Exact/rounded missing lengths | Year 9

01 / LEARN

Understand the idea.

Exact means no information is lost

An exact length such as $\sqrt{74} cm$ retains the full result of the calculation. Its decimal expansion does not terminate, so any finite calculator display is an approximation. A square-root expression can itself be a finished answer.

For a right triangle, find the unknown side's square using Pythagoras, then take the positive square root. Retain that root or the full calculator value if the length is needed in a later calculation.

c = \sqrt{a^{2} + b^{2}}

Choose the requested precision

To report two decimal places, examine the third decimal digit. A digit of five or more rounds the hundredths up. Use $\approx$ when giving a rounded decimal, and keep a trailing zero when it shows the requested precision.

Round at the end rather than at every stage. Repeated rounding can shift a later result. Exact arithmetic also does not remove measurement uncertainty: a length calculated from measured inputs is only as reliable as those inputs.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

An irrational hypotenuse

Find the hypotenuse with legs 5 cm and 7 cm, exactly and to two decimal places.

Worked solution

1
Square and add the perpendicular legs.
$\begin{aligned} c^{2} & = 5^{2} + 7^{2} \\ = 25 + 49 = 74 \end{aligned}$
The missing side is the hypotenuse, so its square is the sum of the leg squares.
2
Keep the exact positive root.
$c = \sqrt{74} cm$
The positive square root gives the physical length without rounding.
3
Round to two decimal places.
$\begin{aligned} c & = 8.602325 \dots cm \\ \approx 8.60 cm \end{aligned}$
The third decimal digit is two, so the hundredths stay unchanged.

Answer

√74 cm; approximately 8.60 cm

EXAMPLE 02

Subtract first

A right triangle has hypotenuse 13 m and one leg 9 m. Find the other leg exactly and to two decimal places.

Worked solution

1
Write Pythagoras with the hypotenuse identified.
$x^{2} + 9^{2} = 13^{2}$
Thirteen metres is the hypotenuse, not another leg.
2
Subtract the known leg square.
$\begin{aligned} x^{2} & = 169 - 81 \\ = 88 \end{aligned}$
The two leg squares must add to the hypotenuse square.
3
State the exact length.
$x = \sqrt{88} m$
A length uses the positive square root.
4
Round the final value.
$\begin{aligned} x & = 9.380831 \dots m \\ \approx 9.38 m \end{aligned}$
The third decimal digit is zero; the result is also shorter than the thirteen-metre hypotenuse.

Answer

√88 m; approximately 9.38 m

03 / YOUR TURN

Try it yourself.

Check your understanding

Find the exact hypotenuse with legs 2 and 3.

Show answer & reasoning

$\sqrt{13}$

Pythagoras gives $c^{2} = 2^{2} + 3^{2} = 4 + 9 = 13$ . Taking the positive root gives the exact hypotenuse $c = \sqrt{13}$ .

Keep practising

6 worksheet questions · 15 marks

Open worksheet Worked solutions

Read the questions online

Find the hypotenuse with legs 4 cm and 6 cm, exactly and to two decimal places.
3 marks
Legs are 9 m and 12 m. Is 15 m an exact hypotenuse? Explain.
2 marks
Find the missing leg with hypotenuse 10 cm and other leg 6 cm.
2 marks
Find the missing leg with hypotenuse 15 m and other leg 10 m, exactly and to two decimal places.
3 marks
Legs are 4.3 m and 6.8 m. Find the hypotenuse to two decimal places.
3 marks
A 2 m by 3 m rectangle has diagonal $\sqrt{13}$ m. Why is 3.61 m only approximate?
2 marks

Read the worked solutions

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

Find the hypotenuse with legs 4 cm and 6 cm, exactly and to two decimal places.
3 marks
$c = \sqrt{16 + 36} = \sqrt{52} \approx 7.2111 \dots$ cm.
$\sqrt{52}$ cm; 7.21 cm
Legs are 9 m and 12 m. Is 15 m an exact hypotenuse? Explain.
2 marks
$c = \sqrt{81 + 144} = \sqrt{225} = 15$ m.
The root is an integer, so no rounding occurred.
Yes
Find the missing leg with hypotenuse 10 cm and other leg 6 cm.
2 marks
$x = \sqrt{100 - 36} = \sqrt{64} = 8$ cm.
8 cm
Find the missing leg with hypotenuse 15 m and other leg 10 m, exactly and to two decimal places.
3 marks
$x = \sqrt{225 - 100} = \sqrt{125} \approx 11.1803 \dots$ m.
$\sqrt{125}$ m; 11.18 m
Legs are 4.3 m and 6.8 m. Find the hypotenuse to two decimal places.
3 marks
$c^{2} = 18.49 + 46.24 = 64.73$ .
$c = \sqrt{64.73} \approx 8.0455 \dots$ m.
8.05 m
A 2 m by 3 m rectangle has diagonal $\sqrt{13}$ m. Why is 3.61 m only approximate?
2 marks
$d^{2} = 2^{2} + 3^{2} = 13$ .
3.61 is rounded; $\sqrt{13}$ retains the exact value.
$\sqrt{13}$ is irrational, so its decimal does not terminate.

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Curriculum reference & lesson scope

Australian Curriculum v9 and SA Mathematics

Year 9 similarity, Pythagoras and right-triangle trigonometric ratios, with prerequisite revision. Non-right-triangle sine/cosine rules are not required.

Official curriculum source ↗

In this lesson

Understand the idea.

Exact means no information is lost

Choose the requested precision

See the method in action.

An irrational hypotenuse

Square and add the perpendicular legs.

Keep the exact positive root.

Round to two decimal places.

Subtract first

Write Pythagoras with the hypotenuse identified.

Subtract the known leg square.

State the exact length.

Round the final value.

Try it yourself.

Check your understanding

Keep practising