Free Squares and Square Roots 02 Practice Test - 8th Grade 

Numbers ending with 2, 3, 7, or 8 at unit's place are never perfect squares.
Estimation without actually finding roots:
484 can be the perfect square of a number ending with 2 or 8.
576 can be the perfect square of a number ending with 4 or 6.
529 can be the perfect square of a number ending with 3 or 7.
 

Between the squares of any two consecutive numbers n and (n+1), lie 2n non-perfect squares.
Therefore between squares of 75 and 76 (n = 75 and n + 1 = 76), lie (2 x 75) = 150 non-perfect squares.

${18}^2+{24}^2=324+576=900={30}^2$

Since the squares of 18 and 24 add up to the square of 30, the numbers 18, 24 and 30 form a Pythagorean triplet, and hence, can be used to represent the three sides of a right- angled triangle.

Since the square sheet needs to be divided equally into 4, 5, and 6 equal parts respectively, its area must be a perfect square divisible by 4, 5, and 6; or the area of the square sheet should be a multiple of 4, 5, and 6.
L. C. M. of 4, 5, and 6 = 60. But 60 is not a perfect square.
$60=2\times 2\times 3\times 5=2^2\times 3\times 5$.
If we multiply the LCM by 3 and 5 both, it will become a perfect square. Therefore the minimum area of the square sheet $=60\times 3\times 5=900$ sq. units.
Now 900 is a minimum perfect square, which is divisible by 4, 5, and 6. Therefore the side of the square should be the square root of the area of the square (Area of a square = $sid{e}^2$ sq. units).
Hence, the minimum length of the side of the square = $\sqrt{\left(900\right)}=\sqrt{(2^2\times 3^2\times 5^2)}$ = 30 units.

Number of trees = 1024. Since number of rows = number of columns, this means that the number of trees is a perfect square. Therefore number of rows/columns = square root of 1024 (number of trees).
$\sqrt{1024}$= $\sqrt{2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2}=32$
Therefore, the initial number of rows/columns = 32. After cutting 1 row and one column, number of rows/columns = 32-1 = 31.
The remaining number of trees = ${31}^2$.
Therefore, number of trees that were cut = ${32}^2-{31}^2$ = 1024 - 961  = 63.

Sum of n consecutive odd integers starting from 1 is equal to $n^2$, where 'n' is the number of terms involved in the series.
Therefore,
i. 1+3+5+7+9+11+13+15+17+19+21 =  sum of first 11 odd integers = ${11}^2=121$.

The square root of a number can be found by repeatedly subtracting odd integers from the number starting from 1.
So in this case,
256- 1-3-5-7-9-11-13-15-17-19-21-23-25-27-29-31 = 0.
(16 steps, hence square root of 256 is 16.)

Given:  'p' is the square of 'q'.
$\Rightarrow$ 'q' is the square root of 'p'
p>400
The square root of 400 is 20.
So if "p"$>$400, its square root 'q' can be any number greater than 20.