ICSE Class 10 Maths Constructions Selina Solutions
ICSE Class 10 Maths Chapter 19 Constructions (Circles)
ICSE Class 10 Maths Chapter 19 Constructions (Circles) teaches ruler-and-compass methods for drawing tangents, incircles, circumcircles and regular-hexagon circle constructions. The safe method is to identify the correct locus: angle bisectors for an incircle, perpendicular bisectors for a circumcircle, and the circle on OP as diameter for tangents from an external point P.
The solutions below follow the Selina Concise Mathematics Class 10 ICSE Chapter 19 construction types. Where a result depends on measurement from a drawing, the answer is given as an approximate value and, where possible, checked by calculation.
Methods and formulae for Concise Mathematics Selina Solutions Class 10 ICSE Chapter 19 Constructions (Circles)
Remember this centre rule: a circle through vertices needs a point equally distant from vertices, so use perpendicular bisectors. A circle touching sides needs a point equally distant from sides, so use angle bisectors. A tangent must meet the radius at 90^\circ.
| Construction | Use | Reason |
|---|---|---|
| Tangents from external point P | Midpoint of OP, then circle on OP as diameter | Angle in a semicircle is 90^\circ |
| Incircle of triangle | Two internal angle bisectors | Incenter is equidistant from sides |
| Circumcircle of triangle | Two perpendicular bisectors of sides | Circumcenter is equidistant from vertices |
| Regular hexagon in a circle | Central angle 60^\circ | 360^\circ\div 6=60^\circ |
PT=\sqrt{OP^2-r^2}
\text{Interior angle of a regular }n\text{-gon}=\frac{(n-2)180^\circ}{n}
Teacher-led worked examples
Worked example 1: tangent length when OP=5\ \text{cm} and r=3\ \text{cm}
Step 1: At the contact point T, OT\perp PT.
Step 2: Use right triangle OPT.
PT^2=OP^2-OT^2=5^2-3^2=16
Final answer: PT=4\ \text{cm}.
Worked example 2: angle between tangents is 45^\circ
Step 1: If tangents meet at P, then OA\perp PA and OB\perp PB.
Step 2: In quadrilateral AOBP, \angle AOB+\angle APB=180^\circ.
\angle AOB=180^\circ-45^\circ=135^\circ
Final answer: Draw radii making 135^\circ, then draw perpendicular tangents at the end points.
Worked example 3: circumradius of equilateral triangle of side 4.5\ \text{cm}
Step 1: For an equilateral triangle, R=\frac{a}{\sqrt{3}}.
R=\frac{4.5}{\sqrt{3}}=1.5\sqrt{3}\approx 2.6\ \text{cm}
Final answer: The circumradius is about 2.6\ \text{cm}.
Exercise 19 step-by-step solutions
Use these as model written steps. Keep arcs visible, label all centres and points, and write the final construction statement.
Question 1(a) to 1(h): objective checks
1(a): One angle bisector and one perpendicular bisector do not by themselves define the incenter or circumcenter. Answer: option (d), none of these.
1(b): The incenter is the intersection of angle bisectors. Answer: option (b).
1(c): For a regular hexagon circumscribing a circle, the radius is the apothem \frac{\sqrt{3}}{2}a, which is less than side a. Answer: option (c).
1(d): A regular hexagon inscribed in a circle forms six equilateral triangles, so side equals radius. Answer: option (a).
1(e): Angle bisectors of a triangle are concurrent; therefore PC also bisects \angle C. Answer: option (c).
1(f): \angle B=180^\circ-(35^\circ+55^\circ)=90^\circ, so the circle with AC as diameter passes through B. Answer: option (c).
1(g): From the perpendicular bisectors, PA=PB and PB=PC, so P is the circumcenter. Answer: option (a).
1(h): A general inscribed pentagon need not have equal central angles, but a regular pentagon has \angle AOB=\frac{360^\circ}{5}. Answer: option (d).
Question 2: tangents from P, where radius =3\ \text{cm} and OP=5\ \text{cm}
Step 1: Draw the circle with centre O and radius 3\ \text{cm}; mark P so that OP=5\ \text{cm}.
Step 2: Construct the midpoint M of OP. With centre M and radius MO, draw a circle cutting the given circle at A and B.
Step 3: Join PA and PB. They are tangents.
PA=PB=\sqrt{5^2-3^2}=4\ \text{cm}
Final answer: Each tangent is 4\ \text{cm}.
Question 3: two tangents making 45^\circ
Step 1: Draw the circle of radius 5\ \text{cm}.
Step 2: Mark points A and B on it so that \angle AOB=180^\circ-45^\circ=135^\circ.
Step 3: Draw perpendiculars to OA and OB at A and B; let them meet at P.
