
A rectangle is constructed. The length of this rectangle is double the width. Then a new rectangle is made by increasing each side by 3 meters. The perimeter of the new rectangle is 2 meters greater than four times the length of the old rectangle. Find the dimensions of the original rectangle
first rectangle:
L = 2x
W = x
second rectangle:
L = 2x + 3
W = x + 3
P = 2(2x+3) + 2(x+3) = 4x+6+2x+6 = 6x+12
L = length
W = width
P = perimeter
P (second rectangle) = 4 * (Length of first rectangle) + 2
this becomes:
6x+12 = 4(2x) + 2 which becomes:
6x+12 = 8x+2
subtract 6x from both sides of the equation to get:
12 = 2x + 2
subtract 2 from both sides of the equation to get:
10 = 2x
divide both sides of the equation by 2 to get:
x = 5
when x = 5:
length of first rectangle becomes 2x = 10
width of first rectangle becomes x = 5
length of second rectangle becomes 2x + 3 = 13
width of second rectangle becomes x + 3 = 8
perimeter of second rectangle becomes 2*13 + 2*8 = 26+16 = 42
4 times length of first rectangle plus 2 becomes 4*10 + 2 = 42
requirements of problem are satisfied.
answer is that the dimensions of the original rectangle are:
length = 10
width = 5
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