Question

Q6 Find the Pythagoras Triplet whose smallest member is
a) 10
b) 8
c) 12
d) 16​

Answer 1

Answer:

Step-by-step explanation:

2m,$m^2-1,m^2+1$

a). let 2m be 10

so, 2m=10

m=$10/2$

m=5

$m^2-1\\5^2-1\\25-1\\24$

$m^2+1\\5^2+1\\25+1\\26$

The Pythagoras Triplet is 10,24,26.

b). let 2m be 8

2m=8

m=8/2

m=4

$m^2-1\\4^2-1\\16-1\\15$

$m^2+1\\4^2+1\\16+1\\17$

The Pythagoras Triplet is 8,15,17.

c). let 2m be 12

2m=12

m=12/2

m=6

$m^2-1\\6^2-1\\36-1\\35$

$m^2+1\\6^2+1\\36+1\\37$

The Pythagoras Triplet is 12,35,37.

d). let 2m be 16

2m=16

m=16/2

m=8

$m^2-1\\8^2-1\\64-1\\63$

$m^2+1\\8^2+1\\64+1\\65$

The Pythagoras Triplet is 16,63,65.