A rectangle with a size of 1x1 cells is drawn on a checkered paper. Find the length of its longest median line.

The problem is asking to find the length of the longest median line of a right-angled triangle drawn on a grid. The grid cells are 1x1.

Looking at the image, we can see a right-angled triangle drawn on a grid. Let's determine the lengths of the two legs of the right-angled triangle. The horizontal leg spans across 4 cells, so its length is 4 units. The vertical leg spans across 3 cells, so its length is 3 units.

In a right-angled triangle, the median to the hypotenuse is half the length of the hypotenuse. The other two medians are drawn from the vertices of the right angles to the midpoints of the opposite sides.

Let the vertices of the triangle be A=(0,3), B=(4,0), and C=(0,0). The lengths of the legs are AC = 3 and BC = 4.

The hypotenuse AB has a length calculated by the Pythagorean theorem: $$AB = \sqrt{AC^2 + BC^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$.

The midpoint of AC is M_AC = (0, 1.5). The median from B to AC is BM_AC. The length of BM_AC is the distance between B(4,0) and M_AC(0, 1.5), which is $$BM_{AC} = \sqrt{(4-0)^2 + (0-1.5)^2} = \sqrt{16 + 2.25} = \sqrt{18.25}$$.

The midpoint of BC is M_BC = (2, 0). The median from A to BC is AM_BC. The length of AM_BC is the distance between A(0,3) and M_BC(2, 0), which is $$AM_{BC} = \sqrt{(0-2)^2 + (3-0)^2} = \sqrt{4 + 9} = \sqrt{13}$$.

The midpoint of the hypotenuse AB is M_AB = $$(rac{0+4}{2}, rac{3+0}{2}) = (2, 1.5)$$. The median from C to AB is CM_AB. The length of CM_AB is the distance between C(0,0) and M_AB(2, 1.5), which is $$CM_{AB} = \sqrt{(2-0)^2 + (1.5-0)^2} = \sqrt{4 + 2.25} = \sqrt{6.25} = 2.5$$.

Comparing the lengths of the medians: $$\sqrt{18.25}$$, $$\sqrt{13}$$, and $$2.5$$. We need to find the largest one.

$$\sqrt{18.25} ≈ 4.272$$

$$\sqrt{13} ≈ 3.606$$

$$2.5$$

The longest median is $$\sqrt{18.25}$$.

The question asks for the length of the longest median line. The medians of a right triangle have lengths $$m_a = \sqrt{b^2 + (a/2)^2}$$, $$m_b = \sqrt{a^2 + (b/2)^2}$$, and $$m_c = c/2$$, where $$a$$ and $$b$$ are the lengths of the legs and $$c$$ is the length of the hypotenuse.

In our case, $$a=4$$ and $$b=3$$. So, $$c=5$$.

$$m_a = \sqrt{3^2 + (4/2)^2} = \sqrt{9 + 2^2} = \sqrt{9+4} = \sqrt{13}$$.

$$m_b = \sqrt{4^2 + (3/2)^2} = \sqrt{16 + (1.5)^2} = \sqrt{16+2.25} = \sqrt{18.25}$$.

$$m_c = 5/2 = 2.5$$.

Comparing the lengths: $$\sqrt{13} ≈ 3.606$$, $$\sqrt{18.25} ≈ 4.272$$, $$2.5$$.

The longest median is $$\sqrt{18.25}$$. The question might be asking for a value that can be directly measured on the grid, or a simplified radical form, or a decimal approximation. Given the context, it's likely looking for the exact value. The question states