Solution:
Given the side lengths of a triangle ABC;
[tex]a=10,b=14,c=26[/tex]
The side lengths form a Pythagorean triple if ithe square of the longest side is equal to the sum of squares of the remaining two sides.
Thus;
[tex]\begin{gathered} 10^2+14^2=100+196 \\ \\ 10^2+14^2=296 \\ \\ 296\ne29^2 \\ \\ \text{ Thus;} \\ \\ 10^2+14^2\ne26^2 \end{gathered}[/tex]
Hence, they do not form a Pythagorean triple.
CORRECT OPTION: (B) No, they do not form a Pythagorean triple.
[tex]10^{2}+14^{2}\ne26^{2}[/tex]
