Question

Find the pythagorean triplet whose greatest number is 22​

Answer 1

GIVEN :-

• The greatest number of a pythagorean triplet is 22.

TO FIND :-

• The Pythagorean triplet.

SOLUTION :-

➠ As we know that the Pythagorean triplet can be written as ,

$\to \boxed{ \red{ \bf \: 2n \: , \: n ^{2} - 1 \: , \:n ^{2} + 1}}$

➠ Since we have the greatest number 22. so from the above triplet we can say that n² + 1 = 22

$\to \sf \: n ^{2} + 1 = 22 \\ \\ \to \sf \: n ^{2} = 22 - 1 \\ \\ \to \sf \: n ^{2} = 21 \: ..........(1) \\ \\ \to \sf \: n = \sqrt{21} \\ \\ \to \boxed{ \blue{ \sf \: n = 4.58}}$

➠ Now let's calculate for the other two numbers.

$\to \sf \: 2n \\ \\ \to \sf \: 2 \times 4.58 \\ \\ \to \boxed{ \blue{ \sf \: 1st \: number = 9.16}}$

Hence the first number is 9.16.

$\to \sf \: n ^{2} - 1 \\ \\ \to \sf \: 21 - 1 \: \: \: \: \: \: \: \: \: \: \: \: \{ \because \: from \: (1) \}\\ \\ \to \boxed{ \blue{ \sf \: \: 2nd \: number \: = 20}}$

Hence the second number is 20.

☛ Hence the Required Pythagorean triplet is [9.16 , 20 , 22].

VERIFICATION :-

✒ As we know that, In a Pythagorean triplet the sum of square of First two numbers is equal to the square of third number.

$\boxed{\red{\bf a^{2} + b^{2} = c^{2}}} \\ \\ \to \sf (9.16)^{2} + (20)^{2} = (22)^{2} \\ \\ \to \sf 84 (approx) + 400 = 484 \\ \\ \to \sf 484 = 484$

L.H.S = R.H.S

HENCE VERIFIED ✔

Answer 2

Given:-

• Pythagoras triplet of greatest number = 22.

To Find:-

• All the numbers including in that Pythagoras triplet.

Solution :-

In Pythagoras triplet, it includes 3 members:

1. 2x,
2. x² - 1,
3. x² + 1

We've given that Pythagoras triplet of greatest number is 22. In those three members, greatest member is x²+1.

So,

⇒( x² + 1 ) = 22

⇒x² = 22-1

⇒x² = 21

⇒x = √21

Put x = √21 in those three members,

1. 2x = 2(√21) = 2√21

2. x² - 1 = (√21)² - 1 = 20

3. x² + 1 = (√21)² + 1 = 22

Vérification :-

(22)² = (2√21)² + (20)²

484 = 84 + 400

484 = 484