esanchez54 esanchez54
  • 06-08-2018
  • Mathematics
contestada

Find the area and perimeter of the following rectangle. The lengths are 4x+y+2; and 3x+3y. The widths are 2x+y+1 and 12

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nobillionaireNobley
nobillionaireNobley nobillionaireNobley
  • 06-08-2018

Equating the lengths, we have: [tex] 4x+y+2=3x+3y\Rightarrow x-2y=-2 [/tex]

Equating the widths, we have: [tex] 2x+y+1=12\Rightarrow2x+y=11 [/tex]

Solving the two equations simultaneously gives that x = 4 and y = 3.

Thus, the length is 4(4) + 3 + 2 = 21 and the width is 12.

Therefore area = length times width = 21 x 12 = 252 square units

perimeter = 2(length + width) = 2(21 + 12) = 2(33) = 66.

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