Question

for each set of side lengths, state whether it is an acute triangle, right triangle, obtuse triangle, or impossible to make a triangle. select one choice per row. a 11, 60, 61 b 11, 60, 70 c 11, 60, 55 d 11, 60, 75 e 11, 60, 50 f 11, 60, 30

Explanation:

Step1: Check triangle inequality

For three side lengths \(a,b,c\) (where \(c\) is the longest side), if \(a + b>c\), a triangle can be formed.

Case a: \(11,60,61\)

\(11+60 = 71>61\). Then check \(a^{2}+b^{2}\) and \(c^{2}\). \(11^{2}+60^{2}=121 + 3600=3721\), \(61^{2}=3721\). Since \(a^{2}+b^{2}=c^{2}\), it is a right - triangle.

Case b: \(11,60,70\)

\(11 + 60=71>70\). \(11^{2}+60^{2}=3721\), \(70^{2}=4900\). Since \(a^{2}+b^{2}

Case c: \(11,60,55\)

\(11+55 = 66>60\). \(11^{2}+55^{2}=121+3025 = 3146\), \(60^{2}=3600\). Since \(a^{2}+b^{2}

Case d: \(11,60,75\)

\(11+60=71<75\). So it is impossible to make a triangle.

Case e: \(11,60,50\)

\(11 + 50=61>60\). \(11^{2}+50^{2}=121 + 2500=2621\), \(60^{2}=3600\). Since \(a^{2}+b^{2}

Case f: \(11,60,30\)

\(11+30 = 41<60\). So it is impossible to make a triangle.

Answer:

a. right; b. obtuse; c. obtuse; d. impossible; e. obtuse; f. impossible