Final answer: PA and PB are the required tangents.
Question 4: equilateral triangle of side 4.5\ \text{cm} and circumcircle
Step 1: Draw BC=4.5\ \text{cm}. With centres B and C, radius 4.5\ \text{cm}, draw arcs meeting at A.
Step 2: Join AB and AC. Draw perpendicular bisectors of two sides meeting at O.
Step 3: With centre O and radius OA, draw the circumcircle.
Final answer: Radius \approx 2.6\ \text{cm}.
Question 5: triangle with BC=7\ \text{cm}, AB-AC=1\ \text{cm}, \angle ABC=45^\circ, and incircle
Step 1: Draw BC=7\ \text{cm} and ray BX such that \angle CBX=45^\circ.
Step 2: Mark E on BX with BE=1\ \text{cm}. Join EC, then draw the perpendicular bisector of EC meeting BX at A. Join AC.
Step 3: Draw angle bisectors of \angle B and \angle C, meeting at I. Draw IL\perp BC and the incircle with centre I, radius IL.
Final answer: Incircle radius measures about 1.8\ \text{cm}.
Question 6: circumcircle and proof that AD=BD
Step 1: Draw AB=6\ \text{cm}, ray AX with \angle XAB=60^\circ, and cut C on AX so that BC=7.5\ \text{cm}.
Step 2: Draw perpendicular bisectors of AB and AC, meeting at O. Draw the circumcircle and drop OD\perp AB.
Step 3: In right triangles \triangle OAD and \triangle OBD, OA=OB, OD is common and both have a right angle at D.
Hence proved: AD=BD by RHS congruence.
Question 7: triangle with BC=4\ \text{cm}, \angle ACB=45^\circ, altitude 2.5\ \text{cm}
Step 1: Draw BC=4\ \text{cm}. At C, draw a ray at 45^\circ to CB.
Step 2: Draw a line parallel to BC at distance 2.5\ \text{cm}; it meets the ray at A. Join AB.
Step 3: Draw perpendicular bisectors of two sides and draw the circumcircle.
Final answer: Circumradius is about 2.1\ \text{cm}.
Questions 8 and 9: circumcenter and incenter facts
Question 8: If perpendicular bisectors of AB and AC meet at O, then O is the circumcenter, OA=OB=OC, and the perpendicular bisector of BC also passes through O.
Question 9: If angle bisectors of A and B meet at O, then O is the incenter, OR=OQ, and \angle ACO=\angle BCO.
Question 10: circle cutting 2\ \text{cm} chords from each side of triangle 8, 6, 5
Step 1: Construct \triangle ABC with BC=6\ \text{cm}, AB=8\ \text{cm}, AC=5\ \text{cm}.
Step 2: Draw two angle bisectors meeting at incenter I. Draw ID\perp AB.
Step 3: On AB, mark DP=DQ=1\ \text{cm}. Draw the circle with centre I and radius IP.
Final answer: The circle cuts 2\ \text{cm} chords from each side.
Questions 11 and 12: regular hexagon circumcircle and incircle
Question 11: Construct a regular hexagon of side 5\ \text{cm}. Its circumcircle has radius equal to the side, so R=5\ \text{cm}.
Question 12: Construct a regular hexagon of side 5.8\ \text{cm}. Draw two angle bisectors to get centre I; draw IP\perp AB and the incircle.
IP=\frac{\sqrt{3}}{2}\times 5.8\approx 5.0\ \text{cm}
Question 13: cyclic quadrilateral with D equidistant from B and C
Step 1: Construct \triangle ABC with BC=5.5\ \text{cm}, AB=6\ \text{cm}, \angle ABC=120^\circ.
Step 2: Draw its circumcircle using perpendicular bisectors.
Step 3: Draw the perpendicular bisector of BC. Its other intersection with the circumcircle is D. Join BD and CD.
Final answer: ABCD is cyclic and DB=DC.
Question 14: point P equidistant from AB and BC on circle with BC as diameter
Step 1: Construct \triangle ABC with AB=3.5\ \text{cm}, BC=6\ \text{cm}, \angle ABC=120^\circ.
Step 2: Draw the circle with BC as diameter. Draw the angle bisector of \angle ABC, meeting the circle at P.
Final answer: P is equidistant from AB and BC, and \angle BCP=30^\circ.
Question 15: circle of radius 2.5\ \text{cm} through A, C, and tangents from B
Step 1: Draw AB=5\ \text{cm}, mark C so that AC=3\ \text{cm}. Construct centre O using equal arcs of radius 2.5\ \text{cm} from A and C.
Step 2: Draw the circle. Join OB, construct midpoint M, and draw circle with centre M, radius MO, cutting the first circle at P, Q.
