Question 493890
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 4 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