Question

The longest side of an obtuse triangle measures 20 cm. The two shorter sides measure x cm and 3x cm.

Rounded to the nearest tenth, what is the greatest possible value of x?

6.3
6.4
7.0
7.1

Answer 1

Let the longest side be c = 20 and the other sides be x and 3x. For the triangle to be obtuse with 20 as the longest side, we need
20^2 > x^2 + (3x)^2 = 10x^2,
so x^2 < 40 and x < 2√10 ≈ 6.3249. Also the triangle inequality x + 3x > 20 gives x > 5.

Thus x can approach 2√10 from below, and the greatest possible value (rounded to the nearest tenth) is 2√10 ≈ 6.3249 → 6.3.

Answer: 6.3