Question

which set of numbers can represent the side lengths, in inches, of an acute triangle? 4,5,7 5,7,8 6,7,10 7,9,12

Explanation:

Step1: Recall 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 acute - triangle, \(a^{2}+b^{2}>c^{2}\), where \(c\) is the longest side.

Step2: Check option 4, 5, 7

Let \(a = 4\), \(b = 5\), \(c = 7\).
First, check the triangle - inequality: \(4 + 5=9>7\), \(4+7 = 11>5\), \(5 + 7=12>4\).
Then, check for acute - triangle: \(a^{2}+b^{2}=4^{2}+5^{2}=16 + 25=41\), \(c^{2}=7^{2}=49\). Since \(41<49\), it is not an acute - triangle.

Step3: Check option 5, 7, 8

Let \(a = 5\), \(b = 7\), \(c = 8\).
Triangle - inequality: \(5 + 7=12>8\), \(5+8 = 13>7\), \(7 + 8=15>5\).
For acute - triangle: \(a^{2}+b^{2}=5^{2}+7^{2}=25 + 49=74\), \(c^{2}=8^{2}=64\). Since \(74>64\), it is an acute - triangle.

Step4: Check option 6, 7, 10

Let \(a = 6\), \(b = 7\), \(c = 10\).
Triangle - inequality: \(6 + 7=13>10\), \(6+10 = 16>7\), \(7 + 10=17>6\).
For acute - triangle: \(a^{2}+b^{2}=6^{2}+7^{2}=36 + 49=85\), \(c^{2}=10^{2}=100\). Since \(85<100\), it is not an acute - triangle.

Step5: Check option 7, 9, 12

Let \(a = 7\), \(b = 9\), \(c = 12\).
Triangle - inequality: \(7 + 9=16>12\), \(7+12 = 19>9\), \(9 + 12=21>7\).
For acute - triangle: \(a^{2}+b^{2}=7^{2}+9^{2}=49+81 = 130\), \(c^{2}=12^{2}=144\). Since \(130<144\), it is not an acute - triangle.

Answer:

5, 7, 8