Concise Mathematics Selina Solutions Class 10 ICSE Chapter 19 Constructions (Circles)

Our free ICSE Class 10 Constructions Circles Solutions provide a clear, step-by-step guide to mastering one of the most practical chapters in your Maths syllabus. This chapter from the Class – 10 Concise Mathematics Selina textbook focuses on the hands-on skill of geometric drawing, specifically involving circles. You will learn the precise methods for constructing tangents to a circle from an external point, as well as how to draw the incircle and circumcircle for various triangles and regular polygons. Success in this chapter depends on understanding the underlying geometric properties and using your compass and ruler with accuracy, which these solutions will help you perfect.

If you are stuck on a specific construction problem or want to verify the correct sequence of steps, you have come to the right place. This page provides detailed, easy-to-follow solutions for all 26 questions in Exercise 19 of the Constructions (Circles) chapter. Each solution is prepared to match the exact method and presentation that the ICSE board expects in your final exams, ensuring you don’t lose marks for incorrect procedure. Here you will find clear diagrams and explanations for every single problem in your textbook.

Exercise 19

Question 1(a)

In the given figure, AP is bisector of angle A of △ ABC and DP is perpendicular bisector of side AB, then :

• (a) P is incenter of △ ABC
• (b) P is circumcenter of △ ABC
• (c) PB bisects angle B
• (d) none of these

The incenter of a triangle is located where all three angle bisectors meet. Meanwhile, the circumcenter is the point where the perpendicular bisectors of the triangle’s sides intersect.

In the scenario depicted, point P is formed by the intersection of an angle bisector and a perpendicular bisector of a side.

Hence, Option 4 is the correct option.

Question 1(b)

Incenter of a triangle is the point of intersection of the :

• (a) perpendicular bisector of its sides
• (b) bisectors of its angles
• (c) one perpendicular of its side and bisector of any one angle of it
• (d) none of these.

The incenter of a triangle is where the bisectors of its three internal angles meet.

Hence, Option 2 is the correct option.

Question 1(c)

For a regular hexagon, inscribing a circle, the length of the side of the hexagon and the radius of the circle are :

• (a) equal
• (b) not equal
• (c) side of hexagon is bigger than the radius of the circle
• (d) side of hexagon is smaller than the radius of the circle.

Consider that a regular hexagon is made up of 6 equilateral triangles.

∴ Triangle BAC is equilateral.

∴ The sides AB, BC, and CA are all equal.

Referring to the diagram,

⇒ AP is shorter than CA

⇒ AP is shorter than BC

⇒ Therefore, the radius is less than the side of the hexagon.

Hence, Option 3 is the correct option.

Question 1(d)

For a regular hexagon inscribed in a circle, the radius of the circle and the length of a side of the hexagon are :

• (a) equal
• (b) not equal
• (c) equal, if hexagon is regular
• (d) not equal, if hexagon is regular.

A regular hexagon can be divided into 6 identical equilateral triangles.

In the diagram, triangle BAC is equilateral.

∴ AB = BC = CA.

Observe that both AB and AC represent the radius of the circle.

⇒ AB = BC.

⇒ Therefore, the radius of the circle is equal to the side length of the hexagon.

Hence, Option 1 is the correct option.

Question 1(e)

In triangle ABC, bisectors of angles A and B meet at point P.

Assertion (A): PC bisects angle C.

Reason(R): Bisectors of angles of a triangle are concurrent.

• (a) A is true, R is false.
• (b) A is false, R is true.
• (c) Both A and R are true and R is correct reason for A.
• (d) Both A and R are true and R is incorrect reason for A.

The intersection point of the angle bisectors in a triangle is known as the incenter. This implies that the bisectors of a triangle’s angles meet at a single point. Therefore, the reason (R) is indeed true.

Observing the diagram, since point P is the incenter, it follows that the bisector of angle C must also pass through P. Consequently, PC bisects angle C, making the assertion (A) true.

∴ Both the assertion and the reason are true, and the reason correctly explains the assertion.

Hence, option 3 is the correct option.

