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 condition for an acute triangle

For a triangle with side lengths \(a\), \(b\), \(c\) (\(c\) is the longest side), the triangle is acute if \(a^{2}+b^{2}>c^{2}\). Here, \(a = x\), \(b=x + 4\), \(c = 20\). So we need \(x^{2}+(x + 4)^{2}>20^{2}\).
Expand \((x + 4)^{2}=x^{2}+8x + 16\). Then the inequality becomes \(x^{2}+x^{2}+8x + 16>400\), or \(2x^{2}+8x-384>0\), divide by \(2\) to get \(x^{2}+4x-192>0\).
Factor the quadratic: \(x^{2}+4x-192=(x + 16)(x - 12)>0\). The roots are \(x=-16\) and \(x = 12\). The solution of the inequality \(x^{2}+4x-192>0\) for \(x>0\) (since side - length \(x>0\)) is \(x>12\).
Also, recall the triangle - inequality: \(x+(x + 4)>20\) (sum of two shorter sides is greater than the longest side). Simplify \(2x+4>20\) gives \(2x>16\) or \(x > 8\).

Step2: Check each option

• For \(x = 8\):

\(a = 8\), \(b=8 + 4=12\), \(c = 20\). Then \(a^{2}+b^{2}=8^{2}+12^{2}=64 + 144=208\), \(c^{2}=400\). Since \(208<400\), it is an obtuse triangle.

• For \(x = 10\):

\(a = 10\), \(b=10 + 4=14\), \(c = 20\). Then \(a^{2}+b^{2}=10^{2}+14^{2}=100+196 = 296\), \(c^{2}=400\). Since \(296<400\), it is an obtuse triangle.

• For \(x = 12\):

\(a = 12\), \(b=12 + 4=16\), \(c = 20\). Then \(a^{2}+b^{2}=12^{2}+16^{2}=144 + 256=400\), \(c^{2}=400\). Since \(a^{2}+b^{2}=c^{2}\), it is a right - triangle.

• For \(x = 14\):

\(a = 14\), \(b=14 + 4=18\), \(c = 20\). Then \(a^{2}+b^{2}=14^{2}+18^{2}=196+324 = 520\), \(c^{2}=400\). Since \(520>400\), it is an acute triangle.

Answer:

14