Question

creating an acute triangle
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 \(c\) (\(c\) being the longest side), the triangle is acute if \(a^{2}+b^{2}>c^{2}\). Here, \(a = x\), \(b=x + 4\), and \(c = 20\). So we need \(x^{2}+(x + 4)^{2}>20^{2}\).

Step2: Expand the left - hand side

Expand \((x + 4)^{2}=x^{2}+8x + 16\). Then the inequality becomes \(x^{2}+x^{2}+8x + 16>400\), which simplifies to \(2x^{2}+8x+16 - 400>0\), or \(2x^{2}+8x - 384>0\). Divide through by 2 to get \(x^{2}+4x-192>0\).

Step3: Solve the quadratic inequality

Factor the quadratic \(x^{2}+4x - 192=(x + 16)(x - 12)\). The roots of the quadratic equation \(x^{2}+4x - 192 = 0\) are \(x=-16\) and \(x = 12\). The inequality \(x^{2}+4x - 192>0\) is satisfied when \(x<-16\) or \(x>12\). Also, for a triangle, we have the triangle - inequality conditions: \(x+(x + 4)>20\) (i.e., \(2x+4>20\) or \(x > 8\)), \(x+20>x + 4\) (which is always true), and \((x + 4)+20>x\) (which is also always true).

Step4: Check the options

Among the options \(8\), \(10\), \(12\), and \(14\), only \(x = 14\) satisfies \(x>12\).

Answer:

14