Question 1(f)

In triangle ABC, ∠A = 35° and ∠C = 55°.

Assertion (A): Circle with AC as diameter will pass through the vertex B.

Reason(R): ∠ABC = 180° – (35° + 55°) = 90° = angle of semi-circle.

• (a) A is true, R is false.
• (b) A is false, R is true.
• (c) Both A and R are true and R is correct reason for A.
• (d) Both A and R are true and R is incorrect reason for A.

In triangle ABC, we know that ∠A = 35° and ∠C = 55°. Since the sum of angles in any triangle is 180°, we can set up the equation:

∠A + ∠B + ∠C = 180°

Substituting the given angles, we have:

35° + ∠B + 55° = 180°

Adding the known angles gives us:

90° + ∠B = 180°

To find ∠B, subtract 90° from both sides:

∠B = 180° – 90°

∴ ∠B = 90°.

Notice that when an angle is inscribed in a semicircle, it is always a right angle, which is 90°. This confirms that reason (R) is true.

Since ∠ABC is indeed 90°, and AC serves as the diameter, the circle will necessarily pass through vertex B. This is because any angle subtended by a diameter at the circle’s circumference is a right angle.

Thus, the assertion (A) is true as well.

∴ Both A and R are true, and R provides the correct explanation for A.

Hence, option 3 is the correct option.

Question 1(g)

In ΔABC, PQ is perpendicular bisector of side AB and PR is perpendicular bisector of side BC.

Statement (1): Perpendicular bisector of side AC will pass through point P.

Statement (2): Perpendicular bisectors of sides of a triangle are concurrent.

• (a) Both the statements are true.
• (b) Both the statements are false.
• (c) Statement 1 is true, and statement 2 is false.
• (d) Statement 1 is false, and statement 2 is true.

Given that PQ is the perpendicular bisector of AB, any point located on PQ will have equal distances to points A and B.

Similarly, PR being the perpendicular bisector of BC indicates that any point on PR is equidistant from points B and C.

Thus, point P, which is on both PQ and PR, maintains equal distance from A, B, and C.

This implies that P serves as the center of a circle that passes through points A, B, and C.

The perpendicular bisectors of the sides of a triangle converge at a single point known as the circumcenter of ∆ABC, which is equidistant from all three vertices.

Therefore, the statement that the perpendicular bisectors of a triangle’s sides are concurrent is accurate, confirming statement 2 as true.

Consequently, the perpendicular bisector of AC must intersect at point P, given that P is equally distant from A and C.

Thus, statement 1 is also true.

∴ Both the statements are true.

Hence, option 1 is the correct option.

Question 1(h)

Pentagon ABCDE is inscribed in a circle with center O.

Statement (1): ∠AOB = \dfrac{360°}{5}

Statement (2): If pentagon is regular, ∠AOB = \dfrac{360°}{5}

• (a) Both the statements are true.
• (b) Both the statements are false.
• (c) Statement 1 is true, and statement 2 is false.
• (d) Statement 1 is false, and statement 2 is true.

Consider pentagon ABCDE inscribed in a circle with center O.

When a pentagon is regular, each side creates equal angles at the circle’s center. Thus, for a regular pentagon ABCDE, the angle ∠AOB would be \dfrac{360°}{5}.

∴ Statement 1 is false, whereas statement 2 is true.

Hence, option 4 is the correct option.

Question 2

Draw a circle of radius 3 cm. Mark a point P at a distance of 5 cm from the center of the circle drawn. Draw two tangents PA and PB to the given circle and measure the length of each tangent.

Steps for the Construction:

1. Begin by sketching a circle with center labeled as O and a radius measuring 3 cm.
2. Identify a point P such that the distance from O to P is exactly 5 cm.
3. Construct the perpendicular bisector of the line segment OP, ensuring it crosses OP at point M.
4. Using M as the center and OM as the radius, draw a new circle. This circle should intersect the original circle centered at O at points A and B.
5. Connect points A and P, as well as B and P, to form segments AP and BP. Proceed to measure the lengths of AP and BP.

Thus, the segments AP and BP are the tangents you need.

