Calculations in practical contexts using suitable digital tools | Year 9

01 / LEARN

Understand the idea.

Plan before pressing keys

Begin with the relationship between the quantities, then substitute the given values. This creates a mathematical expression you can check before using a calculator. Brackets must enclose a complete denominator or any complete input to a root.

Estimate the expected size first. For an area of $950 m^{2}$ , $30^{2} < 950 < 31^{2}$ , so the square side must be between $30$ and $31$ metres. A result near $300$ metres cannot fit that area.

Interpret the display

Record the formula, substitution and final unit alongside the display. Keep the unrounded result in memory if further calculations follow. Round only when interpreting the final answer at the requested precision.

Check by reversing the calculation: square an estimated side or substitute a radius into the circumference formula. Small discrepancies follow rounding; large ones suggest an entry or unit error. Buying whole items may require rounding up to meet a quantity requirement.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A garden side

A square garden has area $950 m^{2}$ . Find its side to two decimal places.

Worked solution

1
Model the square garden.
$s^{2} = 950 m^{2}, s = \sqrt{950} m$
The side length is the positive square root of the area.
2
Estimate a sensible interval.
$30^{2} = 900 < 950 < 961 = 31^{2}$
The calculator result should be between 30 and 31 metres.
3
Evaluate the complete root input.
$s = 30.822070 \dots m$
The result agrees with the interval estimate.
4
Round the final measurement.
$s \approx 30.82 m$
The third decimal digit is 2, so the hundredths digit remains 2.

Answer

$30.82 m$ .

EXAMPLE 02

A grouped denominator

A circular track has circumference $100 m$ . Use $r = C / (2 π)$ to find the radius to two decimal places.

Worked solution

1
Substitute the circumference.
$r = \frac{C}{2 π} = \frac{100}{2 π} m$
Both 2 and pi belong in the denominator.
2
Group the calculator denominator.
$100 \div (2 \times π) = 15.915494 \dots$
The brackets make division act on the whole product.
3
Round and include the unit.
$r \approx 15.92 m$
The third decimal digit is 5, so the hundredths digit increases.
4
Check against the original circumference.
$2 π (15.92) m \approx 100.03 m$
The small difference from 100 metres is consistent with a rounded radius.

Answer

$15.92 m$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

A square has area $170 m^{2}$ . Is its side closer to $13$ or $17$ metres?

Show answer & reasoning

$13 m$ .

The side is closer to $13 m$ . In fact $13^{2} = 169 < 170 < 182.25 = {13.5}^{2}$ , so $13 < \sqrt{170} < 13.5$ , well below 17.

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

A square court has area $625 m^{2}$ . Find its side and check.
2 marks
A square plot has area $730 m^{2}$ . Find its side to two decimal places.
2 marks
Use $r = \frac{C}{2 π}$ to find the radius for circumference $48 cm$ . Give two decimal places.
2 marks
A supplied formula is $D = \sqrt{a^{2} + b^{2}}$ . For $a = 8.4 cm$ and $b = 12.5 cm$ find $D$ to one decimal place.
2 marks
A square of area $50 m^{2}$ needs edging along all four sides. It is sold in whole metres. Find the least length to buy.
3 marks
A student reports $\sqrt{360} = 189.7$ . Explain a quick check and give the correct value to one decimal place.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

A square court has area $625 m^{2}$ . Find its side and check.
2 marks
$\sqrt{625} = 25$ and $25^{2} = 625$ .
$25 m$ .
A square plot has area $730 m^{2}$ . Find its side to two decimal places.
2 marks
$s = \sqrt{730} = 27.018512 \dots m$ .
$27.02 m$ .
Use $r = \frac{C}{2 π}$ to find the radius for circumference $48 cm$ . Give two decimal places.
2 marks
$r = \frac{48}{2 π} = 7.639437 \dots cm$ .
$7.64 cm$ .
A supplied formula is $D = \sqrt{a^{2} + b^{2}}$ . For $a = 8.4 cm$ and $b = 12.5 cm$ find $D$ to one decimal place.
2 marks
$D = \sqrt{70.56 + 156.25} = \sqrt{226.81} = 15.060212 \dots cm$ .
$15.1 cm$ .
A square of area $50 m^{2}$ needs edging along all four sides. It is sold in whole metres. Find the least length to buy.
3 marks
Length $= 4 \sqrt{50} = 28.284271 \dots m$ .
Round up; twenty-eight metres is too short.
$29 m$ .
A student reports $\sqrt{360} = 189.7$ . Explain a quick check and give the correct value to one decimal place.
3 marks
$18^{2} = 324 < 360 < 361 = 19^{2}$ .
$\sqrt{360} = 18.973665 \dots$ .
The root is between $18$ and $19$ ; it rounds to $19.0$ .

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Curriculum reference & lesson scope

Australian Curriculum v9 and SA Curriculum Mathematics

Year 9 real-number classification and applications; root bounds and sensible digital approximation. Full surd manipulation is developed in Year 10.

Official curriculum source ↗

In this lesson

Understand the idea.

Plan before pressing keys

Interpret the display

See the method in action.

A garden side

Model the square garden.

Estimate a sensible interval.

Evaluate the complete root input.

Round the final measurement.

A grouped denominator

Substitute the circumference.

Group the calculator denominator.

Round and include the unit.

Check against the original circumference.

Try it yourself.

Check your understanding

Keep practising