Question

which sets of side lengths represent pythagorean triples? choose three correct answers. 10, 24, 26 8, 15, 17 28, 45, 53 6, 8, 14 1, 2, 5 15, 20, 30

Explanation:

Step1: Check the first set \(10,24,26\)

According to the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\) (where \(c\) is the longest side).
\(10^{2}+24^{2}=100 + 576=676\), and \(26^{2}=676\). So \(10^{2}+24^{2}=26^{2}\)

Step2: Check the second set \(8,15,17\)

\(8^{2}+15^{2}=64+225 = 289\), and \(17^{2}=289\). So \(8^{2}+15^{2}=17^{2}\)

Step3: Check the third set \(28,45,53\)

\(28^{2}+45^{2}=784 + 2025=2809\), and \(53^{2}=2809\). So \(28^{2}+45^{2}=53^{2}\)

Step4: Check the fourth set \(6,8,14\)

\(6^{2}+8^{2}=36+64 = 100\), and \(14^{2}=196\). Since \(6^{2}+8^{2}
eq14^{2}\)

Step5: Check the fifth set \(1,2,5\)

\(1^{2}+2^{2}=1 + 4=5\), and \(5^{2}=25\). Since \(1^{2}+2^{2}
eq5^{2}\)

Step6: Check the sixth set \(15,20,30\)

\(15^{2}+20^{2}=225+400 = 625\), and \(30^{2}=900\). Since \(15^{2}+20^{2}
eq30^{2}\)

Answer:

\(10,24,26\); \(8,15,17\); \(28,45,53\)