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.