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

Explanation:

Step1: Apply the acute - triangle inequality

For an acute triangle with sides \(a\), \(b\), \(c\) (\(c\) is the longest side), \(a^{2}+b^{2}>c^{2}\). Here \(a = x\), \(b = 2x\), \(c = 15\). So \(x^{2}+(2x)^{2}>15^{2}\).

Step2: Simplify the inequality

Expand the left - hand side: \(x^{2}+4x^{2}>225\), which is \(5x^{2}>225\). Divide both sides by 5: \(x^{2}>45\).

Step3: Solve for \(x\)

Take the square root of both sides. Since \(x>0\), \(x>\sqrt{45}\approx6.71\).

Answer:

7