Construction/spatial algorithms | Year 10

01 / LEARN

Understand the idea.

Inputs, steps and conditions

A construction algorithm needs clear inputs and repeatable actions. For distinct endpoints A and B, choose a compass radius greater than half AB and keep that same radius when drawing from each endpoint.

The two equal-radius circles intersect at C and D. Both intersections are equidistant from A and B, so the line through them is the perpendicular bisector of AB. Its crossing with AB identifies the midpoint.

Test the edge cases

The radius condition guarantees two intersections. If twice the radius is smaller than AB, the circles cannot reach each other; if it equals AB, they touch at just one point.

Check the construction for different segment lengths and orientations, but distinguish testing from justification. Equidistance explains why the exact construction always works under its conditions; drawing accuracy only affects your physical result.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Construct a midpoint

Explain how to construct the midpoint of AB using equal compass radii.

Worked solution

1
Choose and preserve a valid radius.
$r > \frac{A B}{2}$
Equal circles with this radius meet twice because their centres are closer than two radii apart.
2
Draw intersecting arcs from both endpoints.
$C A = C B = r, D A = D B = r$
Call the two intersections C and D; the unchanged compass setting gives both equal-distance pairs.
3
Join C to D.
$C D ⟂ A B$
The locus of points equidistant from A and B is their perpendicular bisector; both C and D lie on it.
4
Mark the midpoint at the crossing.
$M = A B \cap C D, A M = M B$
The perpendicular bisector intersects the segment at its midpoint.

Answer

M=AB∩CD

EXAMPLE 02

Choose a valid radius

AB is 10 cm long. A student selects compass radius 4 cm. Explain failure and give a working radius.

Worked solution

1
Compare the two circles’ combined reach.
$2 r = 2 \times 4 = 8 cm < 10 cm$
The centres A and B are farther apart than the sum of the radii, so the circles do not intersect.
2
Find the strict radius condition.
$r > \frac{10}{2} = 5 cm$
A radius of exactly 5 cm gives only one intersection; this construction needs two.
3
Choose a working setting.
$r = 6 cm, 2 r = 12 cm > 10 cm$
Using the same 6 cm radius at both endpoints produces two intersections in exact geometry.

Answer

Radius 6 cm works; 4 cm does not.

03 / YOUR TURN

Try it yourself.

Check your understanding

For a segment 8 cm long, why is radius 4 cm unsuitable for the two-intersection construction?

Show answer & reasoning

It produces only one intersection.

With $A B = 8 cm$ and $r = 4 cm$ , the circles just touch because $2 r = A B$ . One intersection cannot determine the required line; choose $r > 4 cm$ .

Keep practising

6 worksheet questions · 16 marks

Open worksheet Worked solutions

Read the questions online

Draw a segment AB of length 6 cm. Use ruler and compass to construct its perpendicular bisector. Leave your construction arcs visible.
3 marks
A segment is 12 cm long. Which radii 4, 6 and 8 cm produce two equal-circle intersections?
2 marks
Explain why each arc intersection is equidistant from A and B in this construction.
2 marks
Give a coordinate check for the perpendicular bisector of $A = (- 2, 1), B = (6, 1)$ .
3 marks
A student measures the midpoint with a ruler and then draws an approximate perpendicular. Is this the same as proving a compass construction? Explain.
3 marks
List one input condition and one output condition for an algorithm constructing a square from side AB.
3 marks

Read the worked solutions

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

Draw a segment AB of length 6 cm. Use ruler and compass to construct its perpendicular bisector. Leave your construction arcs visible.
3 marks
Use a radius greater than 3 cm from both endpoints.
Join both arc intersections.
The line through the two equal-radius arc intersections bisects AB at $90^{\circ}$ .
A segment is 12 cm long. Which radii 4, 6 and 8 cm produce two equal-circle intersections?
2 marks
The radius must be greater than 6 cm.
Only 8 cm.
Explain why each arc intersection is equidistant from A and B in this construction.
2 marks
This is how the two circles are constructed.
Both distances equal the unchanged compass radius.
Give a coordinate check for the perpendicular bisector of $A = (- 2, 1), B = (6, 1)$ .
3 marks
The midpoint is $(2, 1)$ .
AB is horizontal, so its perpendicular bisector is vertical.
$x = 2$
A student measures the midpoint with a ruler and then draws an approximate perpendicular. Is this the same as proving a compass construction? Explain.
3 marks
Measurement can approximate the result.
A construction proof uses exact geometric relationships.
No.
List one input condition and one output condition for an algorithm constructing a square from side AB.
3 marks
A zero-length segment cannot form a square.
The output conditions must define the required shape.
Input: A and B are distinct; output: four equal sides and four right angles.

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Curriculum reference & lesson scope

Australian Curriculum v9 and SA Curriculum

Angle and similarity reasoning; congruent/similar triangles in proofs; chords, radii and tangents; angles at the centre and circumference; circle theorems and their applications; connected deductive arguments; construction/spatial algorithms; testing and justifying solutions.

Official curriculum source ↗

In this lesson

Understand the idea.

Inputs, steps and conditions

Test the edge cases

See the method in action.

Construct a midpoint

Choose and preserve a valid radius.

Draw intersecting arcs from both endpoints.

Join C to D.

Mark the midpoint at the crossing.

Choose a valid radius

Compare the two circles’ combined reach.

Find the strict radius condition.

Choose a working setting.

Try it yourself.

Check your understanding

Keep practising