Question

Write a Pythagorean triplet whose one member is 8. i) 8,15,17 ii) (8,17,20) iii) (8,10,15) iv) 8,9,17​

Answer 1

ᴄᴏɴᴄᴇᴘᴛ ᴜsᴇᴅ :-

If a, b are two sides of the triangle and c is the hypotenuse, then, a, b, and c :-

• a = m² - n²
• b = 2mn
• c = m² + n²

Note :-

• m, n are any two positive integers
• m > n
• m and n are coprime and both should not be odd numbers

Example :-

Let m = 2 , n = 1 ( m > n .)

Than ,

→ a = m² - n² = 2² - 1² = 4 - 1 = 3

→ b = 2mn = 2 * 2 * 1 = 4

→ c = m² + n² = 2² + 1² = 4 + 1 = 5.

So,

→ a² + b² = c²

→ 3² + 4² = 5²

→ 9 + 16 = 25

→ 25 = 25 . (Verify).

Sᴏʟᴜᴛɪᴏɴ :-

Now, we have given that, one number is 8 .

→ Even number = 8

So, we can conclude that, it will be Equal to 2mn.

Thus,

→ 2mn = 8 = b

→ mn = 4

Since m > n.

{ m = 4 , n = 1 }

Than,

→ m² - n² = 4² - 1² = 16 - 1 = 15 = a

→ m² + n² = 4² + 1² = 16 + 1 = 17 = c

check :-

→ a² + b² = c²

→ 8² + 15² = 17²

→ 64 + 225 = 289

→ 289 = 289 (verify.)

Hence, we can conclude that, Pythagorean triplet whose one member is 8 will be i) 8,15,17..

Answer 2

${ \underline{ \underline{ \blue{ \tt{ \huge{QUESTION- }}}}}}$




• Write a Pythagorean triplet whose one member is 8



(i) 8, 5 , 17

(ii) 8, 17 , 20

(iii) 8, 10, 15

(iv) 8, 9 , 17



${ \underline{ \underline{ \tt{ \blue{ \huge{SOLUTION- }}}}}}$




${ \pink{ \underline{ \underline{ \sf{Pythagorean \: triplet}}}}}$




✒ A Pythagorean triple consists of three positive integers a, b, and c, such that a2 + b2 = c2. Such a triple is commonly written (a, b, c), and a well-known example is (3, 4, 5). 



where, c > ( a,b )



⛦ here, in this question we have to identify a Pythagorean triplet whose one member is 8.



${ \underline{ \tt{ \green{ (i) \: 8, \: 15 \: , \: 17}}}}$




${ \sf{ let, \: }} \\ \\ { \red{ \sf{a = 8}}} \\ { \sf{ \red{b = 15}}} \\ { \sf{ \red{c = 17}}}$




${ \star{ \sf{ \purple{ \: \: {a}^{2} + {b}^{2} = {8}^{2} + {15}^{2} }}}}$




${ \implies{ \purple{ \sf{ {a}^{2} + {b}^{2} = 64 + 225}}}}$




${ \implies{ \purple{ \sf{ {a}^{2} + {b}^{2} = 289}}}}$



_______________________________________




${ \star{ \purple{ \sf{ \: \: \: {c}^{2} = {17}^{2} }}}}$




${ \implies{ \sf{ \purple{ {c}^{2} = 289}}}}$



_______________________________________




${ \boxed{ \pink{ \sf{ \: \: 289 = 289 \: \: }}}}$




${ \sf{ thus,}} \\ \\ { \boxed{ \boxed{ \blue{ \sf{ \: \: \: {a}^{2} + {b}^{2} = {c}^{2} \: \: \: }}}}}$




${ \sf{ hence,}} \\ \\ { \sf{ \red{(i) \: 8, \: 15, \: 17 \: \: - \: is \: the \: answer}}} \\ \\ { \green{ \sf{it \: is \: the \: pythagorean \: triplet }} }\\ { \sf{ \green{\: that \: has \: 8 \: as \: its \: one \: member.}}}$