Upon measurement, you will find:

AP = BP = 4 cm.

Hence, length of each tangent = 4 cm.

Question 3

Draw a circle of radius 5 cm. Draw two tangents to this circle so that the angle between tangents is 45°.

Steps for the Construction:

1. Begin by drawing a circle with center O and a radius of 5 cm. Use BC as the diameter of this circle.
2. Construct arcs from O that form an angle of 135° (since 180° – 45° = 135°) such that ∠AOB = 135°.
3. At points A and B, draw two rays at an angle of 90° to the radii OA and OB, respectively. These rays should intersect at point P.

Hence, AP and BP are required tangents making an angle of 45° with each other.

Question 4

Using ruler and compasses only, draw an equilateral triangle of side 4.5 cm and draw its circumscribed circle. Measure the radius of the circle.

Construction Steps:

1. Begin by sketching a line segment BC with a length of 4.5 cm.
2. Using a compass, set the radius to 4.5 cm. With the compass point on B, draw an arc. Then, with the compass point on C, draw another arc of the same radius. These arcs intersect at point A.
3. Connect A to B and A to C to form triangle ABC.
4. To find the center of the circumscribed circle, draw the perpendicular bisectors of sides AC and BC. Their intersection is point O.
5. With O as the center, draw a circle using the radius OA, OB, or OC. This circle should pass through points A, B, and C. Measure the length of OC.

Upon measurement, OC is found to be 2.6 cm.

Hence, above is the required circumcircle with radius 2.6 cm.

Question 5

Using ruler and compasses only,

(i) Construct triangle ABC, having given BC = 7 cm, AB – AC = 1 cm and ∠ABC = 45°.

(ii) Inscribe a circle in the △ABC constructed in (i) above.

(i) Steps for constructing the triangle:

1. Begin by drawing a line segment BC of length 7 cm.
2. At point B, construct a ray BX such that ∠CBX = 45° and mark a point E on BX such that BE = 1 cm, which represents AB – AC.
3. Connect points E and C, then construct the perpendicular bisector of EC. Let it intersect ray BX at point A.
4. Finally, join points A and C.

∴ Triangle ABC is the required triangle.

(ii) Steps for inscribing a circle in △ABC:

1. Construct the angle bisectors of ∠ABC and ∠ACB. Let these bisectors intersect at point O.
2. From point O, draw a perpendicular line OL to the side BC.
3. Using O as the center and OL as the radius, draw a circle that touches all sides of △ABC. Measure the length of OL.

Upon measurement, OL = 1.8 cm.

∴ The required incircle of △ABC has a radius of 1.8 cm.

Question 6

Using ruler and compasses only,

(i) Construct a triangle ABC with the following data :

Base AB = 6 cm, BC = 7.5 cm and angle CAB = 60°.

(ii) In same diagram, draw a circle which passes through the points A, B and C and mark its center O.

(iii) Draw a perpendicular from O to AB which meets AB in D.

(iv) Prove that : AD = BD.

Steps for the construction:

1. Begin by drawing a line segment AB = 6 \text{ cm}.
2. At point A, construct a ray AX such that it forms a 60^\circ angle with AB.
3. Using a compass, set B as the center and a radius of 7.5 \text{ cm} to draw an arc that intersects the ray AX at point C.
4. Connect points B and C to form triangle ABC.
5. Construct the perpendicular bisectors of segments AB and AC. Let these bisectors intersect at point O.
6. With O as the center, draw a circle using the radius OA, OB, or OC. This circle will pass through points A, B, and C.
7. From point O, draw a line OD perpendicular to AB.

Proof:

In the right-angled triangles \triangle OAD and \triangle OBD:

⇒ OA = OB (since both are radii of the same circle)

⇒ OD = OD (common side)

⇒ \angle ODA = \angle ODB (each is 90^\circ)

∴ \triangle OAD \cong \triangle OBD (by the R.H.S. congruence criterion)

By C.P.C.T. (Corresponding Parts of Congruent Triangles),

⇒ AD = BD.

Hence, proved that AD = BD.

