Find the possible ones place digit in the square root of the following (apply the properties) a) 2039184 b) 10,004,569
How many natural numbers lie between the squares of 41 and 42?
What will be the value of ‘ x’ in Pythagorean triplet (41, 9, x)?
Check whether 15028 is a perfect square? If not find the smallest number by which 15028 be divided to make it a perfect square. Also find the square root of the new number formed.
A gardener has 5190 plants. He wants to plant in such a way that the number of rows and columns remains the same. Find the minimum number of plants left in this arrangement.

The answer is given below

Step-by-step explanation:

1) Find the possible ones place digit in the square root of the following

a) 2039184

The number 2039184 ends with 4, therefore the square root of the number can either end in 2 or 8

2² = 4, √4 = 2

8² = 64, √64 = 8

b) 10,004,569

The number 10,004,569 ends with 9, therefore the square root is the number will end in 3

3² = 9, √9 = 3

2) How many natural numbers lie between the squares of 41 and 42

42² = 1764 and 41² = 1681

Therefore the numbers that lie between 1764 and 1681 = (1764 - 1681) - 1 = 83 - 1 = 82

3) What will be the value of ‘ x’ in Pythagorean triplet (41, 9, x)

Pythagorean consist of three positive numbers a, b, c such that a² + b² = c². Therefore: x² + 9² = 41²

x² = 41² - 9² = 1681 - 81

x² = 1600

x = √1600 = 40

4) Check whether 15028 is a perfect square

15028 = 2 × 2 × 13 × 17 × 17

15028 = 2² × 13 × 17²

It is not a perfect square. If it is divided by 13 it becomes a perfect square, that is:

15028/13 = 2² × 17²

15028/13 = (2 × 17)² = 34²

34² = 15028/13

34² = 1156

The square root of the new number formed is 34 (i.e √1156)

5) A gardener has 5190 plants. He wants to plant in such a way that the number of rows and columns remains the same.

let the number if rows be x. Since the rows and columns are the same, the number of columns = x.

x² = 5190

x = √5190 = 72² + 6.

Therefore at least six plant would be left out

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

it is probably the graph on the bottom right.