Exact roots versus rounded answers | Year 8

01 / LEARN

Understand the idea.

Not every length is a whole number

A square root such as $\sqrt{5}$ is an exact value, even though its decimal expansion does not terminate. A rounded decimal is an approximation, useful when a length must be reported to a stated accuracy.

Use $=$ for exact equality and $\approx$ for an approximation. Keep the root when exact form is requested; when decimal places are specified, give that many digits after the decimal point and retain the length unit.

Round once at the end

Keep exact squares and roots throughout the calculation, then round once at the end. Rounding an intermediate length and reusing it can change the final digit, especially when calculating a difference of lengths.

Use nearby square numbers to check a calculator result. Since $49 < 50 < 64$ , the positive root satisfies $7 < \sqrt{50} < 8$ . These bounds explain its size before a decimal display refines it.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Exact and measured forms

Find the hypotenuse of a right triangle with legs 1 cm and 2 cm. Give exact form and 2 decimal places.

Worked solution

1
Calculate the hypotenuse square.
$c^{2} = 1^{2} + 2^{2} = 1 + 4 = 5$
Both supplied sides are legs, so the squares add.
2
Keep the positive root as the exact answer.
$c = \sqrt{5} cm$
The root is an exact length, and a side length is positive.
3
Calculate and round to two decimal places.
$\sqrt{5} = 2.236067 \dots \approx 2.24$
The third decimal digit is 6, so the second decimal digit rounds up; the approximate length is 2.24 cm.

Answer

$\sqrt{5}$ cm; approximately 2.24 cm.

EXAMPLE 02

Check a decimal with bounds

A missing leg satisfies $x^{2} = 50$ . Explain its integer bounds and round to 1 decimal place.

Worked solution

1
Compare with consecutive square numbers.
$7^{2} = 49 < 50 < 64 = 8^{2}$
Squaring is increasing for positive numbers, so these squares bracket the positive root.
2
State the length bounds and exact root.
$7 < x < 8, x = \sqrt{50}$
A leg is positive, so the negative root is not a possible length.
3
Round the root to one decimal place.
$x = 7.071067 \dots \approx 7.1$
The hundredths digit is 7, so the tenths digit rounds up from zero to one.

Answer

7 < x < 8; x is approximately 7.1.

03 / YOUR TURN

Try it yourself.

Check your understanding

A hypotenuse has $c^{2} = 20$ . Between which consecutive integers is c?

Show answer & reasoning

4 and 5.

Because $4^{2} = 16 < 20 < 25 = 5^{2}$ , the positive hypotenuse length satisfies $4 < c < 5$ . Thus it lies between consecutive integers $4$ and $5$ .

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

Legs are 2 cm and 3 cm. Give the exact hypotenuse.
2 marks
Hypotenuse 6 cm, leg 4 cm: give the other leg exactly.
2 marks
Between which integers is $\sqrt{41}$ ?
2 marks
A calculator gives 9.433981 for a length in metres. Round to 2 decimal places.
2 marks
A calculator gives 6.0827625 for a length in centimetres. Round to 1 decimal place.
2 marks
For legs 5 cm and 7 cm, a calculator gives $\sqrt{74} = 8.602325 \dots$ . Give the exact answer, the answer to 2 decimal places, and an integer-bound check.
3 marks

Read the worked solutions

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

Legs are 2 cm and 3 cm. Give the exact hypotenuse.
2 marks
Add squares: $4 + 9 = 13$ .
Take the positive root without rounding.
$\sqrt{13}$ cm.
Hypotenuse 6 cm, leg 4 cm: give the other leg exactly.
2 marks
Subtract squares: $36 - 16 = 20$ .
Take the positive root.
$\sqrt{20}$ cm.
Between which integers is $\sqrt{41}$ ?
2 marks
The neighbouring squares are 36 and 49.
41 lies between them.
6 and 7.
A calculator gives 9.433981 for a length in metres. Round to 2 decimal places.
2 marks
The third decimal digit is 3.
Keep the hundredths digit unchanged.
9.43 m.
A calculator gives 6.0827625 for a length in centimetres. Round to 1 decimal place.
2 marks
The second decimal digit is 8.
Round the tenths digit up.
6.1 cm.
For legs 5 cm and 7 cm, a calculator gives $\sqrt{74} = 8.602325 \dots$ . Give the exact answer, the answer to 2 decimal places, and an integer-bound check.
3 marks
The squared sum is 74.
Use $64 < 74 < 81$ for the bounds, then round the supplied value.
$\sqrt{74}$ cm; approximately 8.60 cm; between 8 and 9.

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Curriculum reference & lesson scope

Australian Curriculum v9 and SA Curriculum Mathematics (2026)

Hypotenuse and right-angled triangles; why the theorem works; finding the hypotenuse; finding a shorter side; exact roots versus rounded answers; checking whether a triangle is right-angled; practical length problems.

Official curriculum source ↗

In this lesson

Understand the idea.

Not every length is a whole number

Round once at the end

See the method in action.

Exact and measured forms

Calculate the hypotenuse square.

Keep the positive root as the exact answer.

Calculate and round to two decimal places.

Check a decimal with bounds

Compare with consecutive square numbers.

State the length bounds and exact root.

Round the root to one decimal place.

Try it yourself.

Check your understanding

Keep practising