Question 7

Using ruler and compasses only construct a triangle ABC in which BC = 4 cm, ∠ACB = 45° and perpendicular from A on BC is 2.5 cm. Draw a circle circumscribing the triangle ABC.

Steps for the construction:

1. Begin by sketching the line segment BC with a length of 4\, \text{cm}.
2. At point C, construct a perpendicular line CX. From C, measure and mark point E such that CE = 2.5\, \text{cm}.
3. From point E, draw a line EY that is perpendicular to CX.
4. Construct a ray from C that forms a 45^\circ angle with CB. Let this ray intersect EY at point A.
5. Connect points A and B to complete triangle ABC.
6. Find the perpendicular bisectors of sides AB and BC. Their intersection point is O.
7. With O as the center and OA, OB, or OC as the radius, draw a circle that passes through points A, B, and C. Measure OB.

On measuring,

OB = 2\, \text{cm}.

Hence, above is the required circumcircle of \triangle ABC with radius = 2\, \text{cm}.

Question 8

Perpendicular bisectors of the sides AB and AC of a triangle ABC meet at O.

(i) What do you call the point O ?

(ii) What is the relation between the distances OA, OB and OC ?

(iii) Does the perpendicular bisector of BC pass through O ?

(i) The point O is known as the circumcenter of the triangle.

(ii) Since OA, OB, and OC represent the radii of the circumcircle, we have:

Hence, OA = OB = OC.

(iii) Observing the diagram, we see that:

Yes, the perpendicular bisector of BC does indeed pass through O.

Question 9

The bisectors of angles A and B of a scalene triangle ABC meet at O.

(i) What is the point O called ?

(ii) OR and OQ are perpendiculars drawn to AB and CA respectively. What is the relation between OR and OQ ?

(iii) What is relation between angle ACO and angle BCO ?

(i) The point O is known as the incenter of △ABC.

(ii) Consider the figure where OR and OQ are drawn as perpendiculars to sides AB and CA, respectively.

Both OR and OQ serve as radii of the incircle of the triangle.

Thus, OR = OQ.

(iii) Since OC is the angle bisector of ∠C, it implies that:

∴ ∠ACO = ∠BCO.

Thus, ∠ACO = ∠BCO.

Question 10

(i) Using ruler and compasses only, construct a triangle ABC in which AB = 8 cm, BC = 6 cm and CA = 5 cm.

(ii) Find its incenter and mark it I.

(iii) With I as centre, draw a circle which will cut off 2 cm chords from each side of the triangle.

Steps for constructing the required triangle and circle:

1. Start by drawing a line segment BC of length 6 cm.
2. Using your compass, set the radius to 8 cm. Place the compass point on B and draw an arc.
3. Now, adjust the compass to a radius of 5 cm. Place the compass point on C and draw another arc that intersects the first arc. Label the intersection point as A.
4. Connect A to B and A to C. The triangle ABC is now constructed.
5. To locate the incenter I, draw the angle bisectors of angles B and A. Their intersection is the point I.
6. From I, draw a perpendicular line to AB, and let the foot of the perpendicular be D.
7. With D as the center, mark off points P and Q on AB such that DP = DQ = \frac{2}{2} = 1 cm.
8. Finally, using I as the center and radius IP (or IQ), draw a circle that will create 2 cm chords on each side of the triangle.

Question 11

Draw a circle circumscribing a regular hexagon with side = 5 cm.

For a regular hexagon, the formula for each interior angle is given by \Big(\dfrac{2n - 5}{n}\Big) \times 90°. Here, n = 6 for a hexagon.

Calculating, we have:

Construction steps:

1. Create a regular hexagon labeled ABCDEF, ensuring each side measures 5 cm and each interior angle is 120°.
2. Find the perpendicular bisectors of sides AB and AF. These bisectors will meet at point O.
3. Use O as the center and OA as the radius to draw a circle. This circle will intersect all the vertices of the hexagon.

Thus, the circle drawn is the circumcircle of the regular hexagon.

Question 12

Draw an inscribing circle of regular hexagon of side 5.8 cm.

