Game wrote:
If the three sides of a triangle are 3, 4, and 5 (or 5, 12, and 13) respectively:
Is it safe/correct to assume that the triangle is a right angled triangle?
I am just wondering if there is a way to prove that if the sides are as mentioned above, then one of the angles MUST be 90 degree
Yes, any triangle whose sides are in the ratio 3:4:5 is a right triangle. Basically, any triangle whose sides are in the ratio of Pythagorean Triples is a right triangle.
A Pythagorean triple consists of three positive integers a, b, and c, such that a^2 + b^2 = c^2, a well-known example is (3, 4, 5). If (a, b, c) is a Pythagorean triple, then so is (ka, kb, kc) for any positive integer k. There are 16 primitive Pythagorean triples with c ≤ 100:
(3, 4, 5) (5, 12, 13) (7, 24, 25) (8, 15, 17) (9, 40, 41) (11, 60, 61) (12, 35, 37) (13, 84, 85) (16, 63, 65) (20, 21, 29) (28, 45, 53) (33, 56, 65) (36, 77, 85) (39, 80, 89) (48, 55, 73) (65, 72, 97).
Also, say the lengths of the sides of a triangle are a, b, and c, where the largest side is c.
For a right triangle: \(a^2 +b^2= c^2\).
For an acute (a triangle that has all angles less than 90°) triangle: \(a^2 +b^2>c^2\).
For an obtuse (a triangle that has an angle greater than 90°) triangle: \(a^2 +b^2<c^2\).
Hope this helps.