Step 3: Join BP, BQ.
BP=BQ=\sqrt{10}\approx 3.2\ \text{cm}
Questions 16 to 19: remaining construction problems
Question 16: For AB=7\ \text{cm}, AC=5\ \text{cm}, \angle CAB=60^\circ, construct the triangle. The locus equidistant from AB and AC is the bisector of \angle A; the locus equidistant from BA and BC is the bisector of \angle B. Their intersection is the incenter; draw the incircle.
Question 17: Draw BC=6.8\ \text{cm} and its midpoint D. Locate A using arcs AB=5\ \text{cm} and AD=4.4\ \text{cm}. Draw angle bisectors to get the incenter and draw the incircle.
Question 18: Draw concentric circles of radii 4\ \text{cm} and 6\ \text{cm}. From point P on the outer circle, use the midpoint circle on OP to get tangent points on the inner circle.
PA=PB=\sqrt{6^2-4^2}=2\sqrt{5}\approx 4.5\ \text{cm}
Question 19: In right triangle ABC, draw BD\perp AC. The circumcircle of \triangle BDC has diameter BC=7.2\ \text{cm}, because \angle BDC=90^\circ.
\text{Radius}=\frac{7.2}{2}=3.6\ \text{cm}
Quick answer index
| Item | Answer |
|---|---|
| 1(a)-1(h) | (d),(b),(c),(a),(c),(c),(a),(d) |
| 2 | 4\ \text{cm} |
| 3 | Use centre angle 135^\circ |
| 4 | 2.6\ \text{cm} |
| 5 | About 1.8\ \text{cm} |
| 6 | AD=BD |
| 7 | About 2.1\ \text{cm} |
| 8 | Circumcenter; OA=OB=OC |
| 9 | Incenter; OR=OQ |
| 10 | Circle cuts 2\ \text{cm} chords |
| 11 | R=5\ \text{cm} |
| 12 | r\approx 5.0\ \text{cm} |
| 13 | Draw D on perpendicular bisector of BC |
| 14 | \angle BCP=30^\circ |
| 15 | \sqrt{10}\approx 3.2\ \text{cm} |
| 18 | 2\sqrt{5}\approx 4.5\ \text{cm} |
| 19 | 3.6\ \text{cm} |
Examiner’s mindset
In constructions, marks usually come from the correct base drawing, correct auxiliary locus, correct intersections, neat labelling and a final statement. For proof parts, state the theorem or congruence rule. For example, in Question 6, simply writing AD=BD is not enough; the RHS congruence reason completes the proof.
Common mistakes students make
- Using the wrong centre: use angle bisectors for incircle, perpendicular bisectors for circumcircle.
- Forgetting the tangent rule: radius and tangent meet at 90^\circ.
- Hiding construction arcs: keep arcs light but visible.
- Treating measurements as exact: ruler readings are approximate unless a calculation proves the value.
- Not labelling points: labels such as O, I, M, P, Q help the examiner follow your method.
How to practise this chapter
Practise by family: tangents first, then circumcircles, then incircles, then regular hexagons. After each diagram, write one reason: radius perpendicular to tangent, angle-bisector locus, or perpendicular-bisector locus.
Related study pages: ICSE Class 10 solutions, Selina Circles solutions for Class 10, Tangents and Intersecting Chords solutions, and Loci solutions for Class 10 Maths.
Sources used for accuracy
This page is based on the ICSE Class 10 Mathematics treatment of ruler-and-compass constructions, Selina Concise Mathematics Class 10 Chapter 19, and standard circle, tangent, locus, incircle and circumcircle facts. For official syllabus context, use the CISCE official website. For overlapping geometry basics, see NCERT.
Frequently Asked Questions
How should I write construction steps in ICSE Class 10 Maths Chapter 19?
Write the given measurements first, then list each ruler-and-compass step in order, name the centre or intersection point clearly, and end by stating the required circle, tangent, incircle or circumcircle.
Why do we draw a circle on OP as diameter while constructing tangents from an external point?
If the circle on OP as diameter meets the given circle at T, then \angle OTP=90^\circ. Since a tangent is perpendicular to the radius at the point of contact, PT is a tangent.
What is the difference between an incircle and a circumcircle?
An incircle touches the sides, so use angle bisectors. A circumcircle passes through vertices, so use perpendicular bisectors of sides.
Do measured construction answers have to be exact?
Measured construction answers are approximate unless an exact calculation supports them. Write the calculated check when it is available, such as PT=\sqrt{OP^2-r^2}.
Which theorem is used most often in this chapter?
The most used facts are: radius is perpendicular to tangent, angle bisectors meet at the incenter, and perpendicular bisectors of sides meet at the circumcenter.