Question

Write a Pythagorean triplet whose one member is:
A) 6
B) 14
C) 16
D) 18

Answer 1

Hint: Let us first understand the term Pythagorean triplet and then we will use a formula to determine all the numbers in a Pythagorean triplet of the given options one by one. First, we will divide the given number by two, then we will use $$n^2-1$$ and $$n^2+1$$ to determine the other two numbers.

Complete step-by-step solution:
A Pythagorean Triplet is one in which the square of one of the integers in the triplet equals the sum of the squares of the other two numbers. Pythagorean triplets are the result of Pythagoras' right-angled triangle theorem, which states that the sum of the squares of the lengths of the sides other than the hypotenuse in a right-angled triangle is equal to the square of the hypotenuse's length.
Now, we will look at the method that we are going to apply to derive the triplet. The three numbers that are used as a Pythagorean triplet can be represented as $$2n$$, $$n^2-1$$ and $$n^2+1$$. That means we have to assume our number as $$2n$$ and hence we will be able to determine the other two.
A) 6
Let’s us assume that
$$2n=6$$
Therefore, by dividing 2, we will get the value of n as,
$$\Rightarrow n={\displaystyle \frac{6}{2}}$$
$$\Rightarrow n=3$$
Now, by putting the value of n on the other formulas, we get
$$\Rightarrow n^2-1=3^2-1$$
$$\Rightarrow n^2-1=8$$
 and
$$\Rightarrow n^2+1=3^2+1$$
$$\Rightarrow n^2+1=10$$
Therefore, the Pythagorean triplet containing 6 is 6, 8 and 10.
B) 14
Again, let’s assume that
$$2n=14$$
$$\Rightarrow n=7$$
And by putting the value of n, we get
$$\Rightarrow n^2-1=7^2-1$$
$$\Rightarrow n^2-1=48$$
And,
$$\Rightarrow n^2+1=7^2+1$$
$$\Rightarrow n^2+1=50$$
Hence, the Pythagorean triplet containing 14 is 14, 48 and 50.
C) 16
Now, let’s assume that
$$2n=16$$
$$\Rightarrow n=8$$
By putting the value of n, we get
$$\Rightarrow n^2-1=8^2-1$$
$$\Rightarrow n^2-1=63$$
And,
$$\Rightarrow n^2+1=8^2+1$$
$$\Rightarrow n^2+1=65$$
Therefore, the Pythagorean triplet containing 16 is 16, 63 and 65
D) 18
Finally, let’s assume
$$2n=18$$
$$\Rightarrow n=9$$
By putting the value of n, we get
$$\Rightarrow n^2-1=9^2-1$$
$$\Rightarrow n^2-1=80$$
And
$$\Rightarrow n^2+1=9^2+1$$
$$\Rightarrow n^2+1=82$$
Hence, the Pythagorean triplet containing 18 is 18, 80 and 82.

Note: We can clearly say that the length of the hypotenuse is the largest. Hence, $$n^2+1$$ denotes the largest side i.e. hypotenuse. Also, make sure that after getting the Pythagorean triplet, the length should fulfill the condition of the Pythagoras theorem i.e. $$H^2=P^2+B^2$$.