In an isosceles right-angled triangle, the median drawn to the hypotenuse is 2 cm less than the leg. Find the hypotenuse.

Let the length of the leg AC = X cm, then BC = X cm.

Let us determine the length of the hypotenuse AB.

AB ^ 2 = AC ^ 2 + BC ^ 2 = X ^ 2 + X ^ 2 = 2 * X ^ 2.

AB = X * √2 cm.

The median CH drawn to the hypotenuse AB is also the height of the triangle ABC, then the triangles ASN and BCH are rectangular, and AH = BH = AH / 2 = X * √2 / 2 cm.

In a right-angled triangle ACN, AC ^ 2 = AH ^ 2 + CH ^ 2.

X ^ 2 = (X * √2 / 2) ^ 2 + (X – 2) ^ 2.

X ^ 2 = X ^ 2/2 + X ^ 2 – 4 * X + 4.

X ^ 2/2 – 4 * X + 4 = 0.

X ^ 2 – 8 * X + 8 = 0.

Having solved the quadratic equation, we get:

X = AC = 4 + √2 cm.

Then AB = (4 + √2) * √2 = 2 + 4 * √2 cm.

Answer: The length of the hypotenuse AB is (2 + 4 * √2) cm.