sqrt()

sqrt(x) returns squqre root of input number as float data type.
import math
print(math.sqrt(4))    # 2.0 
print(math.sqrt(16))   # 4.0 
print(math.sqrt(20))   # 4.47213595499958
print(math.sqrt(0))    # 0.0
Using negative number
For any value less than 0 , we will get ValueError
import math
print(math.sqrt(-15)) 
Above code will generate error.

Output of math.sqrt() is float data type. We can use type() to get the data type.
import math
x=math.sqrt(25)
print(x)        # 5.0
print(type(x))  # <class 'float'>
We can use int() to convert output to integer data type.
import math
x=math.sqrt(25)
y=int(x)
print(type(y))  # <class 'int'>

Length from X Y coordinates

Length of the line from coordinates

If X and Y coordinates of starting and ending points of a line is given then the length can be calculated.
x1=13639.4
y1=1554.8
x2=13637.5
y2=1570.6
import math 
h=math.sqrt((x2-x1)**2 + (y2-y1)**2) 
print(h)
We will use round() to limit our output to 2 decimal places.
h=round(math.sqrt((x2-x1)**2 + (y2-y1)**2),2) 
Assuming that the input coordinates are in meter , we can get the output in feet and Kadi.
in_feet=h*3.2808
in_kadi=in_feet/0.66
print("Length meeter : ",round(h,2)," Feet:",round(in_feet,2)," 
	Kadi:",round(in_kadi,2))
Conversion formula is here
1 Feet = 0.3048 Meter 
1 Meter = 3.2808 Feet
1 Kadi = 0.66 Feet 
Tkinter Application to convert Feet to Meter and vice versa JavaScript Application to convert Feet to Meter and vice versa
log10 modf()
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    04-07-2020

    Why the square root of 16 is 4.0 and why not 4???

    05-07-2020

    The output of sqrt is float always. Added the part to convert float data type to integer.

    Post your comments , suggestion , error , requirements etc here





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