Construct triangles from sufficient data
Construct triangles from side or angle data using retained compass arcs and protractor rays, checking feasible intersections and the specified side of the base.
Section 4.5 · Construction
Exam scope and assessment notes
Shared Foundation/Higher Construction covers F4.5A–D: measurement and drawing, triangles and other 2D shapes, scale drawings, and both straight-edge-and-compasses bisectors. These six original lessons are linked to those syllabus rows; the links do not establish complete textbook coverage or specialist approval. Use the named instruments and retain your marks, arcs and joins. A numerical answer or displayed reference alone does not show instrument skill. Physical millimetre work needs a real ruler and a verified object or calibrated print; recheck calibration after printing, resizing or zooming. Measure only diagrams explicitly supplied as accurate for measurement. Virtual drawing units do not certify physical millimetres. Written answers and model comparisons are for self-checking, with no marks, mastery or formal assessment approval. The source is linear 4MA1 Issue 2, checked against November 2026/June 2027 entry context; regional options depend on centre eligibility. Modular qualifications remain separate. Publisher chapter links are navigation aids: the relevant final chapter body and Teaching Hub content remain unverified. Other syllabus sections remain separate.
Before you begin
- Measure and draw calibrated line segments and use compasses to transfer a length.
- Measure angles with a protractor and interpret three-letter angle names.
- Use the triangle angle total and compare sums of positive lengths.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A triangle is to be constructed from three side lengths. After drawing its base, which operation locates the third vertex?
What you will learn
- Construct a triangle from three positive feasible side lengths using two compass arcs.
- Use a named included angle or two base angles with correctly directed protractor rays.
- Retain construction evidence and distinguish mirrored, degenerate and impossible data.
Place your own instruments and retain the joins, rays and arcs. The givens are shown first; your construction stays at the positions you choose.
Worked construction
Using a ruler and compasses, construct triangle ABC above AB with AB=10 cm, AC=6 cm and BC=8 cm. Draw the base yourself, keep both compass arcs and explain which intersection is chosen. Do not begin from a supplied completed third vertex.
Drawing unit: cm. Printed graduations calibrate the virtual drawing; they do not certify physical paper or monitor size.
All submitted joins and arcs stay visible. Select a recorded ruler or protractor to inspect its graduations; other instruments retain their actual footprints. Long lines and rays are cropped to this viewing frame.
Coordinates describe the virtual drawing in cm; they are placement controls, not endpoint readings. Positive angles turn anticlockwise; negative sweeps turn clockwise. Clicking chooses a visible position. It does not move an instrument onto a reference answer.
Read or use my recorded ruler
These readings come from your actual ruler placement. A misplaced ruler keeps its offset; this tool does not report the ideal endpoint reading.
Find intersections of my retained traces
Intersections use the actual arc sweeps and line extents. A full circle or supporting line cannot supply a crossing outside a traced arc, ray or segment.
This is a fixed mathematical comparison. A complete written response records an attempt; a teacher still needs to review the reasoning. Virtual graduations do not establish physical ruler, protractor or compass skill, and no marks or mastery are awarded. Verify physical calibration and recheck it after any print or zoom change.
Use compass circles to satisfy two distances
For three given sides, draw and label the base with a ruler. Set one compass opening to the distance from the first endpoint to the new vertex and draw an arc. From the other endpoint draw an arc with the remaining side length. Each intersection satisfies both distance conditions, so join the selected intersection to the base endpoints.
Two arcs can give mirror-image intersections on opposite sides of the base. If the question specifies above or below the base, use that condition to select the vertex; otherwise both may satisfy the side data. Retain the arcs rather than presenting only a finished triangle. A coordinate plot or a calculated third side does not show that the compass construction was performed.
Place angles at the named vertices
For side-angle-side data, draw the base, centre a protractor at the named endpoint and construct the included-angle ray from the correct baseline direction. Use the compass to transfer the second given side length onto that ray, then join to the remaining endpoint. The middle letter in an angle name identifies its vertex, so an angle at the wrong end does not meet the conditions.
For two base angles, construct one ray from each endpoint on the required side of the base. At the second endpoint the baseline points back along the base, not forward beyond it. The two inward-facing rays meet at the third vertex when the data are feasible. Keep the angle marks and rays, and measure requested unknown sides from the actual finished drawing rather than replacing the task with trigonometry.
Check that a genuine triangle exists
Side lengths must be positive and must satisfy the strict triangle inequalities: any two sides together must exceed the third. Equality gives a straight, degenerate arrangement, not a triangle. If two required compass circles do not meet, do not invent an intersection. For two-angle data, the positive angles must leave a positive third angle in the 180° total.
Check the completed side and angle conditions with suitable instruments, and state approximate measured results with the requested precision. The private exact geometry is a reference for consistency, not a required coordinate or surd method. A roughly correct appearance does not justify an unmarked angle, and an exact numerical answer alone does not provide the requested construction evidence.
Sources for this lesson
Pause and explain
Which condition rejects a non-collinear triangle from two given positive base angles?
Worked example
Using a ruler and compasses, construct triangle ABC above AB with AB=10 cm, AC=6 cm and BC=8 cm. Draw the base yourself, keep both compass arcs and explain which intersection is chosen. Do not begin from a supplied completed third vertex.
Show the worked solution
- Draw and label the 10 cm base AB with a ruler. Set the compass to 6 cm and draw an arc centred at A, then set it to 8 cm and draw an arc centred at B.
