Answer:
The first set: 8, 15, and 17.
Step-by-step explanation:
Pair: 8, 15, 17
By the pythagorean theorem, a triangle is a right triangle if and only if
[tex]\text{longest side}^2 = \text{first shorter side}^2 + \text{second shorter side}^2[/tex].
In this case,
[tex]\text{longest side}^2 = 17^2 = 289[/tex].
[tex]\begin{aligned}&\text{first shortest side}^2 + \text{second shortest side}^2 \\ &= 8^2 + 15^2\\ &=64 + 225 = 289 \end{aligned}[/tex].
In other words, indeed [tex]\text{hypotenuse}^2 = \text{first leg}^2 + \text{second leg}^2[/tex]. Hence, 8, 15, 17 does form a right triangle.
Similarly, check the other pairs. Keep in mind that the square of the longest side should be equal to the sum of the square of the two
Pair: 10, 15, 20
Factor out the common factor [tex]2[/tex] to simplify the calculations.
[tex]\text{longest side}^2 = 20^2 = 400[/tex]
[tex]\begin{aligned}&\text{first shortest side}^2 + \text{second shortest side}^2 \\ &= 10^2 + 15^2\\ &=100 + 225 = 325 \end{aligned}[/tex].
[tex]\text{longest side}^2 \ne \text{first shorter side}^2 + \text{second shorter side}^2[/tex].
Hence, by the pythagorean theorem, these three sides don't form a right triangle.
Pair: 12, 18, 22
[tex]\text{longest side}^2 = (2\times 11)^2 = 2^2 \times 121[/tex].
[tex]\begin{aligned}&\text{first shortest side}^2 + \text{second shortest side}^2 \\ &= (2 \times 6)^2 + (2 \times 9)^2\\ &=2^2 \times(36 + 81) = 2^2 \times 117 \end{aligned}[/tex].
[tex]\text{longest side}^2 \ne \text{first shorter side}^2 + \text{second shorter side}^2[/tex].
Hence, by the pythagorean theorem, these three sides don't form a right triangle.
Pair: 7, 9, 11
[tex]\text{longest side}^2 = 11^2 = 121[/tex].
[tex]\begin{aligned}&\text{first shortest side}^2 + \text{second shortest side}^2 \\ &= 7^2 + 9^2\\ &=49+ 81 = 130 \end{aligned}[/tex].
[tex]\text{longest side}^2 \ne \text{first shorter side}^2 + \text{second shorter side}^2[/tex].
Hence, by the pythagorean theorem, these three sides don't form a right triangle.
