MathsICSE Class 10 · Constructions

Constructing the Circumcircle of a Triangle and of a Regular Hexagon (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

To circumscribe a circle about a triangle, construct the perpendicular bisectors of two sides. They meet at the circumcentre $O$, which is equidistant from all three vertices. With centre $O$ and radius $OA$, draw the circle through $A$, $B$ and $C$. For a regular hexagon, the radius equals the side.

Understand Circumcircle of a Triangle and a Regular Hexagon

The CISCE syllabus asks for 'circumscribing … a circle on a triangle and a regular hexagon'. All constructions are with ruler and compasses only.

Why perpendicular bisectors? A point on the perpendicular bisector of $A B$ is equidistant from $A$ and $B$ (a locus from the Loci chapter). The point where the bisectors of $A B$ and $B C$ meet is equidistant from $A$ , $B$ and $C$ , so a circle centred there passes through all three. That point is the circumcentre, and the radius is the circumradius $R$ .

Steps for a triangle.

1. Draw a rough sketch with the data marked.
2. Construct the triangle with ruler and compasses (angles of $6 0^{\circ}$ , $12 0^{\circ}$ , $9 0^{\circ}$ , $4 5^{\circ}$ , $3 0^{\circ}$ by compass, not protractor).
3. Construct the perpendicular bisectors of any two sides; keep all arcs.
4. Mark their meeting point $O$ .
5. With centre $O$ and radius $O A$ , draw the circle; check it passes through $B$ and $C$ .
6. Measure and state $R$ (to the nearest mm).

Where the centre lies.

Acute triangle: inside.
Right-angled triangle: at the mid-point of the hypotenuse (angle in a semicircle), so $R = \frac{hypotenuse}{2}$ .
Obtuse triangle: outside, beyond the longest side.

Checking a measurement. You can verify $R$ with $R = \frac{ab c}{4Δ}$ , where $Δ$ is the area (from Heron's formula or $\frac{1}{2} ab sin C$ ). This is only a check; the exam wants the construction and a measured value.

Regular hexagon. A regular hexagon of side $a$ is six equilateral triangles meeting at its centre, so its circumradius is $a$ .

*Circle about a hexagon:* construct the perpendicular bisectors of two adjacent sides (or join opposite vertices); they meet at the centre; the radius equals the side.
*Hexagon in a circle:* with the compasses set to the radius, step round the circle six times; join the points.

In the papers. 2025 Q3(i) (4 marks) asked for $△ A B C$ with $A B = 6$ cm, $A C = 4.5$ cm, $∠ B A C = 12 0^{\circ}$ and its circumcircle; the expected radius was 5.3 cm (±0.2). 2024 Q1(x) and 2025 Q1(ix) tested that the circumcentre is equidistant from the vertices.

Circumcircle of a Triangle and a Regular Hexagon: formulas to remember

Circumcentre: $O A = O B = O C = R$; Meeting point of the perpendicular bisectors of the sides.
Right-angled triangle: $R = \frac{hypotenuse}{2}$; Centre at the mid-point of the hypotenuse.
Check formula: $R = \frac{ab c}{4Δ}$; For checking a measured radius only.
Regular hexagon: $R = a$; Circumradius equals the side.

How to solve Circumcircle of a Triangle and a Regular Hexagon questions

Triangle first, then two perpendicular bisectors, then the circle, then measure.

1Sketch the triangle roughly and label it exactly as in the question.
2Construct the triangle accurately; make any special angle with compasses.
3Bisect two sides perpendicularly, with arcs above and below each side.
4Mark the meeting point $O$ and draw the circle through the vertices.
5Measure the radius and write 'Circumradius = … cm'.
6Quick check: is $O$ inside, on or outside the triangle as expected for an acute, right or obtuse triangle?

Circumcircle of a Triangle and a Regular Hexagon in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Right triangle: centre on the hypotenuse

Suppose a clockmaker must cut a round frame that passes exactly through three screws, which form a triangle with sides 8 cm, 15 cm and 17 cm.

Find the radius of the frame and where its centre is.

Try it first, then show the solutionSolution

$8^{2} + 1 5^{2} = 64 + 225 = 289 = 1 7^{2}$ , so the triangle is right-angled.
The hypotenuse is a diameter: $R = \frac{17}{2}$ .

