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katrin2010 [14]
3 years ago
10

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 11

2 m. What is the length of the original rectangle?
Mathematics
2 answers:
KengaRu [80]3 years ago
7 0

<u>Original</u>

Length (L): L

width (w): w

Perimeter (P) = 2L + 2w

               48 = 2(L) + 2(w)

              24 = L + w    

         24 - L = w

*******************************************

<u>New</u>

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



Pani-rosa [81]3 years ago
6 0

Answer:

The answer is 16

Step-by-step 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

<u>Before of changings</u>:

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

<u>After of changings</u>:

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

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