Constructions ICSE Class 10: Tangents, Circumcircles and Incircles

What Constructions is about

Weightage: CISCE publishes no chapter or unit weightage for ICSE Mathematics. Constructions is in the 3. Geometry (c) Circles (iv) Constructions part of the 2027 syllabus and can be asked in both Section A and Section B of the 80-mark paper.Source: CISCE ICSE Mathematics syllabus, examination year 2027

Key concepts

Tangents from an external point

Given a circle with centre $O$ and a point $P$ outside it:
1. Join $OP$.
2. Bisect $OP$ (perpendicular bisector) to find its mid-point $M$.
3. With centre $M$ and radius $MO$, draw a circle cutting the given circle at $A$ and $B$.
4. Join $PA$ and $PB$: these are the tangents.

Why it works: $OP$ is a diameter of the new circle, so $\angle OAP=90^{\circ }$ (angle in a semicircle). A line perpendicular to the radius at its end is a tangent.

Check: $PA=PB=\sqrt{O{P}^2-r^2}$.

Worked example

Tangents are drawn from $P$, 7 cm from the centre of a circle of radius 3 cm. What length should you measure for each tangent?

Try it first, then show the solutionSolution

1. $PA=\sqrt{7^2-3^2}=\sqrt{40}$.
2. $\sqrt{40}\approx 6.32$ cm, so expect about 6.3 cm.

Answer: About 6.3 cm

Constructing standard angles with compasses

Triangles in these questions often need these angles, made without a protractor:

• $60^{\circ }$: arc from the vertex; same radius from where it cuts the arm.
• $120^{\circ }$: step the same radius twice along the arc.
• $90^{\circ }$: bisect the angle between $60^{\circ }$ and $120^{\circ }$ (or erect a perpendicular).
• $30^{\circ }$ and $45^{\circ }$: bisect $60^{\circ }$ and $90^{\circ }$.

Always draw a rough sketch first and label the vertices exactly as in the question.

Worked example

How do you construct $\angle ABC=120^{\circ }$ at $B$ on the line $BC$?

Try it first, then show the solutionSolution

1. With centre $B$, draw an arc cutting $BC$ at $X$.
2. With centre $X$ and the same radius, cut the arc at $Y$; with centre $Y$, cut it again at $Z$.
3. Join $BZ$: $\angle ZBC=120^{\circ }$.

Answer: Step the radius twice along the arc from BC.

Circumcircle of a triangle

The circumcircle passes through all three vertices. Its centre, the circumcentre $O$, is equidistant from the vertices, so it lies on the perpendicular bisector of each side.

Steps: construct the triangle; draw the perpendicular bisectors of two sides; their intersection is $O$; with radius $OA$, draw the circle through $A$, $B$ and $C$.

The circumcentre is inside an acute triangle, at the mid-point of the hypotenuse of a right triangle, and outside an obtuse triangle.

Worked example

Triangle $ABC$ has $\angle B=90^{\circ }$, $AB=6$ cm and $BC=8$ cm. Where is the circumcentre and what is the circumradius?

Try it first, then show the solutionSolution

1. $AC=\sqrt{36+64}=10$ cm.
2. The circumcentre is the mid-point of $AC$; $R=5$ cm.

Answer: Mid-point of AC; R = 5 cm

Incircle of a triangle

The incircle touches all three sides. Its centre, the incentre $I$, is equidistant from the sides, so it lies on the bisector of each angle.

Steps: construct the triangle; bisect two angles; their intersection is $I$; from $I$ draw a perpendicular $ID$ to one side; with radius $ID$, draw the circle.

Check the radius with $r=\frac{\text{area}}{s}$, where $s$ is the semi-perimeter (for a right triangle, $r=\frac{a+b-c}{2}$).

Worked example

Find the inradius you should measure for a triangle with sides 7 cm, 8 cm and 9 cm.

Try it first, then show the solutionSolution

1. $s=12$; area $=\sqrt{12\times 5\times 4\times 3}=\sqrt{720}\approx 26.83$ cm².
2. $r=26.83÷12\approx 2.24$ cm.

Answer: About 2.2 cm

Regular hexagon and its circumscribed circle

A regular hexagon is made of six equilateral triangles meeting at the centre, so its side equals the radius of its circumscribed circle.

