Can You Find The Minimum And Maximum Possible Areas For A Rectangle Measuring 4.15 Cm By 7.34 Cm?


3 Answers

Selena Antell Profile
Selena Antell answered
To find the area of a rectangle just times the length by the width so 4.15 times by 7.34 =    30.461 rounded to the nearest hundredth that would usually be 30.46   as to round up or down to hundredths you look at the number after the hundredths column in this case the 1

if it is 5 or above you would round up making the answer 30.47 and less than 5 round down making the answer 30.46.... In this case would be rounded down as the next number is 1.....

so answer should be 30.46

am sorry am not sure why it says find minimum and maximum tho. Maybe someone else will know.

Hope this helps a little to get you started.

Oddman Profile
Oddman answered
A measured value of 4.15 cm may have a couple of different errors associated with it. One of these has to do with the precision of the number. 4.15 is the 2-decimal digit representation of any number between 4.145 and 4.1549+. Similarly, the measurement 7.34 cm can be used to represent any actual value between 7.335 and 7.3449+.

The minimum rectangle area would be the product of the two minimum values the measurements represent: 4.145*7.335 = 30.403575 ≈ 30.40.

The maximum rectangle area would be the product of the two maximum values: 4.1549*7.3449 = 30.517+ ≈ 30.52.

The minimum and maximum values for the area of the rectangle are
30.40 cm2 and 30.52 cm2, respectively.
The other error associated with such measurements has to do with the accuracy to which the measurement can be made. If you are counting wavelengths of a certain red light, then your accuracy is probably as good as it can get. Otherwise, there are errors in translation between the measurement standards we have defined and the ruler used to perform the measurement. These can amount to a small fraction of 1%, or to several %, depending.

To get a feel for this, you can lay a couple of (different) "dime store" rulers against each other. Quite often, the scales are different by an observable amount even over the length of a foot or 15 inches. If rulers can differ by 1/16 of an inch or so over 12 inches, the same rulers will differ by about .02 cm over a distance of 4.15 cm. Then, the area max/min are not determined by the 4.15±.005 cm precision of the number, but by the 4.15±.02 cm accuracy. A similar accuracy error percentage (±.5%) may apply to the 7.34 cm measurement.

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