Question

exercises
determine whether each set of measures can be the measures of the sides of a triangle. if so, classify the triangle as acute, obtuse, or right. justify your answer.

1. 30, 40, 50
2. 20, 30, 40
3. 18, 24, 30
4. 6, 8, 9
5. 6, 12, 18
6. 10, 15, 20
7. \\( \sqrt { 5 } , \sqrt { 12 } , \sqrt { 13 } \\)
8. \\( 2 , \sqrt { 8 } , \sqrt { 12 } \\)
9. 9, 40, 41

Explanation:

Step1: Check the triangle inequality theorem

For a set of side lengths \(a,b,c\) (\(c\) is the longest side), the triangle inequality theorem states that \(a + b>c\).

• For \(30,40,50\): \(30 + 40=70>50\), \(30+50 = 80>40\), \(40 + 50=90>30\)
• For \(20,30,40\): \(20+30 = 50>40\), \(20 + 40=60>30\), \(30+40 = 70>20\)
• For \(18,24,30\): \(18+24=42>30\), \(18+30 = 48>24\), \(24+30=54>18\)
• For \(6,8,9\): \(6 + 8=14>9\), \(6+9 = 15>8\), \(8 + 9=17>6\)
• For \(6,12,18\): \(6+12=18\) (violates \(a + b>c\) as \(6 + 12

ot>18\))

• For \(10,15,20\): \(10+15=25>20\), \(10+20 = 30>15\), \(15+20=35>10\)
• For \(\sqrt{5},\sqrt{12},\sqrt{13}\): \(\sqrt{5}+\sqrt{12}\approx2.24+3.46 = 5.7>3.61\approx\sqrt{13}\), \(\sqrt{5}+\sqrt{13}\approx2.24 + 3.61=5.85>3.46\approx\sqrt{12}\), \(\sqrt{12}+\sqrt{13}\approx3.46+3.61 = 7.07>2.24\approx\sqrt{5}\)
• For \(2,\sqrt{8},\sqrt{12}\): \(2+\sqrt{8}\approx2 + 2.83=4.83>3.46\approx\sqrt{12}\), \(2+\sqrt{12}\approx2+3.46 = 5.46>2.83\approx\sqrt{8}\), \(\sqrt{8}+\sqrt{12}\approx2.83+3.46=6.29>2\)
• For \(9,40,41\): \(9+40=49>41\), \(9+41=50>40\), \(40+41=81>9\)

Step2: Use the Pythagorean theorem and its converse (\(c^{2}=a^{2}+b^{2}\) for right - triangle, \(c^{2}a^{2}+b^{2}\) for obtuse, where \(c\) is the longest side)

• For \(30,40,50\): \(30^{2}+40^{2}=900 + 1600=2500=50^{2}\), so it is a right - triangle.
• For \(20,30,40\): \(20^{2}+30^{2}=400+900 = 1300\), \(40^{2}=1600\), since \(1300<1600\), it is an obtuse - triangle.
• For \(18,24,30\): \(18^{2}+24^{2}=324+576 = 900=30^{2}\), so it is a right - triangle.
• For \(6,8,9\): \(6^{2}+8^{2}=36 + 64=100\), \(9^{2}=81\), since \(100>81\), it is an acute - triangle.
• \(6,12,18\) cannot form a triangle.
• For \(10,15,20\): \(10^{2}+15^{2}=100+225 = 325\), \(20^{2}=400\), since \(325<400\), it is an obtuse - triangle.
• For \(\sqrt{5},\sqrt{12},\sqrt{13}\): \((\sqrt{5})^{2}+(\sqrt{12})^{2}=5 + 12=17\), \((\sqrt{13})^{2}=13\), since \(17>13\), it is an acute - triangle.
• For \(2,\sqrt{8},\sqrt{12}\): \(2^{2}+(\sqrt{8})^{2}=4 + 8=12=(\sqrt{12})^{2}\), so it is a right - triangle.
• For \(9,40,41\): \(9^{2}+40^{2}=81+1600 = 1681=41^{2}\), so it is a right - triangle.

Answer:

1. Can be a triangle, right - triangle.
2. Can be a triangle, obtuse - triangle.
3. Can be a triangle, right - triangle.
4. Can be a triangle, acute - triangle.
5. Cannot be a triangle.
6. Can be a triangle, obtuse - triangle.
7. Can be a triangle, acute - triangle.
8. Can be a triangle, right - triangle.
9. Can be a triangle, right - triangle.