Tag: finding out the diagonal of cube and cuboid

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Questions Related to finding out the diagonal of cube and cuboid

Multiple choice maths surface area and volume of cube and cuboid finding out the diagonal of cube and cuboid length of the diagonal of cube diagonal of cube and cuboid

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
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

Multiple choice maths surface area and volume of cube and cuboid finding out the diagonal of cube and cuboid length of the diagonal of cube diagonal of cube and cuboid

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

  1. $22, 183, 185$
  2. $22, 483, 485$
  3. $22, 23, 25$
  4. $22, 120, 122$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.$

Multiple choice maths surface area and volume of cube and cuboid finding out the diagonal of cube and cuboid length of the diagonal of cube diagonal of cube and cuboid

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

  1. $34, 278, 290$
  2. $34, 288, 291$
  3. $34, 288, 290$
  4. $35, 288, 290$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $34, 288, 290.$

Multiple choice maths surface area and volume of cube and cuboid finding out the diagonal of cube and cuboid length of the diagonal of cube diagonal of cube and cuboid

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

  1. $8, 15, 18$
  2. $8, 13, 16$
  3. $8, 14, 17$
  4. $8, 15, 17$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $D.$