MathsICSE Class 10 · Constructions

Constructing the Incircle (Inscribed Circle) of a Triangle and of a Regular Hexagon (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

To inscribe a circle in a triangle, bisect any two of its angles. The bisectors meet at the incentre $I$, which is equidistant from the three sides. Drop a perpendicular from $I$ to one side; its length is the inradius. With centre $I$ and that radius, draw the circle touching all three sides.

Understand Incircle of a Triangle and a Regular Hexagon

The CISCE syllabus asks for 'inscribing a circle on a triangle and a regular hexagon', with ruler and compasses only.

Why angle bisectors? Every point on the bisector of an angle is equidistant from the two arms (a locus from the Loci chapter). The point $I$ where the bisectors of $∠ B$ and $∠ C$ meet is equidistant from all three sides. That point is the incentre; its distance from a side is the inradius $r$ .

Steps for a triangle.

1. Sketch, then construct the triangle accurately (compass angles only).
2. Bisect two of the angles; mark their meeting point $I$ .
3. From $I$ drop a perpendicular $I D$ to one side (arcs from $I$ cutting the side, then bisect that chord).
4. With centre $I$ and radius $I D$ , draw the circle. It should just touch all three sides.
5. Measure and state $r$ .

The perpendicular matters. The radius is the perpendicular distance from $I$ to a side, not the distance from $I$ to a vertex.

Facts that help you check.

The incentre is always inside the triangle, whatever its shape.
$r = \frac{Δ}{s}$ , where $Δ$ is the area and $s$ the semi-perimeter.
For a right triangle with legs $a$ , $b$ and hypotenuse $c$ : $r = \frac{a + b - c}{2}$ .
The tangent lengths from a vertex to the incircle are equal; from $A$ the length is $s - a$ (where $a = B C$ ).

Regular hexagon. The centre of a regular hexagon is also its incentre. Find it by bisecting two adjacent angles (or perpendicularly bisecting two sides). The inradius is the perpendicular from the centre to a side: $r = \frac{3}{2} a$ for side $a$ , the height of one of the six equilateral triangles.

In the papers. In 2025 Q3(i), asked for the circumcircle, some candidates drew the incircle instead; 2024 Q1(x) and 2025 Q1(ix) tested the difference between the circumcentre and the incentre, the meeting point of the angle bisectors.

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

Incentre: $distance from I to each side = r$; Meeting point of the angle bisectors.
Inradius (check): $r = \frac{Δ}{s}$; $s$ = semi-perimeter.
Right-angled triangle: $r = \frac{a + b - c}{2}$; $c$ the hypotenuse.
Regular hexagon: $r = \frac{3}{2} a$; Side $a$ .

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

Triangle, two angle bisectors, one perpendicular, then the circle.

1Construct the triangle with the given data, labelled as in the question.
2Bisect two angles with compasses; keep the arcs.
3Mark the incentre $I$ where the bisectors meet.
4Construct the perpendicular from $I$ to any side; the foot is the point of contact.
5Draw the circle with radius equal to that perpendicular, and check it touches the other two sides.
6Measure and write the radius; check with $r = \frac{Δ}{s}$ if time allows.

Incircle 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.

Largest circle inside a triangle

Suppose a triangular room has walls 6 m, 8 m and 10 m long, and you want the largest round rug that fits flat on the floor.

Find the radius of the rug.

Try it first, then show the solutionSolution

The largest circle inside a triangle is its incircle.
$6^{2} + 8^{2} = 1 0^{2}$ , so the room is right-angled: $r = \frac{6 + 8 - 10}{2}$ .

Answer: 2 m

Memory hook: Biggest circle inside = incircle.

Incircle of a regular hexagon

Imagine a hexagonal metal nut with sides of 1 cm. The round hole drilled through it must be smaller than the largest circle that fits inside the hexagon.

Find the radius of that largest circle.

Try it first, then show the solutionSolution

The largest circle inside a regular hexagon is its incircle.
$r = \frac{3}{2} \times 1 \approx 0.87$ cm.

Answer: About 0.87 cm

Memory hook: Hexagon inradius = height of one equilateral slice.

Tricks to remember Incircle of a Triangle and a Regular Hexagon

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

IN = angles INside
INcircle: bisect the angles at the corners INside the triangle. CIRCUMcircle: bisect the sides.
Use it when: Choosing the right construction.
Perpendicular, not to the corner
The incircle's radius is dropped square on to a side.
Use it when: Getting the radius right.
Area = r × s
Three triangles from $I$ with height $r$ make up the whole area.
Use it when: Checking a measured inradius.

Incircle 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: What the incentre is equidistant from · 1 mark · MCQBoard level

The incentre of a triangle is equidistant from its:
(a) vertices
(b) sides
(c) mid-points of the sides
(d) medians

Try it first, then show the solutionSolution

Each angle bisector is the locus of points equidistant from two sides; the incentre is on all three.

Answer: (b)

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

Construct $△ A B C$ with $B C = 7$ cm, $C A = 7.5$ cm and $A B = 6.5$ cm. Construct its incircle and measure its radius.

