A Tale of Three Intersecting Lines · Lesson 2 of 10
Constructing a Triangle When its Sides are Given
“Use two distance constraints to construct a triangle from three side lengths.”
• Construct a triangle when all three side lengths are given. • Explain why the third vertex must lie on two arcs. • Choose the correct compass opening for each endpoint of the base. • Identify isosceles triangles using equal radii. • Check a construction against its given measurements.
From Equal Sides to Different Sides
The equilateral construction used the same length for the base and both arcs. Now consider a triangle with sides 4 cm, 5 cm, and 6 cm. The idea stays the same, but the two compass openings differ. We locate a point that satisfies two distances at once, instead of adjusting a drawing until it looks plausible.
Choose AB = 4 cm as the base, and decide that AC = 5 cm and BC = 6 cm. Keep those assignments clear. C must be 5 cm from A and 6 cm from B. A circle centred at A with radius 5 cm contains every possible point satisfying the first distance; a circle centred at B with radius 6 cm does the same for the second.
The distance from the centre of a circle to any point on the circle. An arc is a portion of a circle.
Locating the Third Vertex
A point on just one of the circles may fail the other distance requirement. An intersection of the two circles lies on both and therefore satisfies both. In practice, we draw sufficiently long arcs near the expected third vertex; drawing the full circles is unnecessary.
- Draw AB = 4 cm.
- With centre A and radius 5 cm, draw an arc.
- With centre B and radius 6 cm, draw another arc that intersects the first.
- Name an intersection C and join AC and BC.
The construction gives AB = 4 cm because we drew that length directly. It gives AC = 5 cm because C lies on the first arc, and BC = 6 cm because it lies on the second. Notice that this explanation refers to the tools and distances, not to the apparent shape of the picture.
You could choose the 5 cm or the 6 cm side as the base instead, provided you correctly assign the remaining lengths to the two arcs. The placement on the page may change. After construction, measure every side as a check, and keep the arcs visible so the method is clear.
Match each arc to the required distance from its own centre. If AC is specified as 5 cm, the arc centred at A needs radius 5 cm. A rough sketch helps prevent switching labels. Slight measurement errors call for careful redrawing, not a change to the given lengths.
Problem
Construct a triangle with sides 3 cm, 4 cm, and 5 cm.
- 1.Draw PQ = 5 cm as the base. Assign PR = 3 cm and QR = 4 cm.
- 2.Draw a 3 cm arc from P and a 4 cm arc from Q. Name an intersection R.
- 3.Join PR and QR. The two radii and the drawn base give the required three lengths.
Equal Sides and Circles
The same method also constructs triangles with two equal sides. When the two arcs have equal radii, the third vertex is the same distance from both base endpoints. Circles let us recognise other examples too: join a centre to two different points on its circle, and the two radii have equal lengths.
A triangle with two equal sides. Here we describe the all-three-equal case separately as equilateral.
Problem
Construct a triangle with sides 4 cm, 4 cm, and 6 cm.
- 1.Draw AB = 6 cm. The remaining distances are AC = 4 cm and BC = 4 cm.
- 2.Draw 4 cm arcs from A and B, choose an intersection C, and join the two remaining sides.
- 3.AC and BC are equal radii, so the constructed triangle has two equal sides and is isosceles.
Suppose two circles of equal radius pass through each other’s centres A and B. If C is an intersection, AC and BC are radii, and AB is also a radius. Then all three lengths are equal, so ΔABC is equilateral. If AB is not a radius, equal circle sizes alone do not make the triangle equilateral.
Problem
O is the centre of a circle of radius 5 cm. P and Q are points on it, with O, P, Q not on one line. What can you say about ΔOPQ?
- 1.OP = 5 cm and OQ = 5 cm because both are radii.
- 2.Therefore ΔOPQ has two equal sides and is isosceles.
- 3.PQ is not automatically 5 cm. The triangle is equilateral only if PQ is also 5 cm.
Draw a circle and select two points on it that, together with the centre, are not on one straight line. Join the three points. Then draw two equal circles through each other’s centres and join the centres to an intersection. Compare the equal lengths in the two drawings.
You can extend the investigation to three equal circles. Choose their centres A, B, and C so that each centre lies on the other two circles. The triangle of centres is equilateral because each side is a radius of the same size. Use the centres and suitable circumference or intersection points to locate more isosceles triangles, justifying equal sides as radii rather than relying on the appearance.
We have successfully constructed several sets of lengths. Does that mean every set works? A drawing attempt can fail even when the compass settings are correct. The next lesson investigates what the three distances themselves tell us before we begin construction.
Quiz
AB = 4 cm, AC = 5 cm, and BC = 6 cm. What opening is needed for the arc centred at A?
Why does an intersection of the two arcs locate C?
Which construction makes a triangle with sides 4 cm, 4 cm, and 6 cm?
O is a circle’s centre and P and Q are points on it. Why are OP and OQ equal?
Two equal circles pass through each other’s centres A and B and intersect at C. What are AB, AC, and BC?
Practice Problems
- Construct triangles with sides 4, 6, 8 cm and 1, 5, 5 cm. State each base and the two compass openings.
- Construct a triangle with sides 3.5, 3.5, 3.5 cm and justify its side classification.
- Construct the 4, 5, 6 cm triangle using 6 cm as the base. Explain which arc belongs to each endpoint.
- Draw a circle and use its centre and two suitable circumference points to form an isosceles triangle. Explain the equality.
- Draw two equal circles, each passing through the other’s centre. Locate equilateral triangles using their intersections.
- Draw three equal circles with each centre on the other two circles. Use centres and circle points to identify an equilateral triangle and two isosceles triangles, explaining each equal length.
- A learner swaps the 5 cm and 6 cm arcs in a labelled construction with AC = 5 cm and BC = 6 cm. Which labelled requirements are wrong?
Key Takeaways
• Draw one given side as the base and use the other two lengths as arc radii. • An arc intersection satisfies both distance requirements for the third vertex. • Full circles are unnecessary if the drawn arcs intersect clearly. • Equal radii can justify equal sides and identify an isosceles triangle. • Verify all three labelled lengths after construction.