The geometric construction of a golden mean rectangle is shown in the figure below. It starts with a square ABCD of 1 cm side. Continuing the development of the golden mean rectangles, the derived length of IJ (in cm) is (rounded off to three decimal places).
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Each new golden rectangle forms by adding a square on the longer side of the previous one; use phi^2 = phi + 1 to track the growth, or track it stage by stage with coordinates.
Step 1: Place the square on a grid.
Let $D = (0,0)$, $C = (1,0)$, $A = (0,1)$, $B = (1,1)$, matching the marked 1 cm side $BC$. The midpoint of $AB$ is $O = (0.5, 1)$.
Step 2: Compute the arc length OD directly.
$$ OD = \sqrt{0.5^2 + 1^2} = \sqrt{1.25} \approx 1.118034 $$
The construction swings this length from $D$ around to a point $F$ on the line through $A$ and $B$, extended past $A$, at the same height as $A$ and $B$.
Step 3: Get the width of rectangle FBCE numerically.
$$ AF = OD - OA = 1.118034 - 0.5 = 0.618034 $$
$$ FB = AF + AB = 0.618034 + 1 = 1.618034 $$
So rectangle $FBCE$ measures $1.618034$ cm by $1$ cm.
Step 4: Add the next square numerically (on top, to G and H).
A square of side equal to the longer side of $FBCE$, $1.618034$ cm, sits on top of it. The new rectangle's height becomes
$$ 1 + 1.618034 = 2.618034 \text{ cm} $$
while its width stays $1.618034$ cm.
Step 5: Add the final square numerically (to the right, to I and J).
Now the longer side is the height, $2.618034$ cm, so a square of that side is added to the right. This new square's own side, $2.618034$ cm, becomes the vertical distance between its top corner $I$ and bottom corner $J$, since the square's full height matches the existing rectangle's height.
Step 6: Final answer.
$$ IJ = 2.618034 \text{ cm} \approx 2.618 \text{ cm (to three decimal places)} $$
$$ \boxed{IJ = 2.618} $$
