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 triangle - inequality for obtuse - angled triangle

In an obtuse - angled triangle with sides \(a\), \(b\), and \(c\) (\(c\) being the longest side), \(a^{2}+b^{2}20\) (the sum of two shorter sides must be greater than the longest side), so \(4x>20\) or \(x > 5\).

Step2: Solve the inequality \(x^{2}<40\)

Taking the square root of both sides of \(x^{2}<40\), we get \(-\sqrt{40}0\) (as it represents the length of a side), \(0 < x<\sqrt{40}\approx6.32\).

Step3: Combine with \(x > 5\)

Combining \(x>5\) and \(x<\sqrt{40}\approx6.32\), when rounded to the nearest tenth, the greatest possible value of \(x\) is \(6.3\).

Answer:

\(6.3\)