Answer: 8.5 cm, centred at the mid-point of the 17 cm side.

Memory hook: Right angle → hypotenuse is the diameter.

Circle about a regular hexagon

Imagine a hexagonal floor tile with sides of 7 cm. A worker needs the smallest circular cutter that covers the whole tile.

What radius should the cutter have?

Try it first, then show the solutionSolution

The smallest covering circle passes through all six corners: it is the circumcircle.
For a regular hexagon, circumradius = side.

Answer: 7 cm

Memory hook: Hexagon: the radius walks round six times.

Tricks to remember Circumcircle of a Triangle and a Regular Hexagon

Short memory aids. Say each one aloud once, then use it on a question the same day.

Circum = corners
The CIRCUMcircle goes round the corners; build it from the SIDES (perpendicular bisectors). The INcircle sits inside; build it from the ANGLES.
Use it when: Never mixing up the two constructions.
Two bisectors are enough
The third perpendicular bisector passes through the same point; use it only as a check.
Use it when: Saving time in the exam.
Acute in, right on, obtuse out
The circumcentre is inside, on the hypotenuse, or outside the triangle.
Use it when: Spotting a wrong construction at a glance.

Circumcircle of a Triangle and a Regular Hexagon: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Where is the circumcentre? · 1 mark · MCQBoard level

The circumcentre of a triangle with an obtuse angle lies:
(a) inside the triangle
(b) on the longest side
(c) outside the triangle
(d) at the obtuse-angled vertex

Try it first, then show the solutionSolution

For an obtuse triangle, the perpendicular bisectors meet beyond the longest side, outside the triangle.

Answer: (c)

Example 2: Circumcircle from three sides · 3 marks · Short answerBoard level

Construct $△ A B C$ with $B C = 4$ cm, $C A = 5$ cm and $A B = 6$ cm. Construct its circumcircle and measure the circumradius.

Try it first, then show the solutionSolution

Draw $B C = 4$ cm. With centre $B$ radius 6 cm and centre $C$ radius 5 cm, draw arcs meeting at $A$ . Join $A B$ , $A C$ .
Construct the perpendicular bisectors of $A B$ and $B C$ ; they meet at $O$ (inside, as the triangle is acute).
With centre $O$ and radius $O A$ , draw the circle through $A$ , $B$ , $C$ .
$R \approx 3.0$ cm (check: $Δ \approx 9.92$ cm², $R = \frac{4 \times 5 \times 6}{4 \times 9.92} \approx 3.02$ cm).

Answer: $R \approx 3.0$ cm

Example 3: A 45° angle with compasses · 4 marks · Long answerHarder board level

Construct $△ A B C$ with $A B = 7$ cm, $B C = 5$ cm and $∠ A B C = 4 5^{\circ}$ , using ruler and compasses only. Construct its circumcircle and measure its radius.

Try it first, then show the solutionSolution

Draw $B C = 5$ cm. At $B$ construct $9 0^{\circ}$ and bisect it to get $4 5^{\circ}$ . Cut $B A = 7$ cm on this arm; join $A C$ (about 4.95 cm).
Construct the perpendicular bisectors of $B C$ and $A B$ ; they meet at $O$ .
Draw the circle with centre $O$ and radius $O B$ ; it passes through $A$ and $C$ .
$R \approx 3.5$ cm (check: $R = \frac{A C}{2 s i n 4 5 ^{\circ}} = \frac{4.95}{1.414} \approx 3.5$ cm).

Answer: $R \approx 3.5$ cm

Example 4: Hexagon inside a circle · 4 marks · Long answerHarder board level

Draw a circle of radius 3.2 cm. Inscribe a regular hexagon $A B C D E F$ in it using compasses only, and draw its circumscribed circle. Find the perimeter of the hexagon and its area to one decimal place.

Try it first, then show the solutionSolution

Mark $A$ on the circle. Keeping the compasses at 3.2 cm, step round: $B$ , $C$ , $D$ , $E$ , $F$ . Join them in order.
The given circle passes through all six vertices, so it is the circumscribed circle; its radius equals the side, 3.2 cm.
Perimeter $= 6 \times 3.2 = 19.2$ cm.
Area $= 6 \times \frac{3}{4} \times 3. 2^{2} = \frac{3 3}{2} \times 10.24 \approx 26.6$ cm².