Try it first, then show the solutionSolution

Draw $B C = 7$ cm; arcs of 6.5 cm from $B$ and 7.5 cm from $C$ meet at $A$ .
Bisect $∠ B$ and $∠ C$ ; they meet at $I$ . Drop $I D ⊥ B C$ .
Draw the circle with centre $I$ and radius $I D$ .
$r \approx 2.0$ cm. Check: $s = 10.5$ , $Δ = 10.5 \times 3.5 \times 3 \times 4 = 441 = 21$ , $r = \frac{21}{10.5} = 2$ cm.

Answer: $r \approx 2.0$ cm

Example 3: Incircle with a 60° angle · 4 marks · Long answerHarder board level

Using ruler and compasses only, construct $△ A B C$ with $A B = 5$ cm, $B C = 8$ cm and $∠ A B C = 6 0^{\circ}$ . Construct its incircle, measure its radius, and verify.

Try it first, then show the solutionSolution

Draw $B C = 8$ cm, construct $6 0^{\circ}$ at $B$ and cut $B A = 5$ cm; join $A C$ (it measures 7 cm).
Bisect $∠ B$ and $∠ C$ to get $I$ ; drop $I D ⊥ B C$ ; draw the circle with radius $I D$ .
$r \approx 1.7$ cm.
Check: $Δ = \frac{1}{2} \times 5 \times 8 \times \frac{3}{2} = 103$ , $s = 10$ , so $r = 3 \approx 1.73$ cm.

Answer: $r \approx 1.7$ cm

Example 4: Incircle of a regular hexagon · 4 marks · Long answerHarder board level

Construct a regular hexagon of side 3 cm and its inscribed circle.
(a) Measure the inradius and check it.
(b) Taking $π = 3.14$ , find the area of the hexagon outside the circle, to 2 decimal places.

Try it first, then show the solutionSolution

Draw a circle of radius 3 cm, step off the radius six times and join the points to get the hexagon.
Bisect two adjacent angles (or two sides); they meet at the centre $O$ . Drop $O M ⊥$ one side and draw the circle with radius $O M$ .
(a) $r \approx 2.6$ cm; check $\frac{3}{2} \times 3 \approx 2.598$ cm.
(b) Hexagon area $= \frac{3 3}{2} \times 9 \approx 23.383$ cm²; circle $= 3.14 \times 6.75 = 21.195$ cm². Difference $\approx 2.19$ cm².

Answer: (a) $r \approx 2.6$ cm (b) about 2.19 cm²

Example 5: Points of contact in a right triangle · 4 marks · Long answerHOTS / competency

Construct $△ A B C$ with $A B = 5$ cm, $B C = 12$ cm and $∠ A B C = 9 0^{\circ}$ . Construct its incircle, touching $B C$ , $C A$ and $A B$ at $D$ , $E$ and $F$ .
(a) Find $r$ without measuring.
(b) Find $B D$ , $A F$ and $C E$ , and use them to check your construction.

Try it first, then show the solutionSolution

Construct the right angle at $B$ with compasses; $A C = 13$ cm. Bisect $∠ B$ and $∠ C$ , drop the perpendicular, draw the circle.
(a) $r = \frac{5 + 12 - 13}{2} = 2$ cm.
(b) $I D B F$ is a square, so $B D = B F = 2$ cm. Then $A F = 5 - 2 = 3$ cm and $C D = 12 - 2 = 10$ cm.
Equal tangents: $A E = A F = 3$ cm and $C E = C D = 10$ cm; $A E + C E = 13 = A C$ ✓. Measure $B D$ on your drawing: it should be 2 cm.

Answer: (a) 2 cm (b) $B D = 2$ cm, $A F = 3$ cm, $C E = 10$ cm

Example 6: Case study: a fountain in a triangular plot · 4 marks · Case-basedHOTS / competency

A triangular garden has sides 13 m, 14 m and 15 m. The largest possible circular fountain is to be built inside it.
(a) Which construction locates the fountain's centre?
(b) Find the radius of the fountain.
(c) Taking $π = 3.14$ , find the garden area left around the fountain.

Try it first, then show the solutionSolution

(a) The incircle: bisect two angles of the triangle (drawn to scale, say 1 cm : 2 m); the centre is the incentre.
(b) $s = 21$ , $Δ = 21 \times 8 \times 7 \times 6 = 7056 = 84$ m², so $r = \frac{84}{21} = 4$ m (2 cm on the drawing).
(c) Area left $= 84 - 3.14 \times 16 = 84 - 50.24 = 33.76$ m².

Answer: (b) 4 m (c) 33.76 m²

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

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

Mistake: Bisecting sides instead of angles
Perpendicular bisectors of the sides give the circumcentre.
Wrong working
Perpendicular bisectors of $A B$ and $B C$ used for an incircle.
Right working
Bisect $∠ B$ and $∠ C$ .
Do this: INcircle ↔ angles; circumcircle ↔ sides.
Usually costs: The whole construction; the 2025 report notes the reverse mix-up in Q3(i).
Mistake: Taking the radius to a vertex
The incircle touches the sides, so its radius is the perpendicular from $I$ to a side.
Wrong working
Circle with centre $I$ through $A$ .
Right working
Drop $I D ⊥ B C$ and use radius $I D$ .
Do this: Construct the perpendicular from $I$ before drawing the circle.
Usually costs: The circle and the measurement.
Mistake: Confusing incentre with circumcentre in theory questions
The incentre is equidistant from the sides; the circumcentre from the vertices.
Wrong working
'The incentre is equidistant from the vertices.'
Right working
'The incentre is the point where the angle bisectors meet; it is equidistant from the sides.'
Do this: Learn both definitions as a pair.
Usually costs: Reported in 2024 Q1(x) and 2025 Q1(ix).

Exam tips

Use compasses for every angle in the triangle; using a protractor was a common error in 2025 Q3(i).
Bisecting two angles is enough; the third bisector is a check.
Show the perpendicular from $I$ to the side with its arcs, then draw the circle.
State the measured radius to the nearest mm, and check with $r = \frac{Δ}{s}$ when the sides are known.

Practice questions on Incircle 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 radius of the circle inscribed in a regular hexagon of side 6 cm is:
(a) 3 cm
(b) $33$ cm
(c) 6 cm
(d) $63$ cm
Hint
$r = \frac{3}{2} a$ .
Answer
(b)
Q2 · 3 marks · Short answerBoard level
Construct a triangle with sides 5 cm, 5 cm and 6 cm. Construct its incircle and measure its radius.
Hint
Area $= \frac{1}{2} \times 6 \times 4$ .
Answer
$r = 1.5$ cm ( $Δ = 12$ cm², $s = 8$ cm).
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The incentre of a triangle always lies inside the triangle.
Reason (R): The incentre is the meeting point of the bisectors of the angles, and each of these bisectors lies inside the triangle.
(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
Compare with the circumcentre, which can lie outside.
Answer
(c)
Q4 · 4 marks · Long answerHarder board level
Using ruler and compasses only, construct $△ A B C$ with $B C = 7$ cm, $∠ A B C = 9 0^{\circ}$ and $∠ A C B = 4 5^{\circ}$ . Construct its incircle and measure its radius. Verify by calculation.
Hint
The triangle is isosceles: $A B = 7$ cm, $A C = 72$ cm.
Answer
$r \approx 2.1$ cm; check $\frac{7 + 7 - 7 2}{2} \approx 2.05$ cm.
Q5 · 3 marks · Short answerHarder board level
Construct a triangle with sides 9 cm, 10 cm and 17 cm. Construct its incircle, measure its radius and verify using $r = \frac{Δ}{s}$ .
Hint
$s = 18$ .
Answer
$Δ = 18 \times 9 \times 8 \times 1 = 36$ cm², so $r = 2$ cm.
Q6 · 3 marks · Short answerHarder board level
Construct $△ A B C$ with $A B = B C = 5$ cm and $∠ A B C = 12 0^{\circ}$ . Construct its incircle and measure its radius. Does the incentre lie inside this obtuse triangle?
Hint
$A C = 53 \approx 8.66$ cm.
Answer
$r \approx 1.2$ cm ( $Δ = \frac{25 3}{4} \approx 10.83$ , $s \approx 9.33$ , $r \approx 1.16$ cm). Yes, the incentre is always inside.
Q7 · 4 marks · Long answerHOTS / competency
The sides of a triangle are $3 x$ , $4 x$ and $5 x$ cm, and its inradius is 3 cm. Find $x$ , then construct the triangle and its incircle.
Hint
The triangle is right-angled.
Answer
$r = \frac{3 x + 4 x - 5 x}{2} = x$ , so $x = 3$ : sides 9 cm, 12 cm, 15 cm. Construct the right angle between the 9 cm and 12 cm sides, bisect two angles and draw the circle of radius 3 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 incentre?Show answer
The bisectors of the angles.
2.How do you get the inradius from the construction?Show answer
Drop a perpendicular from the incentre to a side.
3.Inradius of a right triangle with sides 6, 8, 10?Show answer
2
4.Can the incentre lie outside the triangle?Show answer
No, never.
5.Inradius of a regular hexagon of side $a$ ?Show answer
$\frac{3}{2} a$

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

How do you construct the incircle of a triangle?

Bisect two angles of the triangle; they meet at the incentre. Draw a perpendicular from the incentre to one side. With the incentre as centre and that perpendicular as radius, draw the circle; it touches all three sides.

What is the difference between the incircle and the circumcircle?

The incircle lies inside the triangle and touches its three sides; its centre is where the angle bisectors meet. The circumcircle passes through the three vertices; its centre is where the perpendicular bisectors of the sides meet.

How do you inscribe a circle in a regular hexagon?

Find the centre by bisecting two adjacent angles (or two sides), drop a perpendicular from the centre to a side, and draw the circle with that radius. It equals √3/2 times the side.