To construct a regular hexagon of side $a$: draw a circle of radius $a$, step the same radius six times round it, and join the points in order.

To circumscribe a circle on a given regular hexagon: draw the perpendicular bisectors of two adjacent sides (or join two long diagonals); they meet at the centre $O$; draw the circle with radius $OA$ $(=a)$.

Worked example

A regular hexagon has side 3.5 cm. What is the radius of its circumscribed circle?

Try it first, then show the solutionSolution

1. The hexagon is six equilateral triangles of side 3.5 cm.
2. Each vertex is 3.5 cm from the centre: $R=3.5$ cm.

Answer: 3.5 cm

Inscribed circle of a regular hexagon

The inscribed circle touches all six sides. Its centre is the centre $O$ of the hexagon, found by bisecting two interior angles (each $120^{\circ }$) or by the perpendicular bisectors of two sides.

Its radius is the perpendicular from $O$ to a side, which is the height of an equilateral triangle of side $a$:
$r=\frac{\sqrt{3}}{2}a.$
Steps: find $O$; drop a perpendicular $OM$ to one side; draw the circle with radius $OM$.

Worked example

Find the radius of the circle inscribed in a regular hexagon of side 4 cm.

Try it first, then show the solutionSolution

1. $r=\frac{\sqrt{3}}{2}\times 4=2\sqrt{3}$.
2. $r\approx 3.46$ cm.

Answer: About 3.5 cm

Accuracy, checks and presentation

• Use a sharp pencil; lengths to the nearest mm.
• Show all arcs; do not rub out construction lines.
• Label every point as in the question.
• Write the measured value clearly, e.g. 'Radius = 3.6 cm'.
• Check by calculation where you can (Pythagoras for tangents and right triangles, $r=\frac{\sqrt{3}}{2}a$ for a hexagon).

Remember which centre is which: circumcentre from perpendicular bisectors of sides (equidistant from vertices); incentre from angle bisectors (equidistant from sides).

Worked example

A question asks for 'a circle passing through the vertices of triangle $PQR$'. Which bisectors do you draw?

Try it first, then show the solutionSolution

1. A circle through the vertices is the circumcircle.
2. Draw the perpendicular bisectors of two sides.

Answer: Perpendicular bisectors of two sides (not angle bisectors)

Maths in real life: Constructions

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.

Topics in this chapter

• Circumcircle of a Triangle and a Regular Hexagon
• Constructing Tangents from an External Point
• Incircle of a Triangle and a Regular Hexagon

Constructions formula sheet

Tangent length from P

$PT=\sqrt{O{P}^2-r^2}$

Check for the tangent construction.

Circumradius of a right triangle

$R=\frac{\text{hypotenuse}}{2}$

Centre at the mid-point of the hypotenuse.

Inradius of any triangle

$r=\frac{\Delta }{s}$

Δ = area, s = semi-perimeter.

Area from three sides (Heron)

$\Delta =\sqrt{s(s-a)(s-b)(s-c)}$

Used here only to check a measured inradius.

Regular hexagon of side a

$R=a,r=\frac{\sqrt{3}}{2}a$

Circumscribed and inscribed circles.

Equilateral triangle of side a

$R=\frac{a}{\sqrt{3}},r=\frac{a}{2\sqrt{3}}$

Both circles share the same centre.

How the board asks Constructions

The question patterns to expect, and the method that works for each.

Tricks to remember Constructions

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

• CIRCUM–sides–PERP, IN–angles–BISECT

  Circumcircle: perpendicular bisectors of sides. Incircle: bisectors of angles.

  Use it when: Never drawing the wrong circle.

• Half of OP

  Tangents from P: bisect OP, and use the half as a radius.

  Use it when: Tangent construction.

• Hexagon = six equilateral triangles

  Side = radius of the circle through its vertices.

  Use it when: Hexagon constructions.

• Sketch, construct, measure, check

  Four steps for every construction question.

  Use it when: Not losing easy marks.

