In triangle ABC, two medians AD and BE intersect at G at right angles. If AD = 12 cm and BE = 9 cm, find the length of AB.

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Let the triangle be ABC, with medians AD and BE intersecting at G (the centroid) at right angles. Let D be the midpoint of BC, and E be the midpoint of AC. The centroid G divides each median in the ratio 2:1, counting from the vertex.

Let AG = 2x, GD = x, so AD = AG + GD = 3x = 12 cm ⇒ x = 4 cm. So AG = 8 cm, GD = 4 cm.
Similarly, BE = 9 cm. Let BG = 2y, GE = y, so BE = 3y = 9 cm ⇒ y = 3 cm. So BG = 6 cm, GE = 3 cm.

At G, the medians are perpendicular. Consider the triangle formed by the medians at G:
Let’s place G at the origin (0,0), AG along the x-axis, and BG along the y-axis. Let’s denote:
- A at (8, 0)
- B at (0, 6)
- D and E are midpoints of BC and AC, respectively.

Let’s find the coordinates of C.
Let C be at (x, y).

Since D is the midpoint of BC, and G divides AD in 2:1 from A to D:
- D = midpoint of B (0,6) and C (x, y) = ((0+x)/2, (6+y)/2)
- AG = 8, GD = 4, so AD = 12, and G divides AD in 2:1 from A.
- The centroid G divides AD in the ratio AG:GD = 2:1, so
  G = (2D + A)/3
  But since G is at (0,0),
  (0,0) = (2D + A)/3
  ⇒ 2D + A = (0,0)
  ⇒ 2D = (-A)
  ⇒ D = (-A)/2 = (-8/2, 0/2) = (-4, 0)

But D is also the midpoint of B and C:
  D = ((0+x)/2, (6+y)/2) = (-4, 0)
  ⇒ (0+x)/2 = -4 ⇒ x = -8
     (6+y)/2 = 0 ⇒ y = -6
So, C is at (-8, -6)

Now, AB = distance between A (8,0) and B (0,6):
AB = sqrt[(8-0)^2 + (0-6)^2] = sqrt[64 + 36] = sqrt[100] = 10 cm

Thus, the length of AB is 10 cm.

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