Errors in Measurement
Error? No ... you didn't measure it wrong ... this is about accuracy. 
Measuring instruments are not exact!
Degree of Accuracy
Accuracy depends on the instrument you are measuring with. But as a general rule:
Examples:
When your instrument measures in "1"s then any value between 6½ and 7½ is measured as "7" 

When your instrument measures in "2"s then any value between 7 and 9 is measured as "8" 
Plus or Minus
We can show the error using the "Plus or Minus" sign:  ± 
When the value could be between 6½ and 7½ 7 ±0.5 The error is ±0.5 

When the value could be between 7 and 9 8 ±1 The error is ±1 
Example: a fence is measured as 12.5 meters long, accurate to 0.1 of a meter
Accurate to 0.1 m means it could be up to 0.05 m either way:
Length = 12.5 ±0.05 m
So it could really be anywhere between 12.45 m and 12.55 m long.
Absolute, Relative and Percentage Error
But ... when measuring we don't know the actual value! So we use the maximum possible error.
In the example above the Absolute Error is 0.05 m
What happened to the ± ... ? Well, we just want the size (the absolute value) of the difference.
We don't know the actual measurement, so the best we can do is use the measured value:
Relative Error =  Absolute Error 
Measured Value 
Let us see them in an example:
Example: fence (continued)
Length = 12.5 ±0.05 m
So:
Absolute Error = 0.05 m
And:
Relative Error =  0.05 m  = 0.004 
12.5 m 
And:
Percentage Error = 0.4%
More examples:
Example: The thermometer measures to the nearest 2 degrees. The temperature was measured as 38° C
The temperature could be up to 1° either side of 38° (i.e. between 37° and 39°)
Temperature = 38 ±1°
So:
Absolute Error = 1°
And:
Relative Error =  1°  = 0.0263... 
38° 
And:
Percentage Error = 2.63...%
Example: You measure the plant to be 80 cm high (to the nearest cm)
This means you could be up to 0.5 cm wrong (the plant could be between 79.5 and 80.5 cm high)
Height = 80 ±0.5 cm
So:
Absolute Error = 0.5 cm
And:
Relative Error =  0.5 cm  = 0.00625 
80 cm 
And:
Percentage Error = 0.625%
Area
When working out areas you need to think about both the width and length ... they could both be the smallest possible measure, or both the largest.
Example: Alex measured the field to the nearest meter, and got a width of 6 m and a length of 8 m.
Measuring to the nearest meter means the true value could be up to half a meter smaller or larger.
The width (w) could be from 5.5m to 6.5m:
The length (l) could be from 7.5m to 8.5m:
The area is width × length:
The smallest possible area is: 5.5m × 7.5m = 41.25 m^{2}
The measured area is: 6m × 8m = 48 m^{2}
And the largest possible area is: 6.5m × 8.5m = 55.25 m^{2}
Absolute, Relative and Percentage Error for Areas
The only tricky thing here is ... which is the absolute error?
 From 41.25 to 48 = 6.75
 From 48 to 55.25 = 7.25
Answer: pick the biggest one! So:
Absolute Error = 7.25 m^{2}
Relative Error =  7.25 m^{2}  = 0.151... 
48 m^{2} 
Percentage Error = 15.1%
(Which is not very accurate, is it?)
Volume
And volume has three measurements: width, length and height!
Example: Sam measured the box to the nearest 2 cm, and got 24 cm × 24 cm × 20 cm
Measuring to the nearest 2 cm means the true value could be up to 1 cm smaller or larger.
The three measurements are:
 24 ±1 cm
 24 ±1 cm
 20 ±1 cm
Volume is width × length × height:
The smallest possible Volume is: 23cm × 23cm × 19cm = 10051 cm^{3}
The measured Volume is: 24cm × 24cm × 20cm = 11520 cm^{3}
The largest possible Volume is: 25cm × 25cm × 21cm = 13125 cm^{3}
And so we get:
Absolute, Relative and Percentage Error for Volumes
Absolute error:
 From 10051 to 11520 = 1469
 From 11520 to 13125 = 1605
Pick the biggest one:
Absolute Error = 1605 cm^{3}
Relative Error =  1605 cm^{3}  = 0.139... 
11520 cm^{3} 
Percentage Error = 13.9%