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

Explanation

Let AC = b and BC = a. Thus CP = a/2. In right \triangle ACP, b^2 + a^2/4 = (4\sqrt{13})^2 = 208. In right \triangle ACB, b^2 + a^2 = 20^2 = 400. Subtracting gives \frac{3a^2}{4} = 192, so a = 16. Substituting back, b^2 + 256 = 400, so b = 12. Perimeter = 16 + 12 + 20 = 48 cm.

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