The length of a hypotenuse of a right angled triangle exceeds the length of its base by 2 cm and exceeds twice the length of the altitude by 1 cm. Find the perimeter of the triangle.

Let base = b, altitude = a, hypotenuse = h. Given: h = b + 2 and h = 2a + 1. From Pythagoras: h^2 = b^2 + a^2. Substituting h = b + 2: (b + 2)^2 = b^2 + a^2, giving b^2 + 4b + 4 = b^2 + a^2, so a^2 = 4b + 4. Substituting a = (h - 1)/2: ((h - 1)/2)^2 = 4(h - 2) + 4. This gives h^2 - 2h + 1 = 16h - 64 + 16, or h^2 - 18h + 49 = 0. Testing integer solutions: h = 26 gives 676 - 468 + 49 = 257 ≠ 0. Let's verify: if h = 26, b = 24, a = 10. Then h^2 = 676 and b^2 + a^2 = 576 + 100 = 676. Perimeter = 24 + 10 + 26 = 60 cm. Wait - let me recheck. Actually: h = b + 2, h = 2a + 1 → 2a + 1 = b + 2 → b = 2a - 1. h^2 = b^2 + a^2 → (2a + 1)^2 = (2a - 1)^2 + a^2 → 4a^2 + 4a + 1 = 4a^2 - 4a + 1 + a^2 → 8a = a^2 → a = 8. So h = 17, b = 15. Perimeter = 17 + 15 + 8 = 40 cm ✓