Question

(it) Express 220 as the sum UI IJ UUU I
6. Write a Pythagorean triplet whose smallest member is :
(i) 6
(ii) 14
(iii) 16
(iv) 20​

Answer 1

$\huge {\fcolorbox {purple}{pink} {\red {\bf {Answer}}}}$

i) let 2m = 6

$\longrightarrow {m = 3}$

m²+ 1 = 3² + 1= 9 + 1 = 10

m²- 1 = 3 2 - 1 = 9 - 1 = 8

check:

6 ² + 8² = 36 + 64 = 100 = 10²

Hence, the triplet is 6, 8 & 10

ii)Let 2 m = 14

$\longrightarrow {m = 7}$

m² + 1 = 7²+ 1 = 49 + 1 = 50

m²- 1 = 7² - 1 = 49 - 1 = 48

check:

14² + 48² = 196 + 1304 = 2500 = 50²

Hence, the triplet is 14, 48, and 50

iii)Let 2 m = 16,

$\longrightarrow {m = 8}$

m² + 1 = 8² + 1 = 64 + 1 = 65

m²- 1 = 8 2 - 1 = 64 - 1 = 63

check:

16² + 63² = 256 + 3969 = 4225 = 65²

Hence, the triplet is 16, 63 & 65

iv)Let 2 m = 18

$\longrightarrow {m = 9}$

m² + 1 = 9² + 1 = 81 + 1 = 82

m² - 1 = 9² - 1 = 81 - 1 = 80

check:

18²+ 80² = 324 + 6400 = 6724 = 82²

Hence, the triplet is 18, 80 & 82

Hope it helps....