Question

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

Explanation:

Step1: Recall the acute - triangle inequality

For a triangle with side lengths \(a\), \(b\), and \(c\) (where \(c\) is the longest side), the triangle is acute if \(a^{2}+b^{2}>c^{2}\).

Step2: Check the set \(5,7,8\)

Let \(a = 5\), \(b = 7\), \(c = 8\). Then \(a^{2}+b^{2}=5^{2}+7^{2}=25 + 49=74\) and \(c^{2}=8^{2}=64\). Since \(74>64\), it could be an acute - triangle.

Step3: Check the set \(4,5,7\)

Let \(a = 4\), \(b = 5\), \(c = 7\). Then \(a^{2}+b^{2}=4^{2}+5^{2}=16 + 25=41\) and \(c^{2}=7^{2}=49\). Since \(41<49\), it is an obtuse - triangle.

Step4: Check the set \(7,9,12\)

Let \(a = 7\), \(b = 9\), \(c = 12\). Then \(a^{2}+b^{2}=7^{2}+9^{2}=49+81 = 130\) and \(c^{2}=12^{2}=144\). Since \(130<144\), it is an obtuse - triangle.

Step5: Check the set \(6,7,10\)

Let \(a = 6\), \(b = 7\), \(c = 10\). Then \(a^{2}+b^{2}=6^{2}+7^{2}=36 + 49=85\) and \(c^{2}=10^{2}=100\). Since \(85<100\), it is an obtuse - triangle.

Answer:

5, 7, 8