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: Recall the acute - triangle inequality

For a triangle with side lengths \(a\), \(b\), and \(c\) where \(c\) is the longest side, the triangle is acute if \(a^{2}+b^{2}>c^{2}\). Here, \(c = 20\), \(a=x\), and \(b=x + 4\).

Step2: Substitute into the inequality

We get \(x^{2}+(x + 4)^{2}>20^{2}\). Expand \((x + 4)^{2}=x^{2}+8x + 16\). So the inequality becomes \(x^{2}+x^{2}+8x + 16>400\).

Step3: Simplify the inequality

Combine like - terms: \(2x^{2}+8x+16>400\), then \(2x^{2}+8x - 384>0\). Divide through by 2: \(x^{2}+4x-192>0\).

Step4: Solve the quadratic inequality

Factor the quadratic: \(x^{2}+4x - 192=(x + 16)(x - 12)>0\). The roots of the quadratic equation \(x^{2}+4x - 192 = 0\) are \(x=-16\) and \(x = 12\). The solution to the inequality \((x + 16)(x - 12)>0\) is \(x<-16\) or \(x>12\). Also, considering the triangle - side - length condition \(x>0\) and \(x+(x + 4)>20\) (the sum of two shorter sides must be greater than the longest side), \(2x+4>20\), \(2x>16\), \(x > 8\).

Step5: Check the values from the list

From the values \(8\), \(10\), \(12\), and \(14\), when \(x = 14\), \(a = 14\), \(b=14 + 4=18\), and \(a^{2}+b^{2}=14^{2}+18^{2}=196 + 324=520\) and \(c^{2}=20^{2}=400\), and \(a^{2}+b^{2}>c^{2}\).

Answer:

14