\\2*L=64-32\\2*L=32\\L=32/2=16\\`

Then, the length of the original rectangle (before changings) is 16

Answer 2

Original

Length (L): L

width (w): w

Perimeter (P) = 2L + 2w

48 = 2(L) + 2(w)

24 = L + w

24 - L = w

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New

Length (L): L + 24

width (w): 2w ⇒2(24 - L) = 48 - 2L

Perimeter (P) = 2L + 2w

112 = 2(L + 24) + 2(48 - 2L)

112 = 2L + 48 + 96 - 4L

112 = -2L + 140

-28 = -2L

14 = L

Answer: 14 m