Answer: Perimeter 19.2 cm; area about 26.6 cm²

Example 5: Spot the right angle · 3 marks · Short answerHOTS / competency

Construct $△ A B C$ with $B C = 6$ cm, $∠ A B C = 6 0^{\circ}$ and $∠ A C B = 3 0^{\circ}$ . Without bisecting any side, locate the circumcentre and draw the circumcircle. Justify your method and state the radius.

Try it first, then show the solutionSolution

Draw $B C = 6$ cm; construct $6 0^{\circ}$ at $B$ and $3 0^{\circ}$ at $C$ (bisect a $6 0^{\circ}$ angle). The arms meet at $A$ .
$∠ B A C = 18 0^{\circ} - 6 0^{\circ} - 3 0^{\circ} = 9 0^{\circ}$ , so $B C$ is a diameter of the circumcircle (angle in a semicircle).
The circumcentre is the mid-point of $B C$ (find it by measuring 3 cm); radius $= 3$ cm.

Answer: Centre at the mid-point of $B C$ ; $R = 3$ cm

Example 6: Case study: a signal mast for three towers · 4 marks · Case-basedHOTS / competency

On a map with scale 1 cm : 1 km, three mobile towers are at $A (0, 0)$ , $B (8, 0)$ and $C (2, 6)$ . A relay mast must be equally far from all three.
(a) Which construction locates the mast?
(b) Find its position by calculation.
(c) How far is it from each tower, to 2 decimal places?

Try it first, then show the solutionSolution

(a) The circumcentre: the meeting point of the perpendicular bisectors of $A B$ and $A C$ .
(b) Perpendicular bisector of $A B$ : $x = 4$ . Mid-point of $A C$ is $(1, 3)$ and slope of $A C$ is 3, so its bisector is $y - 3 = - \frac{1}{3} (x - 1)$ . At $x = 4$ : $y = 2$ . Mast at $(4, 2)$ .
(c) $R = 4^{2} + 2^{2} = 20 \approx 4.47$ km; check $O B = 16 + 4$ and $O C = 4 + 16$ ✓

Answer: (b) $(4, 2)$ (c) about 4.47 km

Where students lose marks in Circumcircle of a Triangle and a Regular Hexagon

The slips examiners see most often, and what to write instead.

Mistake: Drawing the incircle instead
Angle bisectors give the incentre, not the circumcentre.
Wrong working
Bisecting $∠ A$ and $∠ B$ to circumscribe a circle.
Right working
Perpendicular bisectors of two sides.
Do this: Circumcircle → sides; incircle → angles.
Usually costs: Reported in 2025 Q3(i): candidates bisected angles and drew the incircle.
Mistake: Using a protractor for the angle
The question says ruler and compasses; a protractor angle loses the construction mark.
Wrong working
Measuring $12 0^{\circ}$ with a protractor.
Right working
Construct $12 0^{\circ}$ by stepping off two $6 0^{\circ}$ arcs.
Do this: Learn the compass constructions for $6 0^{\circ}$ , $12 0^{\circ}$ , $9 0^{\circ}$ , $4 5^{\circ}$ and $3 0^{\circ}$ .
Usually costs: Reported in 2025 Q3(i).
Mistake: Confusing circumcentre and incentre in theory questions
The circumcentre is equidistant from the vertices; the incentre from the sides.
Wrong working
'The circumcentre is equidistant from the three sides.'
Right working
'The circumcentre is equidistant from the three vertices.'
Do this: Circumcentre ↔ corners; incentre ↔ sides.
Usually costs: Reported in 2024 Q1(x) and 2025 Q1(ix).

Exam tips

2025 Q3(i) (4 marks): construct $∠ A = 12 0^{\circ}$ with compasses, the triangle, and the perpendicular bisectors; the expected radius was 5.3 cm (±0.2). The examiners also noted wrongly named vertices: copy the labels from the question.
Keep all construction arcs; do not rub them out.
Write the measured radius in a sentence, to the nearest mm.
Use a sharp pencil and keep the compasses fixed when drawing the pairs of arcs for a bisector.

