Question

right triangle relationships and trigonometry
which set of numbers can represent the side lengths, in millimeters, of an obtuse triangle?
8, 10, 14
12, 15, 19
8, 12, 15
10, 14, 17

Explanation:

Step1: Recall the triangle - inequality theorem

For three side - lengths \(a\), \(b\), and \(c\) of a triangle, \(a + b>c\), \(a + c>b\), and \(b + c>a\). Also, for an obtuse triangle, if \(c\) is the longest side, then \(a^{2}+b^{2}

Step2: Check the first set \(a = 8\), \(b = 10\), \(c = 14\)

First, check the triangle - inequality: \(8 + 10=18>14\), \(8 + 14 = 22>10\), \(10+14 = 24>8\). Then, check the obtuse - triangle condition: \(a^{2}+b^{2}=8^{2}+10^{2}=64 + 100=164\), \(c^{2}=14^{2}=196\). Since \(164<196\), it is an obtuse triangle.

Step3: Check the second set \(a = 12\), \(b = 15\), \(c = 19\)

Triangle - inequality: \(12 + 15=27>19\), \(12+19 = 31>15\), \(15 + 19=34>12\). Obtuse - triangle condition: \(a^{2}+b^{2}=12^{2}+15^{2}=144 + 225=369\), \(c^{2}=19^{2}=361\). Since \(369>361\), it is an acute triangle.

Step4: Check the third set \(a = 8\), \(b = 12\), \(c = 15\)

Triangle - inequality: \(8 + 12=20>15\), \(8 + 15=23>12\), \(12 + 15=27>8\). Obtuse - triangle condition: \(a^{2}+b^{2}=8^{2}+12^{2}=64+144 = 208\), \(c^{2}=15^{2}=225\). Since \(208<225\), it is an obtuse triangle.

Step5: Check the fourth set \(a = 10\), \(b = 14\), \(c = 17\)

Triangle - inequality: \(10 + 14=24>17\), \(10+17 = 27>14\), \(14 + 17=31>10\). Obtuse - triangle condition: \(a^{2}+b^{2}=10^{2}+14^{2}=100 + 196=296\), \(c^{2}=17^{2}=289\). Since \(296>289\), it is an acute triangle.

Answer:

8, 10, 14; 8, 12, 15