Tag: finding out the diagonal of cube and cuboid

Applying the Pythagorean triples rule as $a^{2}+b^{2}= c^{2}$
$5^{2}+x^{2}= 13^{2}$
25 + $x^{2}$ = 169
$x^{2}$ = 169 - 25
$x^{2}$ = 144
Squaring on both the sides, we get
x = 12.

Applying the Pythagorean triples rule as $a^{2}+b^{2}= c^{2}$
Option A: $11^{2}+60^{2}= 61^{2}$
= 121 + 3600 = 3721 is a Pythagorean triples
Option B: $16^{2}+63^{2}= 65^{2}$
= 256 + 3969 = 4225 is a Pythagorean triples
Option C: $28^{2}+45^{2}= 53^{2}$
= 784 + 2,025 = 2809 is a Pythagorean triples
Option D: $30^{2}+80^{2}= 89^{2}$
= 900 + 6400 $\neq$ 7921 is not a Pythagorean triples.

Applying the Pythagorean triples rule as $a^{2}+b^{2}= c^{2}$
Option A: $5^{2}+12^{2}= 13^{2}$
25 + 144 = 169
169 = 169 is a Pythagorean triples.

For any natural numbers m > 1, $2m, m^{2} - 1$, $m^{2} + 1$ form a Pythagorean triplet.
If we take $m^2 + 1 = 22$, then $m^2 = 21$
The value of m will not be an integer.
If we take $m^2 - 1 = 22$, then $m^2 = 23$
Again the value of m will not be an integer.
Let $2m = 22$
$m = \cfrac{22}{2}$
$m = 11$
$2m = 2 \times 11 = 22$
$m^{2} - 1$ = $11^{2} - 1$
$= 121 - 1 = 120$
$m^{2} + 1$ = $11^{2} + 1$
$= 121 + 1 = 122$
Therefore, the Pythagorean triplets are $ 22, 120, 122.$

$14, 12, 13$ is not a Pythagorean triplet.
On squaring, we get
$12^2 + 13^2\ and\ 14^2$
$144 + 169 \ and\ 196$
$313 \neq 196$

For any natural numbers m > 1, $2m, m^{2} - 1$, $m^{2} + 1$ forms a Pythagorean triplet.
If we take $m^2 + 1 = 34$, then $m^2 = 33$
The value of m will not be an integer.
If we take $m^2 - 1 = 34$, then $m^2 = 35$
Again the value of m will not be an integer.
Let $2m = 34$
$m = \dfrac{34}{2}$
$m = 17$
$2m = 2 \times 17 = 34$
$m^{2} - 1$ = $17^{2} - 1$
$= 289 - 1 = 288$
$m^{2} + 1$ = $17^{2} + 1$
$=289 + 1 = 290$
Therefore, the Pythagorean triplets are $34, 288, 290.$

$15, 114, 115$ is not a Pythagorean triplet.
On squaring, we get
$15^2 + 114^2 \ and \ 15^2$
$225 + 12996 \ and \  13225$
$13221 \neq 13225$

We can get Pythagorean triplet by using general form $2m,\ m^{2}-1,\ m^{2}+1 $
Let us first take