Question

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

Explanation:

To determine if a triangle with side lengths \(a\), \(b\), \(c\) (where \(c\) is the longest side) is acute, we use the Pythagorean inequality for acute triangles: \(a^{2}+b^{2}>c^{2}\). We will check each option:

Step 1: Check option 4, 5, 7

Longest side \(c = 7\). Calculate \(a^{2}+b^{2}\) and \(c^{2}\):
\(a^{2}+b^{2}=4^{2}+5^{2}=16 + 25=41\)
\(c^{2}=7^{2} = 49\)
Since \(41<49\) (i.e., \(a^{2}+b^{2}

Step 2: Check option 5, 7, 8

Longest side \(c = 8\). Calculate \(a^{2}+b^{2}\) and \(c^{2}\):
\(a^{2}+b^{2}=5^{2}+7^{2}=25 + 49 = 74\)
\(c^{2}=8^{2}=64\)
Since \(74>64\) (i.e., \(a^{2}+b^{2}>c^{2}\)), the triangle is acute. We can stop here, but let's check the other options for completeness.

Step 3: Check option 6, 7, 10

Longest side \(c = 10\). Calculate \(a^{2}+b^{2}\) and \(c^{2}\):
\(a^{2}+b^{2}=6^{2}+7^{2}=36+49 = 85\)
\(c^{2}=10^{2}=100\)
Since \(85<100\) (i.e., \(a^{2}+b^{2}

Step 4: Check option 7, 9, 12

Longest side \(c = 12\). Calculate \(a^{2}+b^{2}\) and \(c^{2}\):
\(a^{2}+b^{2}=7^{2}+9^{2}=49 + 81=130\)
\(c^{2}=12^{2}=144\)
Since \(130<144\) (i.e., \(a^{2}+b^{2}

Answer:

B. 5, 7, 8