- Their crossings give points at the required distances from both endpoints. Select the crossing above AB as C; keep the lower crossing distinguishable as the unselected mirror possibility.
- Join AC and BC with the ruler edge. Recheck that AC is 6 cm and BC is 8 cm while AB remains 10 cm.
- Both mirror positions would satisfy the three side lengths, but only the upper one satisfies the stated orientation. The retained two-radius arcs explain the construction; private coordinates are unnecessary for the learner method.
Answer Constructed ABC has AB=10 cm, AC=6 cm and BC=8 cm with C above AB. Keep the 6 cm and 8 cm arcs; the lower mirror crossing is not the requested orientation.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Construct triangle ABC with AB=7 cm, AC=5 cm and the included angle BAC=60°, choosing C above AB. Use a protractor for the angle ray and compasses to transfer the 5 cm length. Retain both forms of construction evidence, join BC and measure BC to the nearest millimetre.
Give me a hint
The included angle is at A between AB and AC. Set the ray first, then mark the second side on it.
Compare my reasoning
- Draw AB=7 cm. At A align the protractor with AB and construct the 60° ray on the upper side of the base.
- With compasses transfer a 5 cm distance from A onto that ray to locate C. Keep the transferred-radius arc and the angle mark, then join BC.
- Measure BC on your finished accurately drawn triangle. The ideal comparison is about 6.245 cm, so its nearest-millimetre reference is 62 mm, or about 6.2 cm.
- Check AB, AC and the named included angle again. The measured BC is approximate; an exact surd from private geometry is not a replacement for drawing and measuring.
Construct AB=7 cm, AC=5 cm and ∠BAC=60° above the base. Reference measurement of BC: about 62 mm=6.2 cm, retaining the protractor ray and compass transfer.
Look for these in your work
- Place the 60° included angle at A with the correct baseline.
- Transfer the 5 cm length onto the ray and retain the construction marks.
- Measure BC to the nearest millimetre and identify the result as approximate.
Independently construct triangle ABC above AB with AB=8 cm, ∠CAB=45° and ∠CBA=60°. Draw both protractor rays from the correct baseline directions and explain why their crossing locates C. Measure AC and BC to the nearest millimetre and use the angle total as a check.
Give me a hint
At B the baseline ray is BA pointing back towards A. Both required rays rise towards the interior.
Compare my reasoning
- Draw and label AB=8 cm. At A construct the 45° ray from AB on the upper side.
- At B align with BA and construct the 60° ray on the same upper side. The two inward-facing rays meet at C, satisfying the angles at their named vertices.
- Join the sides and measure them on the completed drawing. Ideal nearest-millimetre references are AC≈72 mm=7.2 cm and BC≈59 mm=5.9 cm.
- The remaining angle is 180−45−60=75°, which checks the triangle angle total. Preserve both protractor rays and distinguish measured side readings from exact model values.
Construction above AB uses the 45° and 60° inward rays. Reference measurements: AC≈72 mm=7.2 cm, BC≈59 mm=5.9 cm; angle C is 75° as a check.
Look for these in your work
- Construct both angles at their named endpoints with inward baseline directions.
- Locate C by the ray intersection and retain the rays.
- Measure and label both side readings and check the 75° remaining angle.
Construct triangle ABC above AB with AB=7 cm, AC=4.5 cm and BC=5.5 cm. Keep the two compass arcs. Separately consider proposed side lengths 3 cm, 4 cm and 8 cm: use a compass-circle sketch and the distance conditions to explain why that proposal cannot form a triangle.
Give me a hint
For the valid case intersect the two endpoint arcs. For the second proposal compare the two smaller lengths with the longest base.
Compare my reasoning
- Draw the 7 cm base. Draw a 4.5 cm arc centred at A and a 5.5 cm arc centred at B; their upper crossing locates C.
- Join AC and BC, keep the arcs and check all three given lengths. The other crossing is a mirror solution to the lengths but is excluded by the instruction to use the upper side.
- For an 8 cm base, circles of radii 3 cm and 4 cm cannot meet because 3+4=7 cm, less than the 8 cm separation of their centres.
- Leave that second sketch visibly uncompleted and explain the gap. Do not supply a false third vertex or turn an impossible construction into a merely approximate triangle.
The 7 cm, 4.5 cm, 5.5 cm triangle can be constructed above AB. The 3 cm, 4 cm, 8 cm proposal is impossible because 3+4<8 and its endpoint circles do not meet.
Look for these in your work
- Construct the valid triangle using both specified radii and the upper intersection.
- Retain a separate honest circle sketch for the impossible data.
- Explain strict feasibility and avoid a fake or degenerate completed triangle.
Common mistakes
- Putting a given angle at the wrong vertex or using the forward extension instead of the base ray at the second endpoint.
- Erasing the compass arcs or using a calculated value as a substitute for the construction.
- Inventing a third vertex for impossible or degenerate side data, or silently discarding a permitted mirror solution.
What can you explain now?
Could side lengths 2 cm, 5 cm and 7 cm form a non-collinear triangle? Explain what the compass-circle boundary case would show.
Compare with the explanation
No. Since 2+5=7, the arrangement is degenerate rather than a triangle.
The two smaller lengths just reach along the longest base. The circles touch once on that line instead of giving a vertex away from the base; strict inequality is needed.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Recall the two-arc SSS method and the named-vertex protractor method. Explain the strict feasibility checks and why the specified side of the base selects between mirror intersections.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.