Question 970
Write down the given information:
A piece of wire is bent into shape of rectangle. So, length of the wire is same as the perimeter of rectangle. 
Recall that the perimeter of a figure is the sum of the measures of its sides.
Let the length of the rectangle be L and the width be W.
The sum of the measures of the sides of rectangle is: L + W + L + W
= 2L + 2W 
= 2(L + W)
But the perimeter of the rectangle is 38m.
So, 2(L + W) = 38.            -------- Equation (1)
Also given that the length is twice the width.
So, L = 2W.                   -------- Equation (2)

Substitute the equation 2 in equation 1:
2(2W + W) = 38
2(3W) = 38
6W = 38
Divide by 6 on both sides:
W is approximately equal to 6.33.

Substitute 6.33 for W in the equation 2:
L = 2(6.33) 
=12.66
L is approximately equal to 12.66.

So, the approximate dimensions of the rectangle are 6.33m and 12.66m.
The final answer can also be expressed as fractions: 17/3 and 38/3