ABC is a triangle right-angled at C. Let P be the midpoint of BC. If AP = 4√13 cm and AB = 20 cm, then what is the perimeter of the triangle ABC?

Let BC = 2a (so PC = a) and AC = b. In right triangle ACP: AP² = AC² + PC², so (4√13)² = b² + a², i.e., 208 = b² + a². In right triangle ACB: AB² = AC² + BC², so 400 = b² + 4a². Subtracting: 400 − 208 = 3a², so 3a² = 192, a² = 64, a = 8. So BC = 16. Then b² = 208 − 64 = 144, b = 12. So AC = 12, BC = 16, AB = 20 — a 3-4-5 multiple. Perimeter = 12 + 16 + 20 = 48 cm.