Estimating square roots and recognising pi | Year 8

01 / LEARN

Understand the idea.

Trap the root between squares

To bound a positive square root, compare its radicand with squares you know. Since $5^{2} < 30 < 6^{2}$ , taking positive square roots gives $5 < \sqrt{30} < 6$ . We use nonnegative numbers because squaring preserves their order.

For a narrower interval, square nearby tenths and compare again. Both endpoints must be checked: finding one close decimal does not establish an interval. A bound tells you where the exact root lies; it does not make either endpoint exact.

Pi is a constant ratio

For every circle, pi is the constant ratio $π = \frac{C}{d}$ , where $C$ is circumference and $d$ is diameter. Since $d = 2 r$ , the same relationship gives $C = π d = 2 π r$ . A larger circle keeps this ratio.

An expression such as $8 π cm$ is exact. A decimal circumference is approximate because $π$ has no finite decimal form. Keep extra digits during calculation, then round once to the precision requested; include the length unit.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A tenths-wide interval

Find consecutive tenths between which $\sqrt{20}$ lies.

Worked solution

1
Locate the root between integers.
$4^{2} = 16 < 20 < 25 = 5^{2}$
The positive root is between four and five, so suitable tenths lie in this interval.
2
Square the nearby tenths.
$\begin{aligned} {4.4}^{2} & = 19.36 \\ {4.5}^{2} & = 20.25 \end{aligned}$
These are consecutive tenths whose squares lie on opposite sides of twenty.
3
Take positive square roots.
$19.36 < 20 < 20.25 ⟹ 4.4 < \sqrt{20} < 4.5$
Squaring is increasing for nonnegative values, so the inequalities keep their order.

Answer

$4.4 < \sqrt{20} < 4.5$

EXAMPLE 02

Keep pi exact

A circle has radius 4 cm. Find its exact circumference and its circumference to two decimal places.

Worked solution

1
Find the diameter.
$d = 2 r = 2 \times 4 = 8 cm$
A diameter crosses the circle through its centre and contains two radii.
2
State the exact circumference.
$C = π d = 8 π cm$
The constant ratio of circumference to diameter is pi.
3
Calculate with extra digits.
$C \approx 8 \times 3.14159 = 25.13272 cm$
A decimal replacement for pi makes the result approximate.
4
Round to two decimal places.
$C \approx 25.13 cm$
The third decimal digit is two, so the hundredths digit stays three.

Answer

$8 π$ cm, approximately 25.13 cm.

03 / YOUR TURN

Try it yourself.

Check your understanding

Between which two consecutive integers is $\sqrt{50}$ ?

Show answer & reasoning

7 and 8.

Since $7^{2} = 49 < 50 < 64 = 8^{2}$ , the positive square root lies strictly between $7$ and $8$ .

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Place $\sqrt{35}$ between consecutive integers.
2 marks
Place $\sqrt{70}$ between consecutive integers.
2 marks
Use ${3.4}^{2} = 11.56$ and ${3.5}^{2} = 12.25$ to bound $\sqrt{12}$ .
2 marks
A circle has diameter 6 cm. Give its exact circumference, then use $π = 3.142$ to estimate it to two decimal places.
3 marks
Explain why a measured circumference-to-diameter ratio of 3.14 does not prove $π$ equals 3.14.
2 marks
Which is larger: $\sqrt{26}$ or 5.1? Justify without finding the root.
3 marks

Read the worked solutions

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

Place $\sqrt{35}$ between consecutive integers.
2 marks
$25 < 35 < 36$ .
$5 < \sqrt{35} < 6$
Place $\sqrt{70}$ between consecutive integers.
2 marks
$64 < 70 < 81$ .
$8 < \sqrt{70} < 9$
Use ${3.4}^{2} = 11.56$ and ${3.5}^{2} = 12.25$ to bound $\sqrt{12}$ .
2 marks
Twelve lies between the supplied squares.
$3.4 < \sqrt{12} < 3.5$
A circle has diameter 6 cm. Give its exact circumference, then use $π = 3.142$ to estimate it to two decimal places.
3 marks
$C = 6 π$ .
Using the supplied approximation, $6 \times 3.142 = 18.852$ , which rounds to 18.85.
$6 π$ cm; 18.85 cm.
Explain why a measured circumference-to-diameter ratio of 3.14 does not prove $π$ equals 3.14.
2 marks
Pi is an exact irrational constant.
Finite measurement precision cannot establish its full value.
Measurements and the reported ratio are approximate.
Which is larger: $\sqrt{26}$ or 5.1? Justify without finding the root.
3 marks
${5.1}^{2} = 26.01 > 26$ .
Both values are positive, so $5.1 > \sqrt{26}$ .
5.1

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Curriculum reference & lesson scope

Australian Curriculum v9 / SA Curriculum Mathematics

Four operations with integers; signed fraction/decimal calculations; order of operations; terminating and recurring decimals; rational and irrational numbers; estimating square roots and recognising pi; choosing efficient exact/decimal representations.

Official curriculum source ↗

In this lesson

Understand the idea.

Trap the root between squares

Pi is a constant ratio

See the method in action.

A tenths-wide interval

Locate the root between integers.

Square the nearby tenths.

Take positive square roots.

Keep pi exact

Find the diameter.

State the exact circumference.

Calculate with extra digits.

Round to two decimal places.

Try it yourself.

Check your understanding

Keep practising