# Which of the following triplets are Pythagorean?(i) (8, 15, 17)(ii) (18, 80, 82)(iii) (14, 48, 51)(iv) (10, 24, 26)(v) (16, 63, 65)(vi) (12, 35, 38).

To do:

We have to find whether the given triplets are Pythagorean.

Solution:

Pythagorean triplets are sets of positive integers $a,b,c$ that satisfy the condition $a^2+b^2=c^2$ where $c>a,b$

(i) Here, the greatest number is $17$.

Therefore,

$17^2=289$

$8^2+15^2=64+225$

$=289$

This implies,

$17^2=8^2+15^2$

Hence, (8, 15, 17) is a Pythagorean triplet.

(ii) Here, the greatest number is $82$.

Therefore,

$82^2=6724$

$18^2+80^2=324+6400$

$=6724$

This implies,

$82^2=18^2+80^2$

Hence, (18, 80, 82) is a Pythagorean triplet.

(iii) Here, the greatest number is $51$.

Therefore,

$51^2=2601$

$14^2+48^2=196+2304$

$=2500$

This implies,

$51^2≠ 14^2+48^2$

Hence, (14, 48, 51) is not a Pythagorean triplet.

(iv) Here, the greatest number is $26$.

Therefore,

$26^2=676$

$10^2+24^2=100+576$

$=676$

This implies,

$26^2=10^2+24^2$

Hence, (10, 24, 26) is a Pythagorean triplet.

(v) Here, the greatest number is $65$.

Therefore,

$65^2=4225$

$16^2+63^2=256+3969$

$=4225$

This implies,

$65^2=16^2+63^2$

Hence, (16, 63, 65) is a Pythagorean triplet.

(vi) Here, the greatest number is $38$.

Therefore,

$38^2=1444$

$12^2+35^2=144+1225$

$=1369$

This implies,

$35^2≠12^2+35^2$

Hence, (12, 35, 38) is not a Pythagorean triplet.

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Updated on: 10-Oct-2022

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