Free Squares and Square Roots 01 Practice Test - 8th Grade 

$\text{The number 39 ends with 9.}\text{And}{9}^2=81\text{which ends with 1.}\text{Hence,}{39}^2\text{should have 1 in the units}\text{place.}$

$\text{We can observe from options that}\text{only 1521 satisfies this.}\text{Hence 1521 is the answer.}$

$\text{Alternatively, 39 should be multiplied}\text{with 39 to obtain the square.}$

$\text{So,}39\times 39=1521.$

The square of 42 should have 4 in the unit place (since the square of 2 is 4). 84 is too small to be square of 42. So, option 3 is correct.
Alternatively, 42 can be multiplied with itself to give the result.

The square root of 729 should have 3 or 7 in its unit place. Because square of 3 and 7 has 9 in it's units place. Only 27 satisfies this.
Thus, Aman's cousin gets 27 coins as a birthday gift.

Also, we can find the square root using prime factorization as

$729=3\times 3\times 3\times 3\times 3\times 3$

$729=3^6$

Square root of 729 is equal to

$\sqrt{729}=\sqrt{3^6}=3^3=27$
 

$\text{Pythagorean triplets are set of three}\text{positive integers a, b, c such that,}{a}^2+b^2=c^2.$

${15}^2+8^2=225+64=289={17}^2.\text{Hence, the numbers 15, 8 and 17}\text{form a Pythagorean triplet.}$

$3^2+4^2=9+16=25=5^2.\text{Hence, the numbers 3, 4 and 5}\text{form a Pythagorean triplet.}$

${12}^2+{16}^2=144+256=400={20}^2\text{Hence, the numbers 12, 16 and 20}\text{form a Pythagorean triplet.}$

$1^1+1^1=1+1=2\ne 2^2\text{Hence, the numbers 1, 1 and 2}\text{does not form a Pythagorean triplet.}$

Let us express what Akanksha had to do and what she had done, mathematically.

To start with, let us start with assuming the number chosen by Akanksha as n.

What Akanksha had to do:
$n(n+1)=n^2+n$

What Akanksha had done:
$n\times n+n=n^2+n$, which is the same as before.   

$2^2$ and $8^2$ always have 4 in the units place. Also, we can see that square of 2 and square of 8 results in 4 and 64 respectively which have 4 in the unit's place.

$\sqrt{2}=1.4142$ (Given)

Multiply both sides by $10$

$10\times \sqrt{2}=14.142$

$\sqrt{100\times 2}=14.142$

Squaring both sides, we get the square of $14.142$  to be approximately equal to $200$.

If m can be expressed as $n^2$, where n is a natural number then, m is called a perfect square. 

${\left({\left(x^2\right)}^2\right)}^2=x^8$ and ${\left(x^3\right)}^3=x^9$​ 
As we can see the square of a square of a square has a total power of 8 whereas the cube of a cube has a total power of 9. Thus, they are not equal.

When an integer is multiplied by itself, it results in a perfect square.

When we square integers like 1, 2, 3, 4, 5, ..., we obtain 1, 4, 9, 16, 25, ... .

Hence, the given numbers which are obtained by squaring the integers are known as perfect squares.