Question

determine whether each set of numbers can be measure of the sides of a triangle. if so, classify the triangle as acute, obtuse, or right. justify your answer. 11. 10, 11, 20 12. 12, 14, 49 13. 5√2, 10, 11 14. 21.5, 24, 55.5 15. 30, 40, 50 16. 65, 72, 97

Explanation:

Step1: Check triangle inequality (for 10, 11, 20)

Triangle inequality: \(a + b>c\), \(a + c>b\), \(b + c>a\).
For \(a = 10\), \(b = 11\), \(c = 20\):
\(10+11=21>20\), \(10 + 20=30>11\), \(11+20 = 31>10\). So, it is a triangle.
Now, check the type. Let \(c\) be the longest side. Use \(a^{2}+b^{2}\) and \(c^{2}\)
\(a^{2}+b^{2}=10^{2}+11^{2}=100 + 121=221\)
\(c^{2}=20^{2}=400\)
Since \(a^{2}+b^{2}

Step2: Check triangle inequality (for 12,14,49)

\(12+14=26<49\). So, it is not a triangle.

Step3: Check triangle inequality (for \(5\sqrt{2}\), 10, 11)

\(5\sqrt{2}\approx5\times1.414 = 7.07\)
\(5\sqrt{2}+10\approx7.07+10 = 17.07>11\), \(5\sqrt{2}+11\approx7.07 + 11=18.07>10\), \(10 + 11=21>5\sqrt{2}\)
Now, check the type. Let \(c = 11\), \(a=5\sqrt{2}\), \(b = 10\)
\(a^{2}+b^{2}=(5\sqrt{2})^{2}+10^{2}=50+100 = 150\)
\(c^{2}=121\)
Since \(a^{2}+b^{2}>c^{2}\), the triangle is acute.

Step4: Check triangle inequality (for 21.5,24,55.5)

\(21.5+24=45.5<55.5\). So, it is not a triangle.

Step5: Check triangle inequality (for 30,40,50)

\(30 + 40=70>50\), \(30+50=80>40\), \(40 + 50=90>30\)
Check the type. Let \(c = 50\), \(a = 30\), \(b=40\)
\(a^{2}+b^{2}=30^{2}+40^{2}=900+1600=2500\)
\(c^{2}=2500\)
Since \(a^{2}+b^{2}=c^{2}\), the triangle is right - angled.

Step6: Check triangle inequality (for 65,72,97)

\(65+72=137>97\), \(65 + 97=162>72\), \(72+97=169>65\)
Check the type. Let \(c = 97\), \(a = 65\), \(b = 72\)
\(a^{2}+b^{2}=65^{2}+72^{2}=4225+5184=9409\)
\(c^{2}=9409\)
Since \(a^{2}+b^{2}=c^{2}\), the triangle is right - angled.

Answer:

1. It is a triangle. It is obtuse.
2. It is not a triangle.
3. It is a triangle. It is acute.
4. It is not a triangle.
5. It is a triangle. It is right - angled.
6. It is a triangle. It is right - angled.