Constructions: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section B style)

Read the situation, then answer every part. In the board paper this is a 10-mark question split into 3 + 3 + 4 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Constructions

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

• ✗Mistake: Using a protractor for 120°

  CISCE examiners' comments, 2025: in a ruler-and-compass question, many candidates used a protractor to make ∠BAC = 120° instead of constructing it.

  ✓Do this: Make 120° by stepping the compass radius twice along an arc from the arm.

  Usually costs: The angle-construction mark.

• ✗Mistake: Drawing the incircle instead of the circumcircle

  CISCE examiners' comments, 2025: several candidates bisected the angles and drew the incircle when the question asked for a circle circumscribing the triangle.

  ✓Do this: Circumscribe = through the vertices = perpendicular bisectors of sides.

  Usually costs: Most of the 4 marks.

• ✗Mistake: Confusing circumcentre and incentre

  CISCE examiners' comments, 2024 and 2025: in MCQ and statement questions, candidates thought the circumcentre was equidistant from the sides, or knew it only as the meeting point of side bisectors and not as the point equidistant from the vertices.

  ✓Do this: Circumcentre: equal distances to vertices. Incentre: equal distances to sides.

  Usually costs: 1 mark.

• ✗Mistake: Construction arcs not shown

  CISCE examiners' general comments, 2024: make all arcs visible and exhibit all traces of construction (steps need not be written unless asked).

  ✓Do this: Keep every arc; draw construction lines lightly but leave them.

  Usually costs: Marks for the construction steps.

• ✗Mistake: Wrong labels on the triangle

  CISCE examiners' comments, 2025: some candidates named the vertices of the triangle incorrectly, so the given sides and angle were in the wrong places.

  ✓Do this: Draw a labelled rough sketch from the question before constructing.

  Usually costs: Accuracy of the measured radius.

Practice questions on Constructions

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

1. Q1 · 1 mark · MCQBoard level

   The incentre of a triangle is the point where:
   (a) the medians meet
   (b) the perpendicular bisectors of the sides meet
   (c) the bisectors of the angles meet
   (d) the altitudes meet

   Hint

   In = inside, touching the sides.

   Answer

   (c)

2. Q2 · 1 mark · MCQBoard level

   The radius of the circle circumscribing a regular hexagon of side 5 cm is:
   (a) 2.5 cm
   (b) 5 cm
   (c) $5\sqrt{3}$ cm
   (d) 10 cm

   Hint

   Six equilateral triangles.

   Answer

   (b)

3. Q3 · 1 mark · MCQHarder board level

   Tangents are drawn from a point 10 cm from the centre of a circle of radius 6 cm. Each tangent measures:
   (a) 4 cm
   (b) 6 cm
   (c) 8 cm
   (d) 16 cm

   Hint

   √(100 − 36).

   Answer

   (c)

4. Q4 · 1 mark · Assertion-reasonBoard level

   Assertion (A): The circumcentre of a right-angled triangle lies on its hypotenuse.
   Reason (R): The angle in a semicircle is a right angle, so the hypotenuse is a diameter of the circumcircle.
   (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

   Converse of the semicircle result.

   Answer

   (c)

5. Q5 · 4 marks · Short answerHarder board level

   Construct triangle $ABC$ with $AB=8$ cm, $AC=6$ cm and $\angle BAC=90^{\circ }$ (use compasses for the right angle). Construct its incircle and measure its radius. Check by calculation.

   Hint

   BC = 10 cm.

   Answer

   Bisect two angles, drop a perpendicular from the incentre to a side, draw the circle. $r=\frac{8+6-10}{2}=2$ cm.

6. Q6 · 4 marks · Short answerHarder board level

   Draw a circle of radius 2.5 cm and a point $P$ 6.5 cm from its centre. Construct the tangents $PA$ and $PB$. Measure $PA$ and calculate the length of the chord $AB$.

   Hint

   Half of AB = OA × PA ÷ OP.

   Answer

   $PA=\sqrt{6.5^2-2.5^2}=6$ cm. Half-chord $=\frac{2.5\times 6}{6.5}\approx 2.31$ cm, so $AB\approx 4.6$ cm.

7. Q7 · 3 marks · Long answerBoard level

   Construct a regular hexagon of side 4 cm and draw its circumscribed circle. Write the radius and one reason.

   Hint

   Start with a circle of radius 4 cm.