To find each interior angle of a regular hexagon, use the formula: \Big(\dfrac{2n - 5}{n}\Big) \times 90°, where n is the number of sides.\
\begin{aligned}= \dfrac{2 \times 6 - 4}{6} \times 90° \\= 8 \times 15° \\= 120°.\end{aligned}

Construction steps:

1. Construct a regular hexagon, labeled ABCDEF, ensuring each side measures 5.8 cm and every interior angle is 120°.
2. Draw angle bisectors for the interior angles at vertices A and B. These bisectors will meet at a point, which we label as I.
3. From point I, draw a line segment IP that is perpendicular to side AB.
4. Using I as the center and IP as the radius, draw a circle. This circle will touch each side of the hexagon.

Hence, above is the required incircle of regular hexagon.

Question 13

Construct a triangle ABC in which base BC = 5.5 cm, AB = 6 cm and ∠ABC = 120°.

(i) Construct a circle circumscribing the triangle ABC.

(ii) Draw a cyclic quadrilateral ABCD so that D is equidistant from B and C.

(i) Steps for constructing the triangle and its circumcircle:

1. Begin by drawing a line segment BC measuring 5.5 cm.
2. At point B, construct a ray BX such that the angle ∠XBC is 120°.
3. Using a compass, draw an arc with a radius of 6 cm from point B, ensuring it intersects ray BX at point A.
4. Connect points A and C to complete triangle ABC.
5. To find the circumcenter, draw the perpendicular bisectors of sides AB and BC. Let these bisectors meet at point O.
6. With point O as the center and radius OA, draw a circle that passes through points A, B, and C.

Hence, above is the required circumcircle of triangle ABC.

(ii) Steps for constructing the cyclic quadrilateral:

1. Draw the perpendicular bisector of BC, ensuring it intersects the previously drawn circle at point D.
2. Connect points C and D, and also A and D, to form quadrilateral ABCD.

Hence, above is the required cyclic quadrilateral.

Question 14

Using a ruler and compasses only :

(i) Construct a triangle ABC with the following data :

AB = 3.5 cm, BC = 6 cm and ∠ABC = 120°.

(ii) In same diagram, draw a circle with BC as diameter. Find a point P on the circumference of the circle which is equidistant from AB and BC.

(iii) Measure ∠BCP.

Steps for the construction:

1. Begin by drawing a straight line segment BC that measures 6 cm.
2. At point B, construct a ray BX such that it forms an angle of 120^\circ with BC. Using a compass, set the width to 3.5 cm, place the compass point at B, and mark off point A on BX such that AB = 3.5 cm.
3. Connect points A and C to form triangle ABC.
4. To draw a circle with BC as the diameter, first find the midpoint O of BC by drawing the perpendicular bisector of BC. Then, using O as the center and OB as the radius, draw the circle.
5. To locate point P which is equidistant from AB and BC, construct the angle bisector of \angle ABC. The point where this bisector intersects the circle is point P.

Measure the angle \angle BCP.

Hence, \angle BCP = 30^\circ.

Question 15

Draw a line AB = 5 cm. Mark a point C on AB such that AC = 3 cm. Using a ruler and a compass only, construct :

(i) a circle of radius 2.5 cm, passing through A and C.

(ii) construct two tangents to the circle from the external point B. Measure and record the length of the tangents.

(i) Steps for constructing the circle:

1. Begin by drawing the line segment AB measuring 5 cm.
2. Position the compass at point A and draw an arc of 3 cm on AB, marking point C.
3. Set the compass to a radius of 2.5 cm and draw an arc centered at A.
4. Similarly, draw another arc of 2.5 cm radius centered at C, ensuring it intersects the previous arc. Label the intersection as point O.
5. Now, using O as the center and a radius of 2.5 cm, sketch the circle.

Thus, the circle passing through points A and C is constructed.

(ii) Steps for constructing the tangents:

1. Connect points O and B with a straight line.
2. Construct the perpendicular bisector of OB, which intersects OB at point M.
3. With M as the center, draw a circle using OM as the radius. This circle will intersect the original circle at points P and Q.
4. Draw straight lines from P to B and from Q to B. These lines, PB and QB, are the tangents.

