Question

Write a Pythagorean triplet whose one member is $14$.

Answer 1

As we know, we have Pythagorean

triplet 7, 24 anf 25. as,

$(25{)}^2=(7{)}^2+(24{)}^2$

$\to$ So, For general Pythagorean triplet, we

have sides of positive number.

$\to$ 25, 24 & 7 and (a) product as sides

Let side

$\overline{AB}=24a$

$\overline{BC}=7a$

where $aϵz^{+}$

So, $A{C}^2=A{B}^2+B{C}^2$

$=(24a{)}^2+(7a{)}^2$

$A{C}^2=576a^2+49a^2$

$A{C}^2=625a^2$

$\overline{AC}=25a$

So, For a = 2, we have, triplet

$BC=7\left(2\right)=14$

$AB=24\left(2\right)=48$

& $AC=25\left(2\right)=50$

Pythagorean triplet 14, 48 & 50.