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 acute - triangle inequality

For an acute triangle with side lengths \(a\), \(b\), \(c\) (\(c\) is the longest side), \(a^{2}+b^{2}>c^{2}\).
Since \(21\) is one of the shorter sides, assume \(a = 21\), \(b=x\), \(c = 2x\). Then \(21^{2}+x^{2}>(2x)^{2}\).

Step2: Solve the inequality

Expand the inequality: \(441+x^{2}>4x^{2}\).
Rearrange terms: \(441>4x^{2}-x^{2}\), so \(441 > 3x^{2}\).
Divide both sides by \(3\): \(x^{2}<147\), then \(x<\sqrt{147}\approx12.12\).
Also, consider the triangle - inequality \(x + 21>2x\) (sum of two sides is greater than the third side), which gives \(x<21\).
Another case: assume \(a = 21\), \(b = 2x\), \(c=x\) (but \(x<2x\) when \(x>0\), so this case is not valid for the longest side).
Now, if we consider the other acute - triangle inequality when the sides are \(21\), \(2x\), \(x\) (with \(2x\) as the longest side). Also, from the triangle inequality \(21 + x>2x\) (holds for \(x < 21\)) and \(21+2x>x\) (holds for all \(x>0\)) and \(x + 2x>21\) (i.e., \(x>7\)).
We use the acute - triangle formula \(21^{2}+(2x)^{2}>x^{2}\) (always true for \(x>0\)).
We also use the fact that from \(21^{2}+x^{2}>(2x)^{2}\), \(x^{2}<147\).
If we consider the upper - bound of \(x\) from \(x + 21>2x\) (triangle inequality) and the acute - triangle inequality.
Let's use the acute - triangle formula \(21^{2}+x^{2}=(2x)^{2}+\epsilon\) (\(\epsilon>0\)). When we want to find the maximum value of \(2x\) (the longest side).
From \(21^{2}+x^{2}>4x^{2}\), we can also use the fact that if we consider the boundary case (right - triangle, since acute triangle is \(a^{2}+b^{2}>c^{2}\), right - triangle is \(a^{2}+b^{2}=c^{2}\)).
Let \(21^{2}+x^{2}=(2x)^{2}\), \(441+x^{2}=4x^{2}\), \(3x^{2}=441\), \(x^{2}=147\), \(x=\sqrt{147}\approx12.12\). But we also know from \(x + 21>2x\) (triangle inequality) \(x<21\).
Let's use another approach. For an acute triangle with sides \(a = 21\), \(b=x\), \(c = 2x\)
\(a^{2}+b^{2}>c^{2}\) and \(a + b>c\), \(a + c>b\), \(b + c>a\)
\(441+x^{2}>4x^{2}\Rightarrow3x^{2}<441\Rightarrow x^{2}<147\)
Also, from \(x + 21>2x\Rightarrow x<21\)
We can check the options:
If the longest side is \(2x\)
Option B: If \(2x = 24.2\), then \(x = 12.1\)
Check \(21^{2}+12.1^{2}=441+146.41 = 587.41\) and \((2\times12.1)^{2}=585.64\)
Since \(21^{2}+12.1^{2}>(2\times12.1)^{2}\) and \(21 + 12.1>24.2\) (\(33.1>24.2\))

Answer:

B. \(24.2\space cm\)