Question

an acute triangle has side lengths 21 cm, x cm, and 2x cm. if 21 is one of the shorter sides of the triangle, what is the greatest possible length of the longest side, rounded to the nearest tenth?
18.8 cm
24.2 cm
42.0 cm
72.7 cm

Explanation:

Step1: Apply the triangle - inequality theorem and the property of acute - angled triangles

For an acute - angled triangle with side lengths \(a\), \(b\), and \(c\) (\(c\) being the longest side), \(a^{2}+b^{2}>c^{2}\). Here, since 21 is one of the shorter sides and the side lengths are 21, \(x\), and \(2x\), and \(2x\) is the longest side, we have two conditions from the triangle - inequality theorem: \(21 + x>2x\) (which simplifies to \(x < 21\)) and from the acute - angle condition \(21^{2}+x^{2}>(2x)^{2}\).

Step2: Solve the inequality \(21^{2}+x^{2}>(2x)^{2}\)

Expand the inequality: \(441+x^{2}>4x^{2}\). Rearrange it to get \(3x^{2}<441\), then \(x^{2}<147\), so \(x < \sqrt{147}\approx12.12\). Also, from \(21 + x>2x\), we have \(x < 21\). The more restrictive condition is \(x < \sqrt{147}\).

Step3: Find the length of the longest side

The longest side is \(2x\). Since \(x < \sqrt{147}\), then \(2x<2\sqrt{147}\). Calculate \(2\sqrt{147}\approx2\times12.12 = 24.24\approx24.2\) cm.

Answer:

24.2 cm