Upon measurement, both PB and QB are found to be 3.2 cm.

Therefore, each tangent measures 3.2 cm.

(This construction is based on question 24 from chapter 19 of the ICSE Class 10 textbook. A reference diagram might be sized 505×494 pixels.)

Question 16

Using a ruler and a compass, construct a triangle ABC in which AB = 7 cm, ∠CAB = 60° and AC = 5 cm. Construct the locus of :

(i) points equidistant from AB and AC.

(ii) points equidistant from BA and BC.

Hence construct a circle touching the three sides of the triangle internally.

Steps for constructing the triangle and loci:

1. Begin by drawing a line segment AB = 7 cm.
2. Create the angle \angle XAB = 60^\circ by constructing AX such that \angle XAB is formed.
3. With A as the center, draw an arc with a radius of 5 cm on AX, and label the intersection as point C.
4. Connect B to C. The triangle ABC is now complete.
5. Construct the angle bisectors AY and BZ from angles at A and B respectively.
6. The point where AY and BZ intersect is labeled as O.
7. From O, draw a perpendicular OD to AB.
8. Use O as the center and OD as the radius to draw a circle.

(i) Thus, AY represents the locus of points that are equidistant from AB and AC.

(ii) Thus, BZ represents the locus of points that are equidistant from BA and BC.

Question 17

Construct a triangle ABC in which AB = 5 cm, BC = 6.8 cm and median AD = 4.4 cm. Draw incircle of this triangle.

Steps to construct the triangle and its incircle:

1. Begin by drawing the line segment BC measuring 6.8 \text{ cm}.
2. Identify the midpoint D on BC.
3. Using B as the center, draw an arc with a radius of 5 \text{ cm}.
4. From D, draw another arc with a radius of 4.4 \text{ cm} to intersect the previous arc at point A.
5. Connect A to C. Now, triangle ABC is formed.
6. To find the incenter, draw the angle bisectors of riangletriangle ABC at points B and C. These bisectors will intersect at point I.
7. From I, draw a perpendicular line IM to BC.
8. With I as the center and IM as the radius, draw the incircle.

Hence, above is the required incircle of the triangle.

Question 18

Draw two concentric circles with radii 4 cm and 6 cm. Taking a point on the outer circle, construct a pair of tangents to inner circle. By measuring the lengths of both the tangents, show that they are equal to each other.

Construction steps:

1. Begin by drawing two concentric circles with a common center O, where the inner circle has a radius of 4 cm and the outer circle has a radius of 6 cm.
2. Select a point P on the circumference of the outer circle.
3. Connect the center O with the point P by drawing the line segment OP.
4. Construct the perpendicular bisector of the line segment OP. Let the point where it intersects OP be denoted as M.
5. Using M as the center, draw arcs with a radius equal to OM. These arcs will intersect the inner circle at points A and B.
6. Connect point P to points A and B by drawing line segments PA and PB. Measure the lengths of PA and PB.

Upon measuring, you will find that:

PA = PB = 4.5 cm.

Therefore, PA and PB are tangents to the inner circle, each having a length of 4.5 cm.

Question 19

In triangle ABC, ∠ABC = 90°, side AB = 6 cm, side BC = 7.2 cm and BD is perpendicular to side AC. Draw circumcircle of triangle BDC and then state the length of the radius of this circumcircle drawn.

Construction steps:

1. Begin by sketching line segment BC with a length of 7.2 cm.
2. From point B, construct a ray BX such that ∠XBC equals 90°.
3. Using B as the center, draw an arc with a radius of 6 cm on ray BX to locate point A.
4. Connect points A and C with a line segment.
5. Identify point D on line AC such that BD is perpendicular to AC.
6. Construct the perpendicular bisectors of segments BD and BC. Let these bisectors intersect at point I.
7. Using I as the center, draw a circle with radius IC, ensuring it passes through points B, D, and C. Measure the radius IC.

Thus, the circumcircle is drawn with a radius of 3.6 cm.