Tag: length of the diagonal of cube

Questions Related to length of the diagonal of cube

What is the value of x, if (5, x, 13) is a Pythagorean triple?

  1. 10

  2. 11

  3. 12

  4. 13


Correct Option: C
Explanation:

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.

Which one of the following is not a Pythagorean triples?

  1. 11, 60, 61

  2. 16, 63, 65

  3. 28, 45, 53

  4. 30, 80, 89


Correct Option: D
Explanation:

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.

Identify the Pythagorean triples.

  1. 5, 12, 13

  2. 8, 24, 25

  3. 11, 61, 61

  4. 20, 70, 10


Correct Option: A
Explanation:

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.

Find the Pythagorean triplets, whose one member is $22.$

  1. $22, 183, 185$

  2. $22, 483, 485$

  3. $22, 23, 25$

  4. $22, 120, 122$


Correct Option: D
Explanation:

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.$

Which one of the following is not a Pythagorean triplets?

  1. $7, 24, 25$

  2. $15, 112, 113$

  3. $10, 24, 26$

  4. $14, 12, 13$


Correct Option: D
Explanation:

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

What is the Pythagorean triplet, whose one member is $34$?

  1. $34, 278, 290$

  2. $34, 288, 291$

  3. $34, 288, 290$

  4. $35, 288, 290$


Correct Option: C
Explanation:

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 $34, 288, 290.$

Which one of the following is not a Pythagorean triplet?

  1. $7, 24, 25$

  2. $15, 114, 115$

  3. $27, 364, 365$

  4. $6, 8 10$


Correct Option: B
Explanation:

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

A Pythagorean triplet whose smallest member is $8$, is:

  1. $8, 15, 18$

  2. $8, 13, 16$

  3. $8, 14, 17$

  4. $8, 15, 17$


Correct Option: D
Explanation:

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

$m^{2}-1=8$
So, $m^{2}=8+1=9$
Which gives $m=3$
Therefore $2m=6$ and  $ \displaystyle m^{2}+1 = 10  $
The triplet is thus $6,8,10$, but $8$ is not the smallest member of this triplet.

So let us try
$2m=8$
then $m=4$
We get $ \displaystyle m^{2}+1 = 16-1=15$
and $ \displaystyle m^{2}+1 =16+1=17$
The triplet is $8,15,17$ with $8$ as the smallest member.

Hence, option $D.$