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

Explanation:

Step1: Apply the triangle inequality theorem

For a triangle with sides \(a\), \(b\), \(c\) (\(c\) is the longest side), \(a + b>c\). Here \(a=x\), \(b = 3x\), \(c = 20\). So \(x+3x>20\), which simplifies to \(4x>20\), and \(x > 5\).

Step2: Apply the obtuse - triangle condition

For an obtuse triangle with the longest side \(c\), \(a^{2}+b^{2}Expand the left - hand side: \(x^{2}+9x^{2}<400\), \(10x^{2}<400\), \(x^{2}<40\), \(x<\sqrt{40}\approx6.32\).

Answer:

A. \(6.3\)