Question

Write a Pythagorean triplet whose one number is 8

• A. $8,10,6$
• B. $15,10,5$
• C. $63,27,3$
• D. $80,40,20$

Answer 1

Answer: (A)

$8,10,6$

$8^2=64$

$\Rightarrow$ $100-64=36$

$\Rightarrow$ $\sqrt{36}=6$

$\Rightarrow$ $\sqrt{100}=10$

Let

Hypotenuse $=10$ [ Since its greater than $6$ and $8$ ]

Base $=8$ and

Perpendicular $=6$

$(Hypotenuse{)}^2=(Perpendicular{)}^2+(Base{)}^2$ [ Pythagoras theorem ]

$\Rightarrow$ $(10{)}^2=(6{)}^2+(8{)}^2$

$\Rightarrow$ $100=36+64$

$\Rightarrow$ $100=100$

$\therefore$ Pythagorean triplet are $6,8,10$