NavGanitMateby NavGreen Compute, AustraliaBased on Australian Curriculum: Mathematics v9.0
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Accuracy and measurement errors

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Required skills (click to reveal)

A community garden in Melbourne plans a rectangular plot with a design length of 75 metres and width of 50 metres. The measuring tape used has a maximum error of ±0.25 metres.

The gardeners calculate the perimeter using the measured values. What is the exact maximum possible perimeter of the plot, in metres, when both length and width are at their maximum measured values?

Express your answer as a simplified fraction.

Rectangle measurementsLength 75 metres; width 50 metres75 metres50 metres
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Concept illustrations

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Measurement compared with a reference. Measured length 49 cm differs from the reference 50 cm by 1 cm. Absolute error is 1 cm and percentage error is 2%. The axis is a close-up from 48 to 51 cm.48 cm49 cm50 cm51 cmMeasured: 49Reference: 50|49 − 50| = 1 cm
Measured length 49 cm differs from the reference 50 cm by 1 cm. Absolute error is 1 cm and percentage error is 2%. The axis is a close-up from 48 to 51 cm.
Nearest hundredth on a number line. 6.347 lies between 6.34 and 6.35. It is beyond the midpoint 6.345, so rounding to the nearest hundredth gives 6.35.6.346.3456.356.3476.347 ≈ 6.35
6.347 lies between 6.34 and 6.35. It is beyond the midpoint 6.345, so rounding to the nearest hundredth gives 6.35.

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Formula

Explanation

Worked solution and answer (click to reveal)

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Worked solution

  1. The garden plot is rectangular with a design length of 75 m and width of 50 m. The measuring tape has a maximum error of ±0.25 m, meaning actual measurements could be up to 0.25 m more or less than the stated values.
  2. To find the maximum possible perimeter, both length and width must be at their maximum measured values. The maximum length is 75 + 0.25 = 75.25 m.
  3. The maximum width is 50 + 0.25 = 50.25 m.
  4. The perimeter of a rectangle is given by P = 2 × (length + width). This formula applies because perimeter is the total distance around the rectangle.
  5. Substitute the maximum values: P = 2 × (75.25 + 50.25).
  6. Add the values inside the parentheses: 75.25 + 50.25 = 125.50.
  7. Multiply by 2: 2 × 125.50 = 251.00.
  8. Convert 251.00 to a simplified fraction. Since 251 is an integer, it is equivalent to 251/1.
  9. Check if 251/1 can be simplified. 251 is a prime number (not divisible by any integer other than 1 and itself), so the fraction is already in simplest form.
  10. The maximum possible perimeter is 251 metres, expressed as a simplified fraction: 251/1.
  11. Verify: 2 × (75.25 + 50.25) = 2 × 125.5 = 251. The calculation is correct.
  12. The exact maximum possible perimeter, when both length and width are at their maximum measured values, is 251/1 metres.

Reference answer: 251/1

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