Question

The shorter sides of an acute triangle are x cm and 2x cm. The longest side of the triangle is 15 cm.

What is the smallest possible whole-number value of x?

6
7
8
9

Answer 1

Let the sides be x, 2x, and 15 with 15 the longest. Triangle inequality gives x + 2x > 15 so x > 5, and 2x < 15 so x < 7.5. For the triangle to be acute, the square of the longest side must be less than the sum of the squares of the other two:
15^2 < x^2 + (2x)^2 = 5x^2, so x^2 > 45 and x > √45 ≈ 6.708.

Thus x must be > 6.708 and < 7.5, so the smallest whole-number value is 7.