Question

which value from the list below, when substituted for x, would represent an acute triangle with side lengths x, x + 4, and 20? assume that the longest side of the triangle is of length 20 units. 8 10 12 14

Explanation:

Step1: Apply the triangle inequality theorem

For a triangle with sides \(a\), \(b\), \(c\) (\(c\) is the longest side), the triangle inequality theorem states \(a + b>c\). Here \(a=x\), \(b = x + 4\), \(c = 20\). So \(x+(x + 4)>20\), which simplifies to \(2x+4>20\), then \(2x>16\), and \(x > 8\).

Step2: Apply the acute - triangle condition

For an acute triangle with sides \(a\), \(b\), \(c\) (\(c\) is the longest side), \(a^{2}+b^{2}>c^{2}\). Substitute \(a=x\), \(b=x + 4\), \(c = 20\) into \(a^{2}+b^{2}>c^{2}\). We get \(x^{2}+(x + 4)^{2}>20^{2}\). Expand \((x + 4)^{2}=x^{2}+8x + 16\). Then \(x^{2}+x^{2}+8x + 16>400\), \(2x^{2}+8x-384>0\), \(x^{2}+4x-192>0\).
Factor the quadratic equation \(x^{2}+4x - 192=(x + 16)(x - 12)>0\). The roots of the quadratic equation \(y=x^{2}+4x-192\) are \(x=-16\) and \(x = 12\). The solution of the inequality \(x^{2}+4x-192>0\) is \(x<-16\) or \(x>12\).
Combining with \(x>8\) (from the triangle - inequality), we consider the values from the list.

Answer:

\(14\)