Question

determine if a triangle can be formed with the given lengths. if so, classify the triangle as acute, right or obtuse

1. 7, 20, and 12 yes or no classify:
2. 15, 8, and 17 yes or no classify:
3. 12, 10, and 8 yes or no classify:
4. 20, 8, and 19 yes or no classify:
5. 16, 30, and 34 yes or no classify:
6. 80, 71, and 5 yes or no classify:

Explanation:

Step1: Triangle Inequality Theorem

For three side lengths \(a\), \(b\), \(c\) (\(c\geq a,b\)), a triangle can be formed if \(a + b>c\).

Problem 1: \(a = 7\), \(b = 12\), \(c = 20\)

\(7+12=19<20\). So, no triangle.

Problem 2: \(a = 8\), \(b = 15\), \(c = 17\)

\(8 + 15=23>17\). Then check type: \(a^{2}+b^{2}=8^{2}+15^{2}=64 + 225=289\), \(c^{2}=17^{2}=289\). Since \(a^{2}+b^{2}=c^{2}\), right - triangle.

Problem 3: \(a = 8\), \(b = 10\), \(c = 12\)

\(8 + 10=18>12\). \(a^{2}+b^{2}=8^{2}+10^{2}=64+100 = 164\), \(c^{2}=12^{2}=144\). Since \(a^{2}+b^{2}>c^{2}\), acute - triangle.

Problem 4: \(a = 8\), \(b = 19\), \(c = 20\)

\(8+19 = 27>20\). \(a^{2}+b^{2}=8^{2}+19^{2}=64 + 361=425\), \(c^{2}=20^{2}=400\). Since \(a^{2}+b^{2}>c^{2}\), acute - triangle.

Problem 5: \(a = 16\), \(b = 30\), \(c = 34\)

\(16+30=46>34\). \(a^{2}+b^{2}=16^{2}+30^{2}=256+900 = 1156\), \(c^{2}=34^{2}=1156\). Since \(a^{2}+b^{2}=c^{2}\), right - triangle.

Problem 6: \(a = 5\), \(b = 71\), \(c = 80\)

\(5+71=76<80\). So, no triangle.

Answer:

1. NO
2. YES, Right
3. YES, Acute
4. YES, Acute
5. YES, Right
6. NO