Estimating non-perfect square roots | Year 9

01 / LEARN

The idea, step by step

Start with nearby squares

A square root asks which nonnegative side length produces the stated square area. Known square numbers give a first estimate. With an area of thirty square units, the side is between five and six units because twenty-five and thirty-six are the nearby squares.

Square roots are not halfway between integers whenever areas are halfway between squares. Squaring is not a linear operation. Improve an estimate by squaring trial decimal values rather than dividing the gap in the areas by eye.

Check the rounding boundary

To round a root to one decimal place, decide which tenth is nearest. The rounded value 5.5 covers values from 5.45 up to, but not including, 5.55 under ordinary rounding. Squaring these positive boundaries shows whether the root lies in that interval.

A calculator can check an estimate, but bounding squares explain it. If an interval is requested, give both bounds and keep the inequality signs in their correct directions. A bound and a rounded answer communicate different information.

a^{2} < n < b^{2} \Rightarrow a < \sqrt{n} < b (a, b \geq 0)

A common trap

The square root of $30$ is exactly $5.5$ because $30$ is near the middle of $25$ and $36$ .

Instead: Check the trial value: ${5.5}^{2} = 30.25$ , not $30$ . Root values do not increase at a constant rate as their inputs increase.

02 / WORKED EXAMPLES

See the thinking

Example 1

A root between five and six

Estimate $\sqrt{30}$ to one decimal place.

$25 < 30 < 36$ , so the root is between $5$ and $6$ .
${5.45}^{2} = 29.7025 < 30 < 30.8025 = {5.55}^{2}$ .
The root is between the rounding boundaries for $5.5$ .

\sqrt{30} \approx 5.5

Example 2

Close to a perfect square

Estimate $\sqrt{80}$ to one decimal place.

$8^{2} < 80 < 9^{2}$ .
${8.85}^{2} = 78.3225 < 80 < 80.1025 = {8.95}^{2}$ .
The root is below $8.95$ , so it rounds to $8.9$ rather than $9.0$ .

\sqrt{80} \approx 8.9

Check your understanding

Is $\sqrt{45}$ between $6.7$ and $6.8$ ?

Show answer

Yes.

${6.7}^{2} = 44.89 < 45 < 46.24 = {6.8}^{2}$ .

03 / WORKSHEET

Your turn

6 questions · 13 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Find consecutive whole-number bounds for $\sqrt{37}$ .
2 marks
Find consecutive whole-number bounds for $\sqrt{150}$ .
2 marks
Use squared trial values to locate $\sqrt{12}$ between tenths.
2 marks
Estimate $\sqrt{18}$ to one decimal place.
2 marks
Estimate $\sqrt{70}$ to one decimal place.
2 marks
A square has area $55 {cm}^{2}$ . Estimate its side to the nearest tenth of a centimetre.
3 marks

Read the worked solutions

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

Find consecutive whole-number bounds for $\sqrt{37}$ .
2 marks
$36 < 37 < 49$ .
$6 < \sqrt{37} < 7$ .
Find consecutive whole-number bounds for $\sqrt{150}$ .
2 marks
$144 < 150 < 169$ .
$12 < \sqrt{150} < 13$ .
Use squared trial values to locate $\sqrt{12}$ between tenths.
2 marks
${3.4}^{2} = 11.56 < 12 < 12.25 = {3.5}^{2}$ .
$3.4 < \sqrt{12} < 3.5$ .
Estimate $\sqrt{18}$ to one decimal place.
2 marks
${4.15}^{2} = 17.2225 < 18 < 18.0625 = {4.25}^{2}$ .
$4.2$ .
Estimate $\sqrt{70}$ to one decimal place.
2 marks
${8.35}^{2} = 69.7225 < 70 < 71.4025 = {8.45}^{2}$ .
$8.4$ .
A square has area $55 {cm}^{2}$ . Estimate its side to the nearest tenth of a centimetre.
3 marks
The side is $\sqrt{55} cm$ .
${7.35}^{2} = 54.0225 < 55 < 55.5025 = {7.45}^{2}$ .
$7.4 cm$ .

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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 ↗

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