Sure, let's go through each case step by step to find the Pythagorean triplet whose one member matches the given values.
### (ii) One member is 6:
To find a Pythagorean triplet where one of the members is 6, we can explore existing Pythagorean triplets or manipulate known triplets.
We know that a common Pythagorean triplet is (3, 4, 5). If we multiply each member of this triplet by 2, we get:
[tex]\[ 2 \times 3 = 6 \][/tex]
[tex]\[ 2 \times 4 = 8 \][/tex]
[tex]\[ 2 \times 5 = 10 \][/tex]
Thus, the triplet (6, 8, 10) is a valid Pythagorean triplet.
### (ii) One member is 14:
To find a Pythagorean triplet where one of the members is 14, we can use a known triplet and adjust it accordingly.
Considering the triplet (7, 24, 25), we can adjust by multiplying by 2:
[tex]\[ 2 \times 7 = 14 \][/tex]
[tex]\[ 2 \times 24 = 48 \][/tex]
[tex]\[ 2 \times 25 = 50 \][/tex]
Thus, the triplet (14, 48, 50) fits our requirement.
### (iii) One member is 16:
To find a Pythagorean triplet where one of the members is 16, we use another known triplet.
Using the triplet (8, 15, 17), and multiplying by 2, we get:
[tex]\[ 2 \times 8 = 16 \][/tex]
[tex]\[ 2 \times 15 = 30 \][/tex]
[tex]\[ 2 \times 17 = 34 \][/tex]
Thus, the triplet (16, 30, 34) is a valid Pythagorean triplet.
In conclusion:
- A Pythagorean triplet with one member as 6 is (6, 8, 10).
- A Pythagorean triplet with one member as 14 is (14, 48, 50).
- A Pythagorean triplet with one member as 16 is (16, 30, 34).
These triplets satisfy the Pythagorean theorem:
[tex]\[ a^2 + b^2 = c^2 \][/tex]
for the respective values of a, b, and c.