Practice questions on Circumcircle of a Triangle and a Regular Hexagon

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The circumradius of a right-angled triangle whose hypotenuse is 13 cm is:
(a) 5 cm
(b) 6.5 cm
(c) 12 cm
(d) 13 cm
Hint
The hypotenuse is a diameter.
Answer
(b)
Q2 · 3 marks · Short answerBoard level
Construct $△ P QR$ with $P Q = 5.5$ cm, $QR = 4.8$ cm and $R P = 7.3$ cm. Construct its circumcircle and measure its radius. What do you notice about the centre?
Hint
Check $5. 5^{2} + 4. 8^{2}$ .
Answer
$R \approx 3.7$ cm (exactly 3.65 cm). $5. 5^{2} + 4. 8^{2} = 7. 3^{2}$ , so $∠ Q = 9 0^{\circ}$ and the centre is the mid-point of $P R$ .
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The circumcentre of every triangle lies inside the triangle.
Reason (R): The circumcentre is equidistant from the three vertices.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
Think of a right-angled or obtuse triangle.
Answer
(b)
Q4 · 4 marks · Long answerHarder board level
Using ruler and compasses only, construct $△ A B C$ with $A B = 6$ cm, $A C = 5$ cm and $∠ B A C = 6 0^{\circ}$ . Construct its circumcircle and measure its radius.
Hint
$B C = 31 \approx 5.57$ cm; all angles are acute.
Answer
$R \approx 3.2$ cm (check: $\frac{31}{3} \approx 3.21$ cm); the centre lies inside the triangle.
Q5 · 3 marks · Short answerHarder board level
Construct a regular hexagon of side 2.8 cm and draw its circumscribed circle. Measure the distance between two opposite vertices and explain the result.
Hint
Opposite vertices and the centre are collinear.
Answer
5.6 cm: opposite vertices are the ends of a diameter of the circumcircle, whose radius equals the side 2.8 cm.
Q6 · 4 marks · Long answerHarder board level
Using ruler and compasses only, construct $△ A B C$ with $A B = 4$ cm, $B C = 7$ cm and $∠ A B C = 12 0^{\circ}$ . Construct its circumcircle, measure its radius and state where the centre lies.
Hint
$A C = 93 \approx 9.64$ cm.
Answer
$R \approx 5.6$ cm (check: $\frac{93}{2 s i n 12 0 ^{\circ}} \approx 5.57$ cm). The triangle is obtuse, so the centre is outside it, beyond $A C$ .
Q7 · 4 marks · Long answerHOTS / competency
$A B = 6$ cm. How many circles of radius 4 cm pass through both $A$ and $B$ ? Construct them, and find the distance of each centre from $A B$ to 2 decimal places.
Hint
The centres lie on the perpendicular bisector of $A B$ and are 4 cm from $A$ .
Answer
Two (one on each side of $A B$ ): cut the perpendicular bisector of $A B$ with an arc of radius 4 cm from $A$ . Distance $= 4^{2} - 3^{2} = 7 \approx 2.65$ cm.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Which lines meet at the circumcentre?Show answer
The perpendicular bisectors of the sides.
2.The circumcentre is equidistant from…?Show answer
The three vertices.
3.Circumcentre of a right-angled triangle?Show answer
The mid-point of the hypotenuse.
4.Circumradius of a regular hexagon of side 4.5 cm?Show answer
4.5 cm
5.How many perpendicular bisectors must you construct?Show answer
Two (the third is only a check).

Circumcircle of a Triangle and a Regular Hexagon: frequently asked questions

How do you construct the circumcircle of a triangle?

Construct the perpendicular bisectors of any two sides. Their meeting point is the circumcentre. With that point as centre and the distance to any vertex as radius, draw the circle; it passes through all three vertices.

Can the circumcentre lie outside the triangle?

Yes. It lies outside for an obtuse-angled triangle, on the hypotenuse for a right-angled triangle, and inside for an acute-angled triangle.

How do you circumscribe a circle about a regular hexagon?

Find the centre by bisecting two sides perpendicularly (or by joining opposite vertices). The radius equals the side of the hexagon.