   Answer

   Radius 4 cm: the hexagon is six equilateral triangles of side 4 cm meeting at the centre.

8. Q8 · 4 marks · Long answerHarder board level

   Construct triangle $PQR$ with $PQ=PR=4$ cm and $\angle QPR=120^{\circ }$. Construct its circumcircle and measure the radius. Where does the circumcentre lie?

   Hint

   QR = 4√3 cm; R = QR ÷ (2 sin 120°).

   Answer

   $R=4$ cm. The triangle is obtuse, so the circumcentre lies outside it, beyond $QR$ (on the bisector of $\angle P$).

9. Q9 · 4 marks · Long answerHOTS / competency

   Prove that the construction for tangents from an external point $P$ gives tangents, and that the two tangents are equal.

   Hint

   Angle in a semicircle, then RHS.

   Answer

   $A$ lies on the circle with diameter $OP$, so $\angle OAP=90^{\circ }$; $PA$ is perpendicular to the radius $OA$ at its end, so it is a tangent. Likewise for $PB$. $△OAP\cong △OBP$ (RHS: $OA=OB$, $OP$ common, right angles), so $PA=PB$.

10. Q10 · 4 marks · Long answerHarder board level

    Construct an equilateral triangle of side 6 cm. Construct its circumcircle and incircle, and measure both radii. What do you notice about the centres?

    Hint

    R = a/√3, r = a/(2√3).

    Answer

    $R\approx 3.5$ cm ($6÷\sqrt{3}\approx 3.46$), $r\approx 1.7$ cm ($\approx 1.73$). The two centres coincide, and $R=2r$.

11. Q11 · 1 mark · MCQHOTS / competency

    For a regular hexagon, the ratio of the radius of the inscribed circle to that of the circumscribed circle is:
    (a) $1:2$
    (b) $\sqrt{3}:2$
    (c) $2:\sqrt{3}$
    (d) $1:\sqrt{3}$

    Hint

    r = (√3/2)a and R = a.

    Answer

    (b)

12. Q12 · 5 marks · Long answerHOTS / competency

    Draw a circle with centre $O$ and radius 2 cm, and mark $P$ with $OP=4$ cm. Construct the tangents $PA$ and $PB$. (a) Measure $\angle APB$ and explain the value. (b) Show that triangle $PAB$ is equilateral and find $PA$. (c) Decide, with a reason, whether $O$ is the incentre of triangle $PAB$.

    Hint

    sin ∠APO = 2/4.

    Answer

    (a) $\sin \angle APO=\frac{2}{4}$, so $\angle APO=30^{\circ }$ and $\angle APB=60^{\circ }$. (b) $PA=PB$ and the apex angle is $60^{\circ }$, so the base angles are $60^{\circ }$: equilateral; $PA=\sqrt{16-4}=2\sqrt{3}\approx 3.46$ cm. (c) It is not: the incentre of $△PAB$ is $\frac{2}{3}$ of the way from $P$ along the median, i.e. $\frac{2}{3}\times 3=2$ cm from $P$, whereas $O$ is 4 cm from $P$ (outside the triangle, since $AB$ crosses $OP$ 3 cm from $P$).

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. 1.First step for tangents from an external point P?Show answer

   Join OP and bisect it.

2. 2.Why is PA a tangent in that construction?Show answer

   ∠OAP = 90° (angle in a semicircle on OP).

3. 3.Lines that locate the circumcentre?Show answer

   Perpendicular bisectors of the sides.

4. 4.Lines that locate the incentre?Show answer

   Bisectors of the angles.

5. 5.How do you get the incircle's radius?Show answer

   Perpendicular distance from the incentre to a side.

6. 6.Circumradius of a regular hexagon of side a?Show answer

   a

7. 7.Inradius of a regular hexagon of side a?Show answer

   (√3/2) a

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you construct 120° and 90° with compasses only?
• ☐Can you draw tangents from P and explain why they are tangents?
• ☐Do you use perpendicular bisectors for a circumcircle?
• ☐Do you use angle bisectors for an incircle?
• ☐Can you construct a regular hexagon from a circle?
• ☐Do you write the measured radius clearly?

Constructions: frequently asked questions