Question

Which of the following are Pythagorean triples ?
(a) $(9,8,10)$
(b) $(4,3,5)$
(c) $(6,8,10)$

Answer 1

Hint: Use the property that (a,b,c) is a Pythagorean triplet if the sum of the squares of the first two terms of the set (a,b,c) is equal to the square of the third one, i.e., $a^2+b^2=c^2$. Using this check each option and check which are Pythagorean triples and which are not.

Complete step-by-step answer:
We know that (a,b,c) is a Pythagorean triplet if the sum of the squares of the first two terms of the set (a,b,c) is equal to the square of the third one, i.e., $a^2+b^2=c^2$. So, let us check the cases given in the question one by one.
First, let us check for (a) $(9,8,10)$ . Now for $(9,8,10)$ , we have a = 9, b = 8 and c = 10 and for this to be a Pythagorean triplet $9^2+8^2={10}^2$ . But when we solve and put $9^2=81,8^2=64\text{ and 1}{\text{0}}^2=100$ , we find LHS is equal to 145 and RHS is 100. So, $(9,8,10)$ not a Pythagorean triplet.
Now, let us check for (b) $(4,3,5)$ . Now for $(4,3,5)$ , we have a = 4, b = 3 and c = 5 and for this to be a Pythagorean triplet $4^2+3^2=5^2$ . But when we solve and put $4^2=16,3^2=9\text{ and }{\text{5}}^2=25$ , we find LHS is equal to 25 and RHS is 25. So, $(4,3,5)$ is a Pythagorean triplet.
Finally, let us check for (c) $(6,8,10)$ . Now for $(6,8,10)$ , we have a = 6, b = 8 and c = 10 and for this to be to be a Pythagorean triplet $6^2+8^2={10}^2$ . But when we solve and put $6^2=36,8^2=64\text{ and 1}{\text{0}}^2=100$ , we find LHS is equal to 100 and RHS is 100. So, $(6,8,10)$ is a Pythagorean triplet.
Therefore, option (b) and option (c) are Pythagorean triplet.

Note: You can think of a Pythagoras triplet as the set of lengths of sides of a right angled triangle such that the first element of the set is the base and the last element is the hypotenuse. In this case the condition for the Pythagorean triplet is the Pythagoras theorem for the triangle with sides in the order as mentioned in the previous line.