Error in Differing Dimensions
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Error increases when dimensions increase.
Consider the following example:

1.  A student measures the side of a cube and accidentally reads a length of 5 inches, when the real length is 6 inches.  Notice how the percentage of error increases as the student uses this measurement to compute surface area and volume.

Measurement Compute Surface Area Compute Volume

 The side of a  cube is measured.

 Measurement:  5 in.

 Actual size:  6 in.
Percent of error =

Surface area computed with measurement: 
         SA = 25 • 6 = 150 sq. in.

Actual surface area:
        SA = 36 • 6 = 216 sq. in.

Percent of error =

Volume computed with measurement:
        V = 5 ³ = 125 cubic in.

Actual volume:
        V = 6 ³ = 216 cubic in.

Percent of error =

       rounded to nearest tenth.

 

2.  A box has the measurements 1.4 cm by 8.2 cm by 12.5 cm.  Find the percent of error in calculating its volume.

ANSWER:  Since no other values are given, we will use the greatest possible error based upon the fact that these measurements were taken to the nearest tenth of a centimeter, which will be 0.05 cm.

Volume as measured:  1.4 x 8.2 x 12.5 = 143.5 cubic cm
Maximum volume (+0.05) :  1.45 x 8.25 x 12.55 = 150.129375 cubic cm
Minimum volume (-0.05):  1.35 x 8.15 x 12.45 = 136.981125 cubic cm

Possible error in volume:  Maximum - measured = 6.629375 cubic cm
                                      Measured - minimum = 6.518875 cubic cm

Use the "greatest" possible error in volume:  6.629375 cubic cm

Remember that percent of error is the relative error times 100%.



The